One of the major activities in probability theory is studying the various statistics that can be produced from a complex system with many components. One of the simplest possible systems one can consider is a finite sequence or an infinite sequence of jointly independent scalar random variables, with the case when the are also identically distributed (i.e. the are iid) being a model case of particular interest. (In some cases one may consider a triangular array of scalar random variables, rather than a finite or infinite sequence.) There are many statistics of such sequences that one can study, but one of the most basic such statistics are the partial sums

The first fundamental result about these sums is the law of large numbers (or LLN for short), which comes in two formulations, weak (WLLN) and strong (SLLN). To state these laws, we first must define the notion of convergence in probability.

Definition 1Let be a sequence of random variables taking values in a separable metric space (e.g. the could be scalar random variables, taking values in or ), and let be another random variable taking values in . We say that converges in probability to if, for every radius , one has as . Thus, if are scalar, we have converging to in probability if as for any given .

The measure-theoretic analogue of convergence in probability is convergence in measure.

It is instructive to compare the notion of convergence in probability with almost sure convergence. it is easy to see that converges almost surely to if and only if, for every radius , one has as ; thus, roughly speaking, convergence in probability is good for controlling how a single random variable is close to its putative limiting value , while almost sure convergence is good for controlling how the entire *tail* of a sequence of random variables is close to its putative limit .

We have the following easy relationships between convergence in probability and almost sure convergence:

Exercise 2Let be a sequence of scalar random variables, and let be another scalar random variable.

- (i) If almost surely, show that in probability. Give a counterexample to show that the converse does not necessarily hold.
- (ii) Suppose that for all . Show that almost surely. Give a counterexample to show that the converse does not necessarily hold.
- (iii) If in probability, show that there is a subsequence of the such that almost surely.
- (iv) If are absolutely integrable and as , show that in probability. Give a counterexample to show that the converse does not necessarily hold.
- (v) (Urysohn subsequence principle) Suppose that every subsequence of has a further subsequence that converges to in probability. Show that also converges to in probability.
- (vi) Does the Urysohn subsequence principle still hold if “in probability” is replaced with “almost surely” throughout?
- (vii) If converges in probability to , and or is continuous, show that converges in probability to . More generally, if for each , is a sequence of scalar random variables that converge in probability to , and or is continuous, show that converges in probability to . (Thus, for instance, if and converge in probability to and respectively, then and converge in probability to and respectively.
- (viii) (Fatou’s lemma for convergence in probability) If are non-negative and converge in probability to , show that .
- (ix) (Dominated convergence in probability) If converge in probability to , and one almost surely has for all and some absolutely integrable , show that converges to .

Exercise 3Let be a sequence of scalar random variables converging in probability to another random variable .

- (i) Suppose that there is a random variable which is independent of for each individual . Show that is also independent of .
- (ii) Suppose that the are jointly independent. Show that is almost surely constant (i.e. there is a deterministic scalar such that almost surely).

We can now state the weak and strong law of large numbers, in the model case of iid random variables.

Theorem 4 (Law of large numbers, model case)Let be an iid sequence of copies of an absolutely integrable random variable (thus the are independent and all have the same distribution as ). Write , and for each natural number , let denote the random variable .

- (i) (Weak law of large numbers) The random variables converge in probability to .
- (ii) (Strong law of large numbers) The random variables converge almost surely to .

Informally: if are iid with mean , then for large. Clearly the strong law of large numbers implies the weak law, but the weak law is easier to prove (and has somewhat better quantitative estimates). There are several variants of the law of large numbers, for instance when one drops the hypothesis of identical distribution, or when the random variable is not absolutely integrable, or if one seeks more quantitative bounds on the rate of convergence; we will discuss some of these variants below the fold.

It is instructive to compare the law of large numbers with what one can obtain from the Kolmogorov zero-one law, discussed in Notes 2. Observe that if the are real-valued, then the limit superior and are tail random variables in the sense that they are not affected if one changes finitely many of the ; in particular, events such as are tail events for any . From this and the zero-one law we see that there must exist deterministic quantities such that and almost surely. The strong law of large numbers can then be viewed as the assertion that when is absolutely integrable. On the other hand, the zero-one law argument does not require absolute integrability (and one can replace the denominator by other functions of that go to infinity as ).

The law of large numbers asserts, roughly speaking, that the theoretical expectation of a random variable can be approximated by taking a large number of independent samples of and then forming the empirical mean . This ability to approximate the theoretical statistics of a probability distribution through empirical data is one of the basic starting points for mathematical statistics, though this is not the focus of the course here. The tendency of statistics such as to cluster closely around their mean value is the simplest instance of the concentration of measure phenomenon, which is of tremendous significance not only within probability, but also in applications of probability to disciplines such as statistics, theoretical computer science, combinatorics, random matrix theory and high dimensional geometry. We will not discuss these topics much in this course, but see this previous blog post for some further discussion.

There are several ways to prove the law of large numbers (in both forms). One basic strategy is to use the *moment method* – controlling statistics such as by computing moments such as the mean , variance , or higher moments such as for . The joint independence of the make such moments fairly easy to compute, requiring only some elementary combinatorics. A direct application of the moment method typically requires one to make a finite moment assumption such as , but as we shall see, one can reduce fairly easily to this case by a truncation argument.

For the strong law of large numbers, one can also use methods relating to the theory of martingales, such as stopping time arguments and maximal inequalities; we present some classical arguments of Kolmogorov in this regard.

** — 1. The moment method — **

We begin by using the moment method to establish both the strong and weak law of large numbers for sums of iid random variables, under additional moment hypotheses.

We first make a very simple observation: in order to prove the weak or strong law of large numbers for complex variables, it suffices to do so for real variables, as the complex case follows from the real case after taking real and imaginary parts. Thus we shall restrict attention henceforth to real random variables, in order to avoid some unnecessarily complications involving complex conjugation.

Let be a sequence of iid copies of a scalar random variable , and define the partial sums . Suppose that is absolutely integrable, with expectation (or *mean*) . Then we can use linearity of expectation to compute the expectation (or *first moment*) of :

In particular, the expectation of is . This looks consistent with the strong and weak law of large numbers, but does not immediately imply these laws. However, thanks to Markov’s inequality, we do at least get the following very weak bound

for any , in the case that is unsigned and absolutely integrable. Thus, in the unsigned case at least, we see that usually doesn’t get much larger than the mean . We will refer to (1) as a *first moment* bound on , as it was obtained primarily through a computation of the first moment of .

