In these notes we presume familiarity with the basic concepts of probability theory, such as random variables (which could take values in the reals, vectors, or other measurable spaces), probability, and expectation. Much of this theory is in turn based on measure theory, which we will also presume familiarity with. See for instance this previous set of lecture notes for a brief review.

The basic objects of study in analytic number theory are deterministic; there is nothing inherently random about the set of prime numbers, for instance. Despite this, one can still interpret many of the averages encountered in analytic number theory in probabilistic terms, by introducing random variables into the subject. Consider for instance the form

of the prime number theorem (where we take the limit ). One can interpret this estimate probabilistically as

where is a random variable drawn uniformly from the natural numbers up to , and denotes the expectation. (In this set of notes we will use boldface symbols to denote random variables, and non-boldface symbols for deterministic objects.) By itself, such an interpretation is little more than a change of notation. However, the power of this interpretation becomes more apparent when one then imports concepts from probability theory (together with all their attendant intuitions and tools), such as independence, conditioning, stationarity, total variation distance, and entropy. For instance, suppose we want to use the prime number theorem (1) to make a prediction for the sum

After dividing by , this is essentially

With probabilistic intuition, one may expect the random variables to be approximately independent (there is no obvious relationship between the number of prime factors of , and of ), and so the above average would be expected to be approximately equal to

which by (2) is equal to . Thus we are led to the prediction

The asymptotic (3) is widely believed (it is a special case of the *Chowla conjecture*, which we will discuss in later notes; while there has been recent progress towards establishing it rigorously, it remains open for now.

How would one try to make these probabilistic intuitions more rigorous? The first thing one needs to do is find a more quantitative measurement of what it means for two random variables to be “approximately” independent. There are several candidates for such measurements, but we will focus in these notes on two particularly convenient measures of approximate independence: the “” measure of independence known as covariance, and the “” measure of independence known as mutual information (actually we will usually need the more general notion of conditional mutual information that measures conditional independence). The use of type methods in analytic number theory is well established, though it is usually not described in probabilistic terms, being referred to instead by such names as the “second moment method”, the “large sieve” or the “method of bilinear sums”. The use of methods (or “entropy methods”) is much more recent, and has been able to control certain types of averages in analytic number theory that were out of reach of previous methods such as methods. For instance, in later notes we will use entropy methods to establish the logarithmically averaged version

of (3), which is implied by (3) but strictly weaker (much as the prime number theorem (1) implies the bound , but the latter bound is much easier to establish than the former).

As with many other situations in analytic number theory, we can exploit the fact that certain assertions (such as approximate independence) can become significantly easier to prove if one only seeks to establish them *on average*, rather than uniformly. For instance, given two random variables and of number-theoretic origin (such as the random variables and mentioned previously), it can often be extremely difficult to determine the extent to which behave “independently” (or “conditionally independently”). However, thanks to second moment tools or entropy based tools, it is often possible to assert results of the following flavour: if are a large collection of “independent” random variables, and is a further random variable that is “not too large” in some sense, then must necessarily be nearly independent (or conditionally independent) to many of the , even if one cannot pinpoint precisely which of the the variable is independent with. In the case of the second moment method, this allows us to compute correlations such as for “most” . The entropy method gives bounds that are significantly weaker quantitatively than the second moment method (and in particular, in its current incarnation at least it is only able to say non-trivial assertions involving interactions with residue classes at small primes), but can control significantly more general quantities for “most” thanks to tools such as the Pinsker inequality.

** — 1. Second moment methods — **

In this section we discuss probabilistic techniques of an “” nature. We fix a probability space to model all of random variables; thus for instance we shall model a complex random variable in these notes by a measurable function . (Strictly speaking, there is a subtle distinction one can maintain between a random variable and its various measure-theoretic models, which becomes relevant if one later decides to modify the probability space , but this distinction will not be so important in these notes and so we shall ignore it. See this previous set of notes for more discussion.)

We will focus here on the space of complex random variables (that is to say, measurable maps ) whose *second moment*

of is finite. In many number-theoretic applications the finiteness of the second moment will be automatic because will only take finitely many values. As is well known, the space has the structure of a complex Hilbert space, with inner product

and norm

for . By slight abuse of notation, the complex numbers can be viewed as a subset of , by viewing any given complex number as a constant (deterministic) random variable. Then is a one-dimensional subspace of , spanned by the unit vector . Given a random variable to , the projection of to is then the *mean*

and we obtain an orthogonal splitting of any into its mean and its mean zero part . By Pythagoras’ theorem, we then have

The first quantity on the right-hand side is the square of the distance from to , and this non-negative quantity is known as the variance

The square root of the variance is known as the standard deviation. The variance controls the distribution of the random variable through Chebyshev’s inequality

for any , which is immediate from observing the inequality and then taking expectations of both sides. Roughly speaking, this inequality asserts that typically deviates from its mean by no more than a bounded multiple of the standard deviation .