Now we turn to second moment bounds on , obtained through computations of second moments such as or . It will be convenient to normalise the mean to equal zero, by replacing each with (and by ), so that gets replaced by (and by ). With this normalisation, we see that to prove the strong or weak law of large numbers, it suffices to do so in the mean zero case . (On the other hand, if is unsigned, then normalising in this fashion will almost certainly destroy the unsigned property, so it is not always desirable to perform this normalisation.)

Suppose that has finite second moment (i.e. , that is to say is square-integrable) and has been normalised to have mean zero. We write the variance as . The first moment calculation then shows that has mean zero. Now we compute the variance of , which in the mean zero case is simply ; note from the triangle inequality that this quantity is finite. By linearity of expectation, we have

(All expressions here are absolutely integrable thanks to the Cauchy-Schwarz inequality.) If , then the term is equal to . If , then by hypothesis and are independent and mean zero, and thus

Putting all this together, we obtain

or equivalently

This bound was established in the mean zero case, but it is clear that it also holds in general, since subtracting a constant from a random variable does not affect its variance. Thus we see that while has the same mean as , it has a much smaller variance: in place of . This is the first demonstration of the concentration of measure effect that comes from combining many independent random variables . (At the opposite extreme to the independent case, suppose we took to all be exactly the same random variable: . Then has exactly the same mean and variance as . Decorrelating the does not affect the mean of , but produces significant cancellations that reduce the variance.)

If we insert this variance bound into Chebyshev’s inequality, we obtain the bound

for any natural number and , whenever has mean and a finite variance . The right-hand side goes to zero as for fixed , so we have in fact established the weak law of large numbers in the case that has finite variance.

Note that (2) implies that

converges to zero in probability whenever is a function of that goes to infinity as . Thus for instance

in probability. Informally, this means that tends to stray by not much more than from typically. This intuition will be reinforced in the next set of notes when we study the central limit theorem and related results such as the Chernoff inequality. (It is also supported by the law of the iterated logarithm, which we will probably not be able to get to in this set of notes.)

One can hope to use (2) and the Borel-Cantelli lemma (Exercise 2(ii)) to also obtain the strong law of large numbers in the second moment case, but unfortunately the quantities are not summable in . To resolve this issue, we will go to higher moments than the second moment. One could calculate third moments such as , but this turns out to not convey too much information (unless is unsigned) because of the signed nature of ; the expression would in principle convey more usable information, but is difficult to compute as is not a polynomial combination of the . Instead, we move on to the fourth moment. Again, we normalise to have mean , and now assume a finite fourth moment (which, by the Hölder or Jensen inequalities, implies that all lower moments such as are finite). We again use to denote the variance of . We can expand

Note that all expectations here are absolutely integrable by Hölder’s inequality and the hypothesis of finite fourth moment. The correlation looks complicated, but fortunately it simplifies greatly in most cases. Suppose for instance that is distinct from , then is independent of (even if some of the are equal to each other) and so

since . Similarly for permutations. This leaves only a few quadruples for which could be non-zero: the three cases , , where each of the indices is paired up with exactly one other index; and the diagonal case . If for instance , then

Similarly for the cases and , which gives a total contribution of to . Finally, when , then , and there are contributions of this form to . We conclude that

and hence by Markov’s inequality

for any . If we remove the normalisation , we conclude that

The right-hand side decays like , which is now summable in (in contrast to (2)). Thus we may now apply the Borel-Cantelli lemma and conclude the *strong* law of large numbers in the case when one has bounded fourth moment .

One can of course continue to compute higher and higher moments of (assuming suitable finite moment hypotheses on ), though as one can already see from the fourth moment calculation, the computations become increasingly combinatorial in nature. We will pursue this analysis more in the next set of notes, when we discuss the central limit theorem. For now, we turn to some applications and variants of the moment method (many of which are taken from Durrett’s book).

We begin with two quick applications of the weak law of large numbers to topics outside of probability. We first give an explicit version of the Weierstrass approximation theorem, which asserts that continuous functions on (say) the unit interval can be approximated by polynomials.

Proposition 5 (Approximation by Bernstein polynomials)Let be a continuous function. Then the Bernstein polynomialsconverges uniformly to as .

*Proof:* We first establish the pointwise bound for each . Fix , and let be iid Bernoulli variables (i.e. variables taking values in ) with each equal to with probability . The mean of the is clearly , and the variance is bounded crudely by (in fact it is ), so if we set then by the weak law of large numbers for random variables of finite second moment, we see that converges in probability to (note that is clearly dominated by ). By the dominated convergence theorem in probability, we conclude that converges to . But from direct computation we see that takes values in , with each being attained with probability (i.e. as a binomial distribution), and so from the definition of the Bernstein polynomials we see that . This concludes the pointwise convergence claim.

To establish the uniform convergence, we use the *proof* of the weak law of large numbers, rather than the *statement* of that law, to get the desired uniformity in the parameter . For a given , we see from (2) that

for any . On the other hand, as is continuous on , it is uniformly continuous, and so for any there exists a such that whenever with . For such an and , we conclude that

On the other hand, being continuous on , must be bounded in magnitude by some bound , so that . This leads to the upper bound

and thus by the triangle inequality and the identity

Since is bounded by (say) , and can be made arbitrarily small, we conclude that converges uniformly to as required.

The other application of the weak law of large numbers is to the geometry of high-dimensional cubes, giving the rather unintuitive conclusion that most of the volume of the high-dimensional cube is contained in a thin annulus.

Proposition 6Let . Then, for sufficiently large , a proportion of at least of the cube (by -dimensional Lebesgue measure) is contained in the annulus .

This proposition already indicates that high-dimensional geometry can behave in a manner quite differently from what one might naively expect from low-dimensional geometric intuition; one needs to develop a rather distinct high-dimensional geometric intuition before one can accurately make predictions in large dimensions.

*Proof:* Let be iid random variables drawn uniformly from . Then the random vector is uniformly distributed on the cube . The variables are also iid, and (by the change of variables formula) have mean

Hence, by the weak law of large numbers, the quantity converges in probability to , so in particular the probability

goes to zero as goes to infinity. But this quantity is precisely the proportion of that lies outside the annulus , and the claim follows.

The first and second moment method are very general, and apply to sums of random variables that do not need to be identically distributed, or even independent (although the bounds can get weaker and more complicated if one strays too far from these hypotheses). For instance, it is clear from linearity of expectation that has mean

(assuming of course that are absolutely integrable) and variance

(assuming now that are square-integrable). (For the latter claim, it is convenient, as before, to first normalise each of the to have mean zero.) If the are pairwise independent in addition to being square-integrable, then all the covariances vanish, and we obtain additivity of the variance:

Remark 7Viewing the variance as the square of the standard deviation, the identity (4) can be interpreted as a rigorous instantiation of the following informalprinciple of square root cancellation: if one has a sum of random (or pseudorandom) variables that “oscillate” in the sense that their mean is either zero or close to zero, and each has an expected magnitude of about (in the sense that a statistic such as the standard deviation of is comparable to ), and the “behave independently”, then the sum is expected to have a magnitude of about . Thus for instance a sum of unbiased signs would be expected to have magnitude about if the do not exhibit strong correlations with each other. This principle turns out to be remarkably broadly applicable (at least as a heuristic, if not as a rigorous argument), even in situations for which no randomness is evident (e.g. in considering the type of exponential sums that occur in analytic number theory). We will see some further instantiations of this principle in later notes.

These identities, together with Chebyshev’s inequality, already gives some useful control on many statistics, including some which are not obviously of the form of a sum of independent random variables. A classic example of this is the coupon collector problem, which we formulate as follows. Let be a natural number, and let be an infinite sequence of “coupons”, which are iid and uniformly distributed from the finite set . Let denote the first time at which one has collected all different types of coupons, thus is the first natural number for which the set attains the maximal cardinality of (or if no such natural number exists, though it is easy to see that this is a null event, indeed note from the strong law of large numbers that almost surely one will collect infinitely many of each coupon over time). The question is then to describe the behaviour of as gets large.

At first glance, does not seem to be easily describable as the sum of many independent random variables. However, if one looks at it the right way, one can see such a structure emerge (and much of the art of probability is in finding useful and different ways of thinking of the same random variable). Namely, for each , let denote the first time one has collected coupons out of , thus is the first non-negative integer such that has cardinality , with

If we then write for , then the take values in the natural numbers, we have the telescoping sum

Remarkably, the random variables have a simple structure:

Proposition 8The random variables are jointly independent, and each has a geometric distribution with parameter , in the sense thatfor and .

The joint independence of the reflects the “Markovian” or “memoryless” nature of a certain process relating to the coupon collector problem, and can be easily established once one has understood the concept of conditional expectation, but the further exploration of these concepts will have to be deferred to the course after this one (which I will not be teaching or writing notes for). But as the coupon collecting problem is so simple, we shall proceed instead by direct computation.

*Proof:* It suffices to show that

for any choice of natural numbers . In order for the event to hold, the first coupon can be arbitrary, but the coupons have to be equal to ; then must be from one the remaining elements of not equal to , and must be from the two-element set ; and so on and so forth up to . The claim then follows from a routine application of elementary combinatorics to count all the possible values for the tuple of the above form and dividing by the total number of such tuples.

Exercise 9Show that if is a geometric distribution with parameter for some (thus for all ) then has mean and variance .

From the above proposition and exercise as well as (3), (4) we see that

and

From the integral test (and crudely bounding ) one can thus obtain the bounds

and

where we use the usual asymptotic notation of denoting by any quantity bounded in magnitude by a constant multiple of . From Chebyshev’s inequality we thus see that

for any (note the bound is trivial unless is large). This implies in particular that converges to in probability as (assuming that our underlying probability space can model a separate coupon collector problem for each choice of ). Thus, roughly speaking, we see that one expects to take about units of time to collect all coupons.

Another application of the weak and strong law of large numbers (even with the moment hypotheses currently imposed on these laws) is a converse to the Borel-Cantelli lemma in the jointly independent case:

Exercise 10 (Second Borel-Cantelli lemma)Let be a sequence of jointly independent events. If , show that almost surely an infinite number of the hold simultaneously. (Hint:compute the mean and variance of . One can also compute the fourth moment if desired, but it is not necessary to do so for this result.)

One application of the second Borel-Cantelli lemma has the colourful name of the “infinite monkey theorem“:

Exercise 11 (Infinite monkey theorem)Let be iid random variables drawn uniformly from a finite alphabet . Show that almost surely, every finite word of letters in the alphabet appears infinitely often in the string .

In the usual formulation of the weak or strong law of large numbers, we draw the sums from a single infinite sequence of iid random variables. One can generalise the situation slightly by working instead with sums from rows of a triangular array, which are jointly independent within rows but not necessarily across rows:

Exercise 12 (Triangular arrays)Let be a triangular array of scalar random variables , such that for each , the row is a collection of independent random variables. For each , we form the partial sums .

- (i) (Weak law) If all the have mean and , show that converges in probability to .
- (ii) (Strong law) If all the have mean and , show that converges almost surely to .

Note that the weak and strong law of large numbers established previously corresponds to the case when the triangular array collapses to a single sequence of iid random variables.

We now illustrate the use of moment method and law of large number methods to two important examples of random structures, namely random graphs and random matrices.

Exercise 13For a natural number and a parameter , define anErdös-Renyi graphon vertices with parameter to be a random graph on a (deterministic) vertex set of vertices (thus is a random variable taking values in the discrete space of all possible graphs one can place on ) such that the events for unordered pairs in are jointly independent and each occur with probability .For each , let be an Erdös-Renyi graph on vertices with parameter (we do not require the graphs to be independent of each other).

- (i) If is the number of edges in , show that converges almost surely to . (
Hint:use Exercise 12.)- (ii) If is the number of triangles in (i.e. the set of unordered triples in such that ), show that converges in probability to . (Note: there is not quite enough joint independence here to directly apply the law of large numbers, however the second moment method still works nicely.)
- (iii) Show in fact that converges almost surely to . (Note: in contrast with the situation with the strong law of large numbers, the fourth moment does not need to be computed here.)

Exercise 14For each , let be a random matrix (i.e. a random variable taking values in the space or of matrices) such that the entries of are jointly independent in and take values in with a probability of each. (Such matrices are known asrandom sign matrices.) We do not assume any independence for the sequence .

- (i) Show that the random variables are deterministically equal to , where denotes the adjoint (which, in this case, is also the transpose) of and denotes the trace (sum of the diagonal entries) of a matrix.
- (ii) Show that for any natural number , the quantities are bounded uniformly in (i.e. they are bounded by a quantity that can depend on but not on ). (You may wish to first work with simple cases like or to gain intuition.)
- (iii) If denotes the operator norm of , and , show that converges almost surely to zero, and that diverges almost surely to infinity. (
Hint:use the spectral theorem to relate with the quantities .)

One can obtain much sharper information on quantities such as the operator norm of a random matrix; see this previous blog post for further discussion.