A slight generalisation of Chebyshev’s inequality that can be convenient to use is

for any and any complex number (which typically will be a simplified approximation to the mean ), which is proven similarly to (6) but noting (from (5)) that .

Informally, (6) is an assertion that a square-integrable random variable will concentrate around its mean if its variance is not too large. See these previous notes for more discussion of the concentration of measure phenomenon. One can often obtain stronger concentration of measure than what is provided by Chebyshev’s inequality if one is able to calculate higher moments than the second moment, such as the fourth moment or exponential moments , but we will not pursue this direction in this set of notes.

Clearly the variance is homogeneous of order two, thus

for any and . In particular, the variance is not always additive: the claim fails in particular when is not almost surely zero. However, there is an important substitute for this formula. Given two random variables , the inner product of the corresponding mean zero parts is a complex number known as the covariance:

As are orthogonal to , it is not difficult to obtain the alternate formula

for the covariance.

The covariance is then a positive semi-definite inner product on (it basically arises from the Hilbert space structure of the space of mean zero variables), and . From the Cauchy-Schwarz inequality we have

If have non-zero variance (that is, they are not almost surely constant), then the ratio

is then known as the correlation between and , and is a complex number of magnitude at most ; for real-valued that are not almost surely constant, the correlation is instead a real number between and . At one extreme, a correlation of magnitude occurs if and only if is a scalar multiple of . At the other extreme, a correlation of zero is an indication (though not a guarantee) of independence. Recall that two random variables are *independent* if one has

for all (Borel) measurable . In particular, setting , for and integrating using Fubini’s theorem, we conclude that

similarly with replaced by , and similarly for . In particular we have

and thus from (8) we thus see that independent random variables have zero covariance (and zero correlation, when they are not almost surely constant). On the other hand, the converse fails:

Exercise 1Provide an example of two random variables which are not independent, but which have zero correlation or covariance with each other. (There are many ways to produce some examples. One comes from exploiting various systems of orthogonal functions, such as sines and cosines. Another comes from working with random variables taking only a small number of values, such as .

for any finite collection of random variables . These identities combine well with Chebyshev-type inequalities such as (6), (7), and this leads to a very common instance of the second moment method in action. For instance, we can use it to understand the distribution of the number of prime factors of a random number that fall within a given set . Given any set of natural numbers, define the *logarithmic size* to be the quantity

Thus for instance Euler’s theorem asserts that the primes have infinite logarithmic size.

Lemma 2 (Turan-Kubilius inequality, special case)Let be an interval of length at least , and let be an integer drawn uniformly at random from this interval, thusfor all . Let be a finite collection of primes, all of which have size at most . Then the random variable has mean

and variance

In particular,

and from (7) we have

for any .

*Proof:* For any natural number , we have

We now write . From (11) we see that each indicator random variable , has mean and variance ; similarly, for any two distinct , we see from (11), (8) the indicators , have covariance

and the claim now follows from (10).

The exponents of in the error terms here are not optimal; but in practice, we apply this inequality when is much larger than any given power of , so factors such as will be negligible. Informally speaking, the above lemma asserts that a typical number in a large interval will have roughly prime factors in a given finite set of primes, as long as the logarithmic size is large.

If we apply the above lemma to for some large , and equal to the primes up to (say) , we have , and hence

Since , we recover the main result

of Section 5 of Notes 1 (indeed this is essentially the same argument as in that section, dressed up in probabilistic language). In particular, we recover the Hardy-Ramanujan law that a proportion of the natural numbers in have prime factors.

Exercise 3 (Turan-Kubilius inequality, general case)Let be an additive function (which means that whenever are coprime. Show thatwhere

(Hint: one may first want to work with the special case when vanishes whenever so that the second moment method can be profitably applied, and then figure out how to address the contributions of prime powers larger than .)

Exercise 4 (Turan-Kubilius inequality, logarithmic version)Let with , and let be a collection of primes of size less than with . Show that

Exercise 5 (Paley-Zygmund inequality)Let be non-negative with positive mean. Show thatThis inequality can sometimes give slightly sharper results than the Chebyshev inequality when using the second moment method.

Now we give a useful lemma that quantifies a heuristic mentioned in the introduction, namely that if several random variables do not correlate with each other, then it is not possible for any further random variable to correlate with many of them simultaneously. We first state an abstract Hilbert space version.

Lemma 6 (Bessel type inequality, Hilbert space version)If are elements of a Hilbert space , and are positive reals, then

*Proof:* We use the duality method. Namely, we can write the left-hand side of (13) as

for some complex numbers with (just take to be normalised by the left-hand side of (13), or zero if that left-hand side vanishes. By Cauchy-Schwarz, it then suffices to establish the dual inequality

The left-hand side can be written as

Using the arithmetic mean-geometric mean inequality and symmetry, this may be bounded by

Since , the claim follows.

Corollary 7 (Bessel type inequality, probabilistic version)If , and are positive reals, then

*Proof:* By subtracting the mean from each of we may assume that these random variables have mean zero. The claim now follows from Lemma 6.

To get a feel for this inequality, suppose for sake of discussion that and all have unit variance and , but that the are pairwise uncorrelated. Then the right-hand side is equal to , and the left-hand side is the sum of squares of the correlations between and each of the . Any individual correlation is then still permitted to be as large as , but it is not possible for multiple correlations to be this large simultaneously. This is geometrically intuitive if one views the random variables as vectors in a Hilbert space (and correlation as a rough proxy for the angle between such vectors). This lemma also shares many commonalities with the large sieve inequality, discussed in this set of notes.

One basic number-theoretic application of this inequality is the following sampling inequality of Elliott, that lets one approximate a sum of an arithmetic function by its values on multiples of primes :

Exercise 8 (Elliott’s inequality)Let be an interval of length at least . Show that for any function , one has(

Hint:Apply Corollary 7 with , , and , where is the uniform variable from Lemma 2.) Conclude in particular that for every , one hasfor all primes outside of a set of exceptional primes of logarithmic size .

Informally, the point of this inequality is that an arbitrary arithmetic function may exhibit correlation with the indicator function of the multiples of for some primes , but cannot exhibit significant correlation with all of these indicators simultaneously, because these indicators are not very correlated to each other. We note however that this inequality only gains a tiny bit over trivial bounds, because the set of primes up to only has logarithmic size by Mertens’ theorems; thus, any asymptotics that are obtained using this inequality will typically have error terms that only improve upon the trivial bound by factors such as .

Exercise 9 (Elliott’s inequality, logarithmic form)Let with . Show that for any function , one hasand thus for every , one has

for all primes outside of an exceptional set of primes of logarithmic size .

Exercise 10Use Exercise (9) and a duality argument to provide an alternate proof of Exercise 4. (Hint:express the left-hand side of (12) as a correlation between and some suitably -normalised arithmetic function .)

As a quick application of Elliott’s inequality, let us establish a weak version of the prime number theorem:

Proposition 11 (Weak prime number theorem)For any we havewhenever are sufficiently large depending on .

This estimate is weaker than what one can obtain by existing methods, such as Exercise 56 of Notes 1. However in the next section we will refine this argument to recover the full prime number theorem.

*Proof:* Fix , and suppose that are sufficiently large. From Exercise 9 one has

for all primes outside of an exceptional set of logarithmic size . If we restrict attention to primes then one sees from the integral test that one can replace the sum by and only incur an additional error of . If we furthermore restrict to primes larger than , then the contribution of those that are divisible by is also . For not divisible by , one has . Putting all this together, we conclude that

for all primes outside of an exceptional set of logarithmic size . In particular, for large enough this statement is true for at least one such . The claim then follows.

As another application of Elliott’s inequality, we present a criterion for orthogonality between multiplicative functions and other sequences, first discovered by Katai (with related results also introduced earlier by Daboussi and Delange), and rediscovered by Bourgain, Sarnak, and Ziegler:

Proposition 12 (Daboussi-Delange-Katai-Bourgain-Sarnak-Ziegler criterion)Let be a multiplicative function with for all , and let be another bounded function. Suppose that one hasas for any two distinct primes . Then one has

as .

*Proof:* Suppose the claim fails, then there exists (which we can assume to be small) and arbitrarily large such that

By Exercise 8, this implies that

for all primes outside of an exceptional set of logarithmic size . Call such primes “good primes”. In particular, by the pigeonhole principle, and assuming large enough, there exists a dyadic range with which contains good primes.