Exercise 15The Cramér random model for the primes is a random subset of the natural numbers with , , and the events for being jointly independent with (the restriction to is to ensure that is less than ). It is a simple, yet reasonably convincing, probabilistic model for the primes , which can be used to provide heuristic confirmations for many conjectures in analytic number theory. (It can be refined to give what are believed to be more accurate predictions; see this previous blog post for further discussion.)

- (i) (Probabilistic prime number theorem) Prove that almost surely, the quantity converges to one as .
- (ii) (Probabilistic Riemann hypothesis) Show that if , then the quantity
converges almost surely to zero as .

- (iii) (Probabilistic twin prime conjecture) Show that almost surely, there are an infinite number of elements of such that also lies in .
- (iv) (Probabilistic Goldbach conjecture) Show that almost surely, all but finitely many natural numbers are expressible as the sum of two elements of .

Probabilistic methods are not only useful for getting *heuristic* predictions about the primes; they can also give *rigorous* results about the primes. We give one basic example, namely a probabilistic proof (due to Turán) of a theorem of Hardy and Ramanujan, which roughly speaking asserts that a typical large number has about distinct prime factors.

Exercise 16 (Hardy-Ramanujan theorem)Let be a natural number (so in particular ), and let be a natural number drawn uniformly at random from to . Assume Mertens’ theoremfor all , where the sum is over primes up to .

- (i) Show that the random variable (where is when divides and otherwise, and the sum is over primes up to ) has mean and variance . (
Hint:compute (up to reasonable error) the means, variances and covariances of the random variables .)- (ii) If denotes the number of distinct prime factors of , show that converges to in probability as . (
Hint:first show that .) More precisely, show thatconverges in probability to zero, whenever is any function such that goes to infinity as .

Exercise 17 (Shannon entropy)Let be a finite non-empty set of some cardinality , and let be a random variable taking values in . Define the Shannon entropy to be the quantitywith the convention that . (In some texts, the logarithm to base is used instead of the natural logarithm .)

- (i) Show that . (
Hint:use Jensen’s inequality.) Determine when the equality holds.- (ii) Let and be a natural number. Let be iid copies of , thus is a random variable taking values in , and the distribution is a probability measure on . Let denote the set
Show that if is sufficiently large, then

and

(

Hint:use the weak law of large numbers to understand the number of times each element of occurs in .)Thus, roughly speaking, while in principle takes values in all of , in practice it is concentrated in a set of size about , and is roughly uniformly distributed on that set. This is the beginning of the microstate interpretation of Shannon entropy, but we will not develop the theory of Shannon entropy further in this course.

** — 2. Truncation — **

The weak and strong laws of large numbers have been proven under additional moment assumptions (of finite second moment and finite fourth moment respectively). To remove these assumptions, we use the simple but effective *truncation method*, decomposing a general scalar random variable into a truncated component such as for some suitable threshold , and a tail . The main term is bounded and thus has all moments finite. The tail will not have much better moment properties than the original variable , but one can still hope to make it “small” in various ways. There is a tradeoff regarding the selection of the truncation parameter : if is too large, then the truncated component has poor estimates, but if is too small then the tail causes trouble.

Let’s first see how this works with the weak law of large numbers. As before, we assume we have an iid sequence copies of a real absolutely integrable with no additional finite moment hypotheses. We write . At present, we cannot take variances of the or and so the second moment method is not directly available. But we now perform a truncation; it turns out that a good choice of threshold here is , thus we write where and , and then similarly decompose where

and

as for any given (not depending on ). By the triangle inequality, we can split

We begin by studying .

The random variables are iid with mean and variance at most

Thus, has mean and variance at most . By dominated convergence, as , so for sufficiently large we can bound

and hence by the Chebyshev inequality

Observe that is bounded surely by the absolutely integrable , and goes to zero as , so by dominated convergence we conclude that

as (keeping fixed).

To handle , we observe that each is only non-zero with probability , and hence by subadditivity

By dominated convergence again, as , and thus

Putting all this together, we conclude (5) as required. This concludes the proof of the weak law of large numbers (in the iid case) for arbitrary absolutely integrable . For future reference, we observe that the above arguments give the bound

whenever is sufficiently large depending on .

Due to the reliance on the dominated convergence theorem, the above argument does not provide any uniform rate of decay in (5). Indeed there is no such uniform rate. Consider for instance the sum where are iid random variables that equal with probability and with probability . Then the are unsigned and all have mean , but vanishes with probability , which converges to as . Thus we see that the probability that stays a distance from the mean value of is bounded away from zero. This is not inconsistent with the weak law of large numbers, because the underlying random variable depends on in this example. However, it rules out an estimate of the form

that holds uniformly whenever obeys the bound , and is a quantity that goes to zero as for a fixed choice of . (Contrast this with (2), which does provide such a uniform bound if one also assumes a bound on the second moment .)

One can ask what happens to the when the underlying random variable is not absolutely integrable. In the unsigned case, we have

Exercise 18Let be iid copies of an unsigned random variable with infinite mean, and write . Show that diverges to infinity in probability, in the sense that as for any fixed .

The above exercise shows that grows faster than (in probability, at least) when is unsigned with infinite mean, but this does not completely settle the question of the precise rate at which does grow. We will not answer this question in full generality here, but content ourselves with analysing a classic example of the unsigned infinite mean setting, namely the Saint Petersburg paradox. The paradox can be formulated as follows. Suppose we have a lottery whose payout takes taking values in the powers of two with

for . The question is to work out what is the “fair” or “breakeven” price to pay for a lottery ticket. If one plays this lottery times, the total payout is where are independent copies of , so the question boils down to asking what one expects the value of to be. If were absolutely integrable, the strong or weak law of large numbers would indicate that is the fair price to pay, but in this case we have

This suggests, paradoxically, that any finite price for this lottery, no matter how high, would be a bargain!

To clarify this paradox, we need to get a better understanding of the random variable . For a given , we let be a truncation parameter to be chosen later, and split where and as before (we no longer need the absolute value signs here as all random variables are unsigned). Since the take values in powers of two, we may as well also set to be a power of two. We split where

and

The random variable can be computed to have mean

and we can upper bound the variance by

and hence has mean and variance at most . By Chebyshev’s inequality, we thus have

for any .

Now we turn to . We cannot use the first or second moment methods here because the are not absolutely integrable. However, we can instead use the following “zeroth moment method” argument. Observe that the random variable is only nonzero with probability (that is to say, the “zeroth moment” is , using the convention ). Thus is nonzero with probability at most . We conclude that

This bound is valid for any natural number and any . Of course for this bound to be useful, we want to select parameters so that the right-hand side is somewhat small. If we pick for instance to be the integer part of , and to be , we see that

which (for large ) implies that

In particular, we see that converges in probability to one. This suggests that the fair price to pay for the Saint Petersburg lottery is a function of the number of tickets one wishes to play, and should be approximately when is large. In particular, the lottery is indeed worth paying out any finite cost , but one needs to buy about tickets before one breaks even!