Fix a good prime in . From (15) we have

We can replace the range by with negligible error. We also have except when is a multiple of , but this latter case only contributes which is also negligible compared to the right-hand side. We conclude that

for every good prime. On the other hand, from Lemma 6 we have

where range over the good primes in . The left-hand side is then , and by hypothesis the right-hand side is for large enough. As and is small, this gives the desired contradiction

Exercise 13 (Daboussi-Delange theorem)Let be irrational, and let be a multiplicative function with for all . Show thatas . If instead is rational, show that there exists be a multiplicative function with for which the statement (16) fails. (Hint: use Dirichlet characters and Plancherel’s theorem for finite abelian groups.)

** — 2. An elementary proof of the prime number theorem — **

Define the Mertens function

As shown in Theorem 58 of Notes 1, the prime number theorem is equivalent to the bound

as . We now give a recent proof of this theorem, due to Redmond McNamara (personal communication), that relies primarily on Elliott’s inequality and the Selberg symmetry formula; it is a relative of the standard elementary proof of this theorem due to Erdös and Selberg. In order to keep the exposition simple, we will not arrange the argument in a fashion that optimises the decay rate (in any event, there are other proofs of the prime number theorem that give significantly stronger bounds).

Firstly we see that Elliott’s inequality gives the following weaker version of (17):

Lemma 14 (Oscillation for Mertens’ function)If and , then we havefor all primes outside of an exceptional set of primes of logarithmic size .

*Proof:* We may assume as the claim is trivial otherwise. From Exercise 8 applied to and , we have

for all outside of an exceptional set of primes of logarithmic size . Since for not divisible by , the right-hand side can be written as

Since outside of an exceptional set of logarithmic size , the claim follows.

Informally, this lemma asserts that for most primes , which morally implies that for most primes . If we can then locate suitable primes with , thus should then lead to , which should then yield the prime number theorem . The manipulations below are intended to make this argument rigorous.

It will be convenient to work with a logarithmically averaged version of this claim.

Corollary 15 (Logarithmically averaged oscillation)If and is sufficiently large depending on , then

*Proof:* For each , we have from the previous lemma that

for all outside of an exceptional set of logarithmic size . We then have

so it suffices by Markov’s inequality to show that

But by Fubini’s theorem, the left-hand side may be bounded by

and the claim follows.

Let be sufficiently small, and let be sufficiently large depending on . Call a prime *good* if the bound (18) holds and *bad* otherwise, thus all primes outside of an exceptional set of bad primes of logarithmic size are good. Now we observe that we can make small as long as we can make two good primes multiply to be close to a third:

*Proof:* By definition of good prime, we have the bounds

We rescale (20) by to conclude that

We can replace the integration range here from to with an error of if is large enough. Also, since , we have . Thus we have

Combining this with (19), (21) and the triangle inequality (writing as a linear combination of , , and ) we conclude that

This is an averaged version of the claim we need. To remove the averaging, we use the identity (see equation (63) of Notes 1) to conclude that

From the triangle inequality one has

and hence by Mertens’ theorem

From the Brun-Titchmarsh inequality (Corollary 61 of Notes 1) we have

and so from the previous estimate and Fubini’s theorem one has

and hence by (22) (using trivial bounds to handle the region outside of )

Since

we conclude (for large enough) that

and the claim follows.

To finish the proof of the prime number theorem, it thus suffices to locate, for sufficiently large, three good primes with . If we already had the prime number theorem, or even the weaker form that every interval of the form contained primes for large enough, then this would be quite easy: pick a large natural number (depending on , but independent of ), so that the primes up to has logarithmic size (so that only of them are bad, as measured by logarithmic size), and let be random numbers and drawn uniformly from (say) . From the prime number theorem, for each , the interval contains primes. In particular, contains primes, but the expected number of bad primes in this interval is . Thus by Markov’s inequality there would be at least a chance (say) of having at least one good prime in ; similarly there is a chance of having a good prime in , and a chance of having a good prime in . Thus (as an application of the probabilistic method), there exist (deterministic) good primes with the required properties.

Of course, using the prime number theorem here to prove the prime number theorem would be circular. However, we can still locate a good triple of primes using the Selberg symmetry formula

as , where is the second von Mangoldt function

see Proposition 60 of Notes 1. We can strip away the contribution of the primes:

Exercise 17Show thatas .

In particular, on evaluating this at and subtracting, we have

whenever is sufficiently large depending on . In particular, for any such , one either has

(or both). Informally, the Selberg symmetry formula shows that the interval contains either a lot of primes, or a lot of semiprimes. The factor of is slightly annoying, so we now remove it. Consider the contribution of those primes to (25) with . This is bounded by

which we can bound crudely using the Chebyshev bound by

which by Mertens theorem is . Thus the contribution of this case can be safely removed from (25). Similarly for those cases when . For the remaining cases we bound . We conclude that for any sufficiently large , either (24) or

holds (or both).

In order to find primes with close to , it would be very convenient if we could find a for which (24) and (26) *both* hold. We can’t quite do this directly, but due to the “connected” nature of the set of scales , but we can do the next best thing:

Proposition 18Suppose is sufficiently large depending on . Then there exists with such that

*Proof:* We know that every in obeys at least one of (27), (28). Our task is to produce an adjacent pair of , one of which obeys (27) and the other obeys (28). Suppose for contradiction that no such pair exists, then whenever fails to obey (27), then any adjacent must also fail to do so, and similarly for (28). Thus either (27) will fail to hold for all , or (28) will fail to hold for all such . If (27) fails for all , then on summing we have

which contradicts Mertens’ theorem if is small enough because the left-hand side is . Similarly, if (28) fails for all , then

and again Mertens’ theorem can be used to lower bound the left-hand side by (in fact one can even gain an additional factor of if one works things through carefully) and obtain a contradiction.

The above proposition does indeed provide a triple of primes with . If is sufficiently large depending on and less than (say) , so that , this would give us what we need as long as one of the triples consisted only of good primes. The only way this can fail is if either

for some , or if

for some . In the first case, we can sum to conclude that

and in the second case we have

Since the total set of bad primes up to has logarithmic size , we conclude from the pigeonhole principle (and the divergence of the harmonic series ) that for any depending only on , and any large enough, there exists such that neither of (29) and (30) hold. Indeed the set of obeying (29) has logarithmic size , and similarly for (30). Choosing a that avoids both of these scenarios, we then find a good and good with , so that , and then by Proposition 16 we conclude that for all sufficiently large . Sending to zero, we obtain the prime number theorem.

** — 3. Entropy methods — **

In the previous section we explored the consequences of the second moment method, which applies to square-integrable random variables taking values in the real or complex numbers. Now we explore entropy methods, which now apply to random variables which take a finite number of values (equipped with the discrete sigma-algebra), but whose range need not be numerical in nature. (One could extend entropy methods to slightly larger classes of random variables, such as ones that attain a countable number of values, but for our applications finitely-valued random variables will suffice.)

The fundamental notion here is that of the Shannon entropy of a random variable. If takes values in a finite set , its Shannon entropy (or *entropy* for short) is defined by the formula

where ranges over all the possible values of , and we adopt the convention , so that values that are almost surely not attained by do not influence the entropy. We choose here to use the natural logarithm to normalise our entropy (in which case a unit of entropy is known as a “nat“); in the information theory literature it is also common to use the base two logarithm to measure entropy (in which case a unit of entropy is known as a “bit“, which is equal to nats). However, the precise choice of normalisation will not be important in our discussion.

It is clear that if two random variables have the same probability distribution, then they have the same entropy. Also, the precise choice of range set is not terribly important: if takes values in , and is an injection, then it is clear that and have the same entropy:

This is in sharp contrast to moment-based statistics such as the mean or variance, which can be radically changed by applying some injective transformation to the range values.

Informally, the entropy informally measures how “spread out” or “disordered” the distribution of is, behaving like a logarithm of the size of the “essential support” of such a variable; from an information-theoretic viewpoint, it measures the amount of “information” one learns when one is told the value of . Here are some basic properties of Shannon entropy that help support this intuition:

Exercise 19 (Basic properties of Shannon entropy)Let be a random variable taking values in a finite set .

- (i) Show that , with equality if and only if is almost surely deterministic (that is to say, it is almost surely equal to a constant ).
- (ii) Show that
with equality if and only if is uniformly distributed on . (Hint: use Jensen’s inequality and the concavity of the map on .)

- (iii) (Shannon-McMillan-Breiman theorem) Let be a natural number, and let be independent copies of . As , show that there is a subset of cardinality with the properties that
and

uniformly for all . (The proof of this theorem will require Stirling’s formula, which you may assume here as a black box; see also this previous blog post.) Informally, we thus see a large tuple of independent samples of approximately behaves like a uniform distribution on values.