In contrast to the absolutely integrable case, in which the weak law can be upgraded to the strong law, there is no strong law for the Saint Petersburg paradox:

Exercise 19With the notation as in the above analysis of the Saint Petersburg paradox, show that is almost surely unbounded. (Hint:it suffices to show that is almost surely unbounded. For this, use the second Borel-Cantelli lemma.)

The following exercise can be viewed as a continuous analogue of the Saint Petersburg paradox.

Exercise 20A real random variable is said to have a standard Cauchy distribution if it has the probability density function .

- (i) Verify that standard Cauchy distributions exist (this boils down to checking that the integral of the probability density function is ).
- (ii) Show that a real random variable with the standard Cauchy distribution is not absolutely integrable.
- (iii) If are iid copies of a random variable with the standard Cauchy distribution, show that converges in probability to but is almost surely unbounded.

Exercise 21 (Weak law of large numbers for triangular arrays)Let be a triangular array of random variables, with the variables jointly independent for each . Let be a sequence going to infinity, and write and . Assume thatand

as . Show that

in probability.

Now we turn to establishing the strong law of large numbers in full generality. A first attempt would be to apply the Borel-Cantelli lemma to the bound (6). However, the decay rates for quantities such as are far too weak to be absolutely summable, in large part due to the reliance on the dominated convergence theorem. To get around this we follow some arguments of Etemadi. We first need to make a few preliminary reductions, aimed at “sparsifying” the set of times that one needs to control. It is here that we will genuinely use the fact that the averages are being drawn from a single sequence of random variables, rather than from a triangular array.

We turn to the details. In previous arguments it was convenient to normalise the underlying random variable to have mean zero. Here we will use a different reduction, namely to the case when is unsigned; the strong law for real absolutely integrable clearly follows from the unsigned case by expressing as the difference of two unsigned absolutely integrable variables and (and splitting similarly).

Henceforth we assume (and hence the ) to be unsigned. Crucially, this now implies that the partial sums are monotone: . While this does not quite imply any monotonicity on the sequence , it does make it significantly easier to show that it converges. The key point is as follows.

Lemma 22Let be an increasing sequence, and let be a real number. Suppose that for any , the sequence converges to as , where . Then converges to .

*Proof:* Let . For any , let be the index such that . Then we have

and (for sufficiently large )

and thus

Taking limit inferior and superior, we conclude that

and then sending we obtain the claim.

An inspection of the above argument shows that we only need to verify the hypothesis for a countable sequence of (e.g. for natural number ). Thus, to show that converges to almost surely, it suffices to show that for any , one has almost surely as .

Fix . The point is that the “lacunary” sequence is much sparser than the sequence of natural numbers , and one now will lose a lot less from the Borel-Cantelli argument. Indeed, for any , we can apply (6) to conclude that

whenever is sufficiently large depending on . Thus, by the Borel-Cantelli lemma, it will suffice to show that the sums

and

are finite. Using the monotone convergence theorem to interchange the sum and expectation, it thus suffices to show the pointwise estimates

and

for some . But this follows from the geometric series formula (the first sum is over the elements of the sequence that are greater than or equal to , while the latter is over those that are less than ). This proves the strong law of large numbers for arbitrary absolutely integrable iid .

We remark that by carefully inspecting the above proof of the strong law of large numbers, we see that the hypothesis of joint independence of the can be relaxed to pairwise independence.

The next exercise shows how one can use the strong law of large numbers to approximate the cumulative distribution function of a random variable by an empirical cumulative distribution function.

Exercise 23Let be iid copies of a real random variable .

- (i) Show that for every real number , one has almost surely that
and

as .

- (ii) Establish the Glivenko-Cantelli theorem: almost surely, one has

uniformly inas . (Hint:For any natural number , let denote the largest integer multiple of less than . Show first that is within of for all when is sufficiently large.)

Exercise 24 (Lack of strong law for triangular arrays)Let be a random variable taking values in the natural numbers with , where (this is an example of a zeta distribution).

- (i) Show that is absolutely integrable.
- (ii) Let be jointly independent copies of . Show that the random variables are almost surely unbounded. (
Hint:for any constant , show that occurs with probability at least for some depending on . Then use the second Borel-Cantelli lemma.)

** — 3. The Kolmogorov maximal inequality — **

Let be a sequence of jointly independent square-integrable real random variables of mean zero; we do not assume the to be identically distributed. As usual, we form the sums , then has mean zero and variance

From Chebyshev’s inequality, we thus have

for any and natural number . Perhaps surprisingly, we have the following improvement to this bound, known as the Kolmogorov maximal inequality:

Theorem 25 (Kolmogorov maximal inequality)With the notation and hypotheses as above, we have

*Proof:* For each , let be the event that , but that for all . It is clear that the event is the disjunction of the disjoint events , thus

On the event , we have , and hence

We will shortly prove the inequality

for all . Assuming this inequality for the moment, we can put together all the above estimates, using the disjointness of the , to conclude that

and the claim follows from (7).

It remains to prove (8). Since

we have

But note that the random variable is completely determined by , while is completely determined by the . Thus and are independent. Since also has mean zero, we have

and the claim (8) follows.

An inspection of the above proof reveals that the key ingredient is the *lack of correlation between past and future* – a variable such as , which is determined by the portion of the sequence to the “past” (and present) of , is uncorrelated with a variable such as that depends only on the “future” of . One can formalise such a lack of correlation through the concept of a martingale, which will be covered in later courses in this sequence but which is beyond the scope of these notes. The use of the first time at which exceeds or attains the threshold is a simple example of a stopping time, which will be a heavily used concept in the theory of martingales (and is also used extensively in harmonic analysis, which is also greatly interested in establishing maximal inequalities).

The Kolmogorov maximal inequality gives the following variant of the strong law of large numbers.

Theorem 26 (Convergence of random series)Let be jointly independent square-integrable real random variables of mean zero withThen the series is almost surely convergent (i.e., the partial sums converge almost surely).

*Proof:* From the Kolmogorov maximal inequality and continuity from below we have

This is already enough to show that the partial sums are almost surely bounded in , but this isn’t quite enough to establish conditional convergence. To finish the job, we apply (9) with replaced by the shifted sequence for a natural number to conclude that

Sending to infinity using continuity from above, we conclude that

for all ; applying this for all rational , we conclude that is almost surely a Cauchy sequence, and the claim follows.