One can view Shannon entropy as a generalisation of the notion of cardinality of a finite set (or equivalently, cardinality of finite sets can be viewed as a special case of Shannon entropy); see this previous blog post for an elaboration of this point.

The concept of Shannon entropy becomes significantly more powerful when combined with that of conditioning. Recall that a random variable taking values in a range set can be modeled by a measurable map from a probability space to the range . If is an event in of positive probability, we can then *condition* to the event to form a new random variable on the conditioned probability space , where

is the restriction of the -algebra to ,

is the conditional probability measure on , and is the restriction of to . This random variable lives on a different probability space than itself, so it does not make sense to directly combine these variables (thus for instance one cannot form the sum even when both random variables are real or complex valued); however, one can still form the Shannon entropy of the conditioned random variable , which is given by the same formula

Given another random variable taking values in another finite set , we can then define the conditional Shannon entropy to be the expected entropy of the level sets , thus

with the convention that the summand here vanishes when . From the law of total probability we have

for any , and hence by Jensen’s inequality

for any ; summing we obtain the Shannon entropy inequality

Informally, this inequality asserts that the new information content of can be decreased, but not increased, if one is first told some additional information .

This inequality (33) can be rewritten in several ways:

Exercise 20Let , be random variables taking values in finite sets respectively.

- (i) Establish the chain rule
where is the joint random variable . In particular, (33) can be expressed as a subadditivity formula

- (ii) If is a function of , in the sense that for some (deterministic) function , show that .
- (iii) Define the mutual information by the formula
Establish the inequalities

with the first inequality holding with equality if and only if are independent, and the latter inequalities holding if and only if is a function of (or vice versa).

From the above exercise we see that the mutual information is a measure of dependence between and , much as correlation or covariance was in the previous sections. There is however one key difference: whereas a zero correlation or covariance is a consequence but not a guarantee of independence, zero mutual information is *logically equivalent* to independence, and is thus a stronger property. To put it another way, zero correlation or covariance allows one to calculate the average in terms of individual averages of , but zero mutual information is stronger because it allows one to calculate the more general averages in terms of individual averages of , for arbitrary functions taking values into the complex numbers. This increased power of the mutual information statistic will allow us to estimate various averages of interest in analytic number theory in ways that do not seem amenable to second moment methods.

The subadditivity property formula can be conditioned to any event occuring with positive probability (replacing the random variables by their conditioned counterparts ), yielding the inequality

Applying this inequality to the level events of some auxiliary random variable taking values in another finite set , multiplying by , and summing, we conclude the inequality

In other words, the conditional mutual information

between and conditioning on is always non-negative:

One has conditional analogues of the above exercise:

Exercise 21Let , , be random variables taking values in finite sets respectively.

- (i) Establish the conditional chain rule
In particular, (36) is equivalent to the inequality

- (ii) Show that equality holds in (36) if and only if are conditionally independent relative to , which means that
for any , , .

- (iii) Show that , with equality if and only if is almost surely a deterministic function of .
- (iv) Show the data processing inequality
for any functions , , and more generally that

- (v) If is an injective function, show that
However, if is not assumed to be injective, show by means of examples that there is no order relation between the left and right-hand side of (40) (in other words, show that either side may be greater than the other). Thus, increasing or decreasing the amount of information that is known may influence the mutual information between two remaining random variables in either direction.

- (vi) If is a function of , and also a function of (thus for some and ), and a further random variable is a function jointly of (thus for some ), establish the submodularity inequality

We now give a key motivating application of the Shannon entropy inequalities. Suppose one has a sequence of random variables, all taking values in a finite set , which are stationary in the sense that the tuples and have the same distribution for every . In particular we will have

and hence by (39)

If we write , we conclude from (34) that we have the concavity property

In particular we have for any , which on summing and telescoping series (noting that ) gives

and hence we have the entropy monotonicity

In particular, the limit exists. This quantity is known as the Kolmogorov-Sinai entropy of the stationary process ; it is an important statistic in the theory of dynamical systems, and roughly speaking measures the amount of entropy produced by this process as a function of a discrete time vairable . We will not directly need the Kolmogorov-Sinai entropy in our notes, but a variant of the entropy monotonicity formula (41) will be important shortly.

In our application we will be dealing with processes that are only asymptotically stationary rather than stationary. To control this we recall the notion of the total variation distance between two random variables taking values in the same finite space , defined by

There is an essentially equivalent notion of this distance which is also often in use:

Exercise 22If two random variables take values in the same finite space , establish the inequalities

Shannon entropy is continuous in total variation distance as long as we keep the range finite. More quantitatively, we have

Lemma 23If two random variables take values in the same finite space , thenwith the convention that the error term vanishes when .

*Proof:* Set . The claim is trivial when (since then have the same distribution) and when (from (32)), so let us assume , and our task is to show that

If we write , , and , then

By dividing into the cases and we see that

since , it thus suffices to show that

But from Jensen’s inequality (32) one has

since , the claim follows.

In the converse direction, if a random variable has entropy close to the maximum , then one can control the total variation:

Lemma 24 (Special case of Pinsker inequality)If takes values in a finite set , and is a uniformly distributed random variable on , then

Of course, we have , so we may also write the above inequality as

The optimal value of the implied constant here is known to equal , but we will not use this sharp version of the inequality here.

*Proof:* If we write and , and , then we can rewrite the claimed inequality as

Observe that the function is concave, and in fact for all . From this and Taylor expansion with remainder we may write

for some between and . Since is independent of , and , we thus have on summing in

By Cauchy-Schwarz we then have

Since and , the claim follows.

The above lemma does not hold when the comparison variable is not assumed to be uniform; in particular, two non-uniform random variables can have precisely the same entropy but yet have different distributions, so that their total variation distance is positive. There is a more general variant, known as the Pinsker inequality, which we will not use in these notes:

Exercise 25 (Pinsker inequality)If take values in a finite set , define the Kullback-Leibler divergence of relative to by the formula(with the convention that the summand vanishes when vanishes).

- (i) Establish the Gibbs inequality .
- (ii) Establish the Pinsker inequality
In particular, vanishes if and only if have identical distribution. Show that this implies Lemma 24 as a special case.

- (iii) Give an example to show that the Kullback-Liebler divergence need not be symmetric, thus there exist such that .
- (iv) If are random variables taking values in finite sets , and are
independentrandom variables taking values in respectively with each having the same distribution of , show that

In our applications we will need a relative version of Lemma 24:

Corollary 26 (Relative Pinsker inequality)If takes values in a finite set , takes values in a finite set , and is a uniformly distributed random variable on that is independent of , then

*Proof:* From direct calculation we have the identity

As is independent of , is uniformly distributed on . From Lemma 24 we conclude

Inserting this bound and using the Cauchy-Schwarz inequality, we obtain the claim.

Now we are ready to apply the above machinery to give a key inequality that is analogous to Elliott’s inequality. Inequalities of this type first appeared in one of my papers, introducing what I called the “entropy decrement argument”; the following arrangement of the inequality and proof is due to Redmond McNamara (personal communication).

Theorem 27 (Entropy decrement inequality)Let be a random variable taking values in a finite set of integers, which obeys the approximate stationarityfor some . Let be a collection of distinct primes less than some threshold , and let be natural numbers that are also bounded by . Let be a function taking values in a finite set . For , let denote the -valued random variable

and let denote the -valued random variable

Also, let be a random variable drawn uniformly from , independently of . Then

The factor (arising from an invocation of the Chinese remainder theorem in the proof) unfortunately restricts the usefulness of this theorem to the regime in which all the primes involved are of “sub-logarithmic size”, but once one is in that regime, the second term on the right-hand side of (45) tends to be negligible in practice. Informally, this theorem asserts that for most small primes , the random variables and behave as if they are independent of each other.

*Proof:* We can assume , as the claim is trivial for (the all have zero entropy). For , we introduce the -valued random variable

The idea is to exploit some monotonicity properties of the quantity , in analogy with (41). By telescoping series we have

where we extend (44) to the case. From (38) we have

Now we lower bound the summand on the right-hand side. From multiple applications of the conditional chain rule (37) we have

We now use the approximate stationarity of to derive an approximate monotonicity property for . If , then from (39) we have

Write and

Note that is a deterministic function of and vice versa. Thus we can replace by in the above formula, and conclude that

The tuple takes values in a set of cardinality thanks to the Chebyshev bounds. Hence by two applications of Lemma 23, (43) we have

The first term on the right-hand side is . Worsening the error term slightly, we conclude that

and hence

for any . In particular

which by (47), (48) rearranges to

From (46) we conclude that

Meanwhile, from Corollary 26, (39), (38) we have

The probability distribution of is a function on , which by the Chinese remainder theorem we can identify with a cyclic group where . From (43) we see that the value of this distribution at adjacent values of this cyclic group varies by , hence the total variation distance between this random variable and the uniform distribution on is by Chebyshev bounds. By Lemma 23 we then have

and thus

The claim follows.

We now compare this result to Elliott’s inequality. If one tries to address precisely the same question that Elliott’s inequality does – namely, to try to compare a sum with sampled subsums – then the results are quantitatively much weaker:

Corollary 28 (Weak Elliott inequality)Let be an interval of length at least . Let be a function with for all , and let . Then one hasfor all primes outside of an exceptional set of primes of logarithmic size .

Comparing this with Exercise 8 we see that we cover a much smaller range of primes ; also the size of the exceptional set is slightly worse. This version of Elliot’s inequality is however still strong enough to recover a proof of the prime number theorem as in the previous section.

*Proof:* We can assume that is small, as the claim is trivial for comparable to . We can also assume that

since the claim is also trivial otherwise (just make all primes up to exceptional, then use Mertens’ theorem). As a consequence of this, any quantity involving in the denominator will end up being completely negligible in practice. We can also restrict attention to primes less than (say) , since the remaining primes between and have logarithmic size .

By rounding the real and imaginary parts of to the nearest multiple of , we may assume that takes values in some finite set of complex numbers of size with cardinality . Let be drawn uniformly at random from . Then (43) holds with , and from Theorem 27 with and (which makes the second term of the right-hand side of (45) negligible) we have

where are the primes up to , arranged in increasing order. By Markov’s inequality, we thus have

for outside of a set of primes of logarithmic size .

Let be as above. Now let be the function

that is to say picks out the unique component of the tuple in which is divisible by . This function is bounded by , and then by (42) we have

The left-hand side is equal to

which on switching the summations and using the large nature of can be rewritten as

Meanwhile, the left-hand side is equal to

which again by switching the summations becomes

The claim follows.

In the above argument we applied (42) with a very specific choice of function . The power of Theorem 27 lies in the ability to select many other such functions , leading to estimates that do not seem to be obtainable purely from the second moment method. In particular we have the following generalisation of the previous estimate:

Proposition 29 (Weak Elliott inequality for multiple correlations)Let be an interval of length at least . Let be a function with for all , and let . Let be integers. Then one hasfor all primes outside of an exceptional set of primes of logarithmic size .

*Proof:* We allow all implied constants to depend on . As before we can assume that is sufficiently small (depending on ), that takes values in a set of bounded complex numbers of cardinality , and that is large in the sense of (49), and restrict attention to primes up to . By shifting the and using the large nature of we can assume that the are all non-negative, taking values in for some . We now apply Theorem 27 with and conclude as before that

for outside of a set of primes of logarithmic size .

Let be as above. Let be the function

This function is still bounded by , so by (42) as before we have

The left-hand side is equal to

which on switching the summations and using the large nature of can be rewritten as

Meanwhile, the right-hand side is equal to

which again by switching the summations becomes

The claim follows.

There is a logarithmically averaged version of the above proposition:

Exercise 30 (Weak Elliott inequality for logarithmically averaged multiple correlations)Let with , let be a function bounded in magnitude by , let , and let be integers. Show thatfor all primes outside of an exceptional set of primes of logarithmic size .

When one specialises to multiplicative functions, this lets us dilate shifts in multiple correlations by primes:

Exercise 31Let with , let be a multiplicative function bounded in magnitude by , let , and let be nonnegative integers. Show thatfor all primes outside of an exceptional set of primes of logarithmic size .

For instance, setting to be the Möbius function, , , and (say), we see that

for all primes outside of an exceptional set of primes of logarithmic size . In particular, for large enough, one can obtain bounds of the form

for various moderately large sets of primes . It turns out that these double sums on the right-hand side can be estimated by methods which we will cover in later series of notes. Among other things, this allows us to establish estimates such as

as , which to date have only been established using these entropy methods (in conjunction with the methods discussed in later notes). This is progress towards an open problem in analytic number theory known as *Chowla’s conjecture*, which we will also discuss in later notes.

## 33 comments

Comments feed for this article

12 November, 2019 at 8:22 pm

asahay22Some minor typos:

1. (8) has a Y where you mean bolded Y.

2. There’s something funny happening with your links/equation references right before (22), where you cite the Brun-Titchmarsh inequality.

3. I think the equation references in Lemma 6 are probably supposed to be (13) not (14).

[Corrected, thanks – T.]13 November, 2019 at 2:49 pm

AnonymousIt seems that the proof of lemma 2 does not require to be a collection of primes.

13 November, 2019 at 4:27 pm

Terence TaoYes, pairwise coprime would suffice in this case. (