We can use this result together with some elementary manipulation of sums to give the following alternative proof of the strong law of large numbers.

Theorem 27 (Strong law of large numbers)Let be iid copies of an absolutely integrable variable of mean , and let . Then converges almost surely to .

*Proof:* We may normalise to be real valued with . Note that that

and hence by the Borel-Cantelli lemma we almost surely have for all but finitely many . Thus if we write and , then the difference between and almost surely goes to zero as . Thus it will suffice to show that goes to zero almost surely.

The random variables are still jointly independent, but are not quite mean zero. However, the normalised random variables are of mean zero and

so by Theorem 26 we see that the sum is almost surely convergent.

Write , thus the sequence is almost surely a Cauchy sequence. From the identity

we conclude that the sequence almost surely converges to zero, that is to say

almost surely. On the other hand, we have

and hence by dominated convergence

as , which implies that

as . By the triangle inequality, we conclude that almost surely, as required.

Exercise 28 (Kronecker lemma)Let be a convergent series of real numbers , and let be a sequence tending to infinity. Show that converges to zero as . (This is known as Kronecker’s lemma; the special case was implicitly used in the above argument.)

Exercise 29 (Kolmogorov three-series theorem, one direction)Let be a sequence of jointly independent real random variables, and let . Suppose that the two series and are absolutely convergent, and the third series is convergent. Show that the series is almost surely convergent. (The converse claim is also true, and will be discussed in later notes; the two claims are known collectively as the Kolmogorov three-series theorem.)

One advantage that the maximal inequality approach to the strong law of large numbers has over the moment method approach is that it tends to offer superior bounds on the (almost sure) *rate* of convergence. We content ourselves with just one example of this:

Exercise 30 (Cheap law of iterated logarithm)Let be a sequence of jointly independent real random variables of mean zero and bounded variance (thus ). Write . Show that converges almost surely to zero as for any given . (Hint:use Theorem 26 and the Kronecker lemma for a suitable weighted sum of the .) There is a more precise version of this fact known as the law of the iterated logarithm, which is beyond the scope of these notes.

The exercises below will be moved to a more appropriate location later, but are currently placed here in order to not disrupt existing numbering.

Exercise 31Let be iid copies of an absolutely integrable random variable with mean . Show that the averages converge in to , that is to say thatas .

Exercise 32A scalar random variable is said to be inweakif one hasThus Markov’s inequality tells us that every absolutely integrable random variable is in weak , but the converse is not true (e.g. random variables with the Cauchy distribution are weak but not absolutely integrable). Show that if are iid copies of an unsigned weak random variable, then there exists quantities such that converges in probability to , where . (Thus: there is a weak law of large numbers for weak random variables, and a strong law for strong (i.e. absolutely integrable) random variables.)

## 57 comments

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23 October, 2015 at 10:55 am

PhilipAnyone working the exercises out? I have been rusty on the epsilon-delta proofs, but it seems that the inequalities can be found from the required conditions of the theorems.

23 October, 2015 at 4:44 pm

pauljungI think you want in the numerator in your discussion of the CLT under (2). In Exercise 18, I think you want that the probability goes to 1 instead of 0. Also, there is no mention of the limit X in your characterization of a.s. convergence under Definition 1. It’s too bad that you won’t be teaching 275B.

[Corrected, thanks -T.]23 October, 2015 at 6:45 pm

Anonymousit would be interesting to show some pseudo random properties (like CLT ) for lacunary trigonometric series.

[I plan to have some exercises on this in the next set of notes – T.]26 October, 2015 at 8:28 am

MaxDear Prof. Tao,

I have a question on exercise 12: I tried to work out what the weakest condition on the moments of the for the strong law of large numbers to hold is. I can show that if for an arbitrary is sufficient, and now I am wondering whether one can even get further down to second moments being sufficient for the strong law to hold.

Best regards,

Max

26 October, 2015 at 8:45 am

Terence TaoSee Exercise 24 for an example showing that one cannot go much below the second moment; it is likely one can adapt this construction to find the optimal threshold. (There are probability textbooks that carefully identify the minimal hypotheses required for various key theorems like the law of large numbers; I’ll try to track down such a reference for this and future inquiries of this nature.)

[EDIT: it seems that textbooks from the Russian school, such as V. V. Petrov, Sums of Independent Random Variables, Springer Ergebnisse Vol 82, V. V. Petrov, Limit Theorems of Probability Theory, Oxford Studies in Probability Vol 4, and B. V. Gnedenko and A. N. Kolmogorov, Limit Distributions for Sums of Independent Random Variables, Addison-Wesley, would be good resources for this sort of thing.]

26 October, 2015 at 5:21 pm

AnonymousSome people asked in a forum the differences between

the weak LLN and strong LLN. This may include different conditions

and the examples in which the weak LLN holds but the strong LLN does not.

2 November, 2015 at 7:06 pm

275A, Notes 4: The central limit theorem | What's new[…] of an absolutely integrable real scalar random variable , and form the partial sums . As we saw in the last set of notes, the law of large numbers ensures that the empirical averages converge (both in probability and […]

3 November, 2015 at 7:28 am

VenkyA small question: In the proof of the LLN (before Prop 5), should the denominator in the right hand side of the Markov inequality have epsilon^4 instead of epsilon? Thanks again for great notes. I really began to appreciate the importance of tail events.

[Corrected, thanks – T.]5 November, 2015 at 10:35 pm

AnonymousIn Exercise 14(i), is the “almost surely” actually necessary? It seems that the trace expression is surely 1 by just expanding what is.

[Fair point – I’ve adjusted the exercise accordingly. For other ensembles, such as gaussian matrices, one only has convergence in probability and in almost sure senses, rather than deterministic equality, but indeed in this simple sign matrix model there is in fact no random fluctuation. -T.]7 November, 2015 at 6:13 am

AnonymousDear Terence Tao,

For the second term in the upper bound in Equation , where does the term come from? It seems that we are still using the inequality below Equation . So maybe we are upper bounding by , for large , which is by Markov inequality upper bounded by . But I am not sure even if for large , we can relate the events and .

Thanks very much.

[Oops, this was a typo arising from a previous version of the argument, now fixed – T.]8 November, 2015 at 1:42 am

DillonIn exercise 17, should it be

instead of

?

[Corrected, thanks – T.]10 November, 2015 at 5:46 am

AnonymousIs there a sign error in the second inequality in part (ii) of Exercise 17? In particular should it really be and (should the negative signs be there)?

[Corrected, thanks – T.]13 November, 2015 at 1:21 pm

AnonymousI think the in the second centered inequality of the proof of Lemma 22 should be reversed.

[Corrected, thanks – T.]16 November, 2015 at 6:35 am

RexDear Terry,

In the proof of the weak law of large numbers without finite moment assumptions, could you say a bit more about how we are using dominated convergence to get

and also

? It is not obvious to me what the dominating functions here are.

16 November, 2015 at 9:10 am

Terence Taoand are dominated in magnitude by the absolutely integrable random variable . (Here it is convenient to use the probabilistic version of the DCT rather than the measure theoretic version, i.e. Theorem 23 of Notes 1.)

19 November, 2015 at 2:51 pm

275A, Notes 5: Variants of the central limit theorem | What's new[…] with Exercise 17 of Notes 3). One can check that is decreasing for , and so one can compute […]

26 November, 2015 at 11:59 am

TMLooks like there are some typos in the proof of the polynomial approximation theorem (estimates for and and $\epsilon$, $\delta$ parameters getting mixed up; in particular, the first and second displayed equations where in the first one should be and the second one doesn’t look like it should be there at all (the Markov inequality estimate seems wrong since the variance of is not .

26 November, 2015 at 12:16 pm

TMUpdate: Sorry for the hasty comment above (couldn’t edit).

I believe I understand the point of the second displayed equation now – the mix-up with and along with the omission of an intermediate step threw me off the first couple times I went through the proof.

I reckon this was the intended statement (somewhat more explicitly):

[Corrected, thanks – T.]29 November, 2015 at 5:27 am

AnonymousIn Proposition 5:

\displaystyle f_n(x) := \sum_{i=0}^n \binom{n}{i} x^i (1-x)^{n-i} f(\frac{x}{n})

should be:

\displaystyle f_n(x) := \sum_{i=0}^n \binom{n}{i} x^i (1-x)^{n-i} f(\frac{i}{n})

(AFAIK AIAM=…and it isn’t much).

[Corrected, thanks – T.]29 November, 2015 at 5:59 am

AnonymousIn the coupon collector exposition: ‘This implies in particular that…’ and then {T_N/N \log N} should be {T_N/N \log N \to 1} ?

[Corrected, thanks – T.]29 November, 2015 at 4:32 pm

AnonymousIn truncation section, law large numbers, an equation:

might have \leq and = transposed i.e.

[Actually, I believe it is correct as stated – note that need not have mean zero. -T.]29 November, 2015 at 4:41 pm

AnonymousIn St. Petersburg expectation formula:

\displaystyle {\bf E} X = \sum_{i=0}^\infty 2^i \times 2^{-i} = \sum_{i=0}^\infty 1 = \infty.

shouldn’t the sum begin at i=1?

[Corrected, thanks – T.]29 November, 2015 at 4:44 pm

AnonymousIn St. Petersburg truncated variance equation:

\displaystyle {\bf Var}(X 1_{X \leq M}) \leq {\bf E} (X 1_{X \leq M})^2

\displaystyle = \sum_{i=1}^m 2^{2i} 2^i

\displaystyle \leq 2^{m+1}

there is a missing minus, 2^i should be 2^{-i}

[Corrected, thanks – T.]30 November, 2015 at 12:40 am

AnonymousHi Terry, theres a small typo in Proposition 5 (Approx by Bernstein polys).

The last term should be f(i/n) rather than f(x/n).

[Corrected, thanks – T.]3 January, 2016 at 8:39 am

Anonymous“…..thus {T_{i,N}} is the first non-negative integer such that {\{Y_1,\dots,Y_{T_i}\}} has cardinality {i}….”

Should the Y_T_j have an additional index N?

[Corrected, thanks – T.]3 January, 2016 at 10:12 am

AnonymousExercise 15, the set A in the first two lines should be P?

[Corrected, thanks – T.]13 January, 2016 at 3:45 pm

AnonymousThe last point wise estimate in the (general) law of large numbers still has an expectation.

[Corrected, thanks – T.]16 May, 2016 at 6:08 am

cosine30Dear Prof. Tao,

I have a question on exercise 24: I tried to first expand the probability and to then get a lower bound on the individual summands of the form . I tried to do this by expanding further , and then I subdivided the sum further, but I am getting nowhere. To make my question more concrete: Is this the way to go, or is there some standard inequality that is applicable here (lower bounds on probabilities seem to be much rarer than upper bounds though)?

If your time permits it, I would be grateful for a short answer.

Best regards,

Max

16 May, 2016 at 8:35 am

Terence TaoNote that the question is not asking for an exact value for the tail probability , but only a lower bound. So one does not need to perform all of these exact computations. Instead, one only needs to think about some simpler events that are contained in the given event. I suggest starting with the events for .

16 May, 2016 at 9:03 am

MrCactu5 (@MonsieurCactus)The more I look at your articles the more they look the same. Don’t you use 4-th moment estimates in the Wigner Semicircle Theorem (for all random variables) as well as certain special cases of Szemerédi’s Theorem (k=3,4 and weak-mixing)?

12 November, 2016 at 11:59 am

AnonymousIn the section of moment methods, if one assumes only finite third moment instead of four, then can one still finish the proof using only what you have for the finite second moment?

[Not easily – in particular, not without something like a truncation argument – T.]In the section of truncation, would you elaborate the following step?

By the triangle inequality, we can splitI only get

[You should have instead of . Then, observe that if , then either is non-zero, or is zero and . -T.]12 November, 2016 at 1:06 pm

AnonymousDo you mean with only the finite first moment assumption (Theorem 4) or anything else in “Now we turn to establishing the strong law of large numbers in full generality.”?

[Yes – T.]In the argument in the section of truncation, I still don’t find where the assumption of independence is used in the proof of the WLLN and SLLN after several times of reading. Am I missing something? Would you please elaborate?

[Independence was used to compute the variance of . -T.]12 November, 2016 at 4:54 pm

AnonymousIf one assumes the finite fourth moment, can one prove the SLLN without the Borel-Cantelli lemma but instead calculate

[Yes – T.]13 November, 2016 at 7:13 am

AnonymousWhen one uses the approach of Borel-Cantelli in the case of finite fourth moment, one gets

for all , which is

morethan what is needed for giving almost sure convergence of . Would you say that this is the “defect” that Etemadi remedies in his argument to prove the SLLN with assuming only finite first moment when one still wishes to use Borel-Cantelli?14 November, 2016 at 4:26 am

AnonymousOops, one only gets

not for all .

17 February, 2017 at 9:57 pm

254A, Notes 2: The central limit theorem | What's new[…] of an absolutely integrable real scalar random variable , and form the partial sums . As we saw in the last set of notes, the law of large numbers ensures that the empirical averages converge (both in probability and […]

7 January, 2019 at 9:51 am

haonanzHello Prof. Tao, I have a question regard the remarks just above Exercise 18. I do not quite see the connection between the counter example you provided leads to the conclusion that it rules out an estimate with decay rate in n. I can see that for the counter example, the moment proof where we applied Chebyshev inequality won’t hold, as the expectation would be constant 1 that does not decay to zero as n grows large. Do you mind elaborate a little bit more on the connection here? Thanks,

7 January, 2019 at 10:03 am

Terence TaoAs stated in the post, the counterexample shows that there does not exist a bound of the form uniform for all with , where as . This is because (setting for instance and ), the counterexample obeys , but the probability converges to rather than 0 as .

29 January, 2019 at 10:14 am

AnonymousHi Prof. Tao. Could you please clarify how you are using the change of variables formula in the proof of proposition 6?

29 January, 2019 at 7:56 pm

Terence TaoSee Theorem 32 of Notes 1.

1 February, 2019 at 9:46 am

AnonymousI still don’t see how the limits of integration changed to [0,1].

2 February, 2019 at 10:06 am

Terence TaoThe distribution of is symmetric around the origin, and is an even function of , so is equal to .

4 February, 2019 at 1:56 am

AnonymousAnd the 1/2? Following Theorem 32 I obtain .

4 February, 2019 at 8:47 am

Terence TaoThe uniform probability measure on is , not (the total measure has to equal 1 to be a probability measure). Similarly, the uniform probability measure on would be .

10 February, 2019 at 2:06 pm

haonanzHello Professor. Tao, I have question regard the result of Saint Petersburg Paradox, as in your note, we have shown converges to 1 in probability, and in Exercise 19 we can show it is almost surely unbounded. I see this as another example to show that convergence in probability is a weaker notion than a.s convergence, However I could not convince myself without going through the proof a sequence of a.s unbounded random variable can converge in probability to a finite value. Thanks,

10 February, 2019 at 6:13 pm

Terence TaoI am not sure what the actual question is here, but you can try proving to yourself the opposite claim “A sequence of a.s. unbounded random variables cannot converge in probability to (say) zero” and see where any arguments you have in mind to justify this break down. (One can also concoct an example by modifying the “typewriter sequence”, see Example 4 of https://terrytao.wordpress.com/2010/10/02/245a-notes-4-modes-of-convergence/ .)

12 February, 2019 at 8:26 am

haonanzThanks for your reply. I think I have found my source of confusion. I think the statement a.s unbounded here means the is a.s unbounded as opposed to every term in the sequence is a.s unbounded. As if the later is the case (for simplicity I will denote my random variables as ). By definition of a.s unbounded, would be true for all M and n, which I can use to conclude that cannot converge to any finite value (more direct one is zero). Using the typer writer example, then one I can think about is some kind of moving train as (n,n+1/n) with positive probability in the interval and zero outside. Is my understanding here correct here? Thanks again,

12 February, 2019 at 8:39 am

Terence TaoYes, in these notes “ is a.s. unbounded” is short for “The sequence is a.s. unbounded”, or equivalently that is almost surely infinite. One can also replace the supremum by a limit superior if desired. (Generally speaking, the terms “bounded” and “unbounded” only apply to functions and sequences, not to scalars, where one instead uses the distinction between “finite” and “infinite”.)

22 March, 2022 at 9:18 am

Aditya Guha RoyProf. Tao, in Definition 1 of convergence in probability you write it for “separable metric spaces”.

Is there any particular reason why studying convergence in probability is more meaningful when the underlying metric space (where the random variables take values) is separable?

22 March, 2022 at 9:24 am

AnonymousIt is the same as studying convergence of sequences in analysis. I don’t think there is anything special about probability spaces. A separable space is much more easier to work with than non-separable spaces. If a space is not separable, convergence property usually fails.

22 March, 2022 at 9:29 am

Aditya Guha RoyThanks!

Could you please tell about some particular examples where speaking about convergence in non-separable spaces can lead to counter-intuitive or misleading directions?

22 March, 2022 at 12:27 pm

Terence TaoBroadly speaking, by working in “countable” settings, one can safely ignore null events, since the countable union of null events is still null. Once one enters the “uncountable” world, though, it can require much more care to be able to view null events as negligible.

In my experience, one can often remove “countability” hypotheses such as separability (or being a Polish space, a standard Borel space, or a standard probability space) from basic tools in analysis and probability as long as everything is defined carefully (for instance, it can be convenient to replace the Borel sigma-algebra by the slightly smaller Baire sigma-algebra), but imposing such hypotheses allows one to ignore these sorts of technical subtleties (and also gains access to some useful technical tools, such as disintegration and measurable selection lemmas). For a typical instance where such hypotheses can be dropped if one works carefully enough, see https://mathoverflow.net/questions/352552/is-the-separability-of-the-space-needed-in-the-proof-of-the-prohorovs-theorem

In the long term, I expect that eventually a fully “uncountable” foundation to analysis and probability will be given, in which countability axioms are rarely needed; but with our current state of literature many of our basic tools are still restricted to the countable setting and so it is most convenient (especially for beginners) to largely restrict the scope of our theory to such countable settings.

22 March, 2022 at 9:26 am

Aditya Guha RoyIn Exercise 2 part (ix) Dominated Convergence theorem: we see that one easy way to prove it is to use Urysohn subsequence principle and then use Dominated Convergence Theorem for almost sure convergence to obtain that every subsequence of the sequence has a further subsequence converging to whence (since these are reals) the sequence itsef converges to .

Is there a more direct way to prove this?

22 March, 2022 at 9:44 am

Aditya Guha RoyThe same reason allows us to prove (vii).

(I think this is a correct and neat way, but it seems like we are cheating somehow by passing on to a subsequence.)

22 March, 2022 at 12:23 pm

Terence TaoAs a warm up, establish the bounded case of the dominated convergence theorem, in which the dominating variable is constant. The general case can then be established by truncating and applying a limiting argument.

23 March, 2022 at 1:26 pm

AnonymousA cute approach is to just change measures, define and apply Egorov’s theorem

23 March, 2022 at 9:41 pm

Convergence of random variables: part 1 | Aditya Guha Roy's weblog[…] reason why we took to be separable is briefly suggested in this comment by Professor Terence Tao. Unless mentioned otherwise, all random variables discussed here will be […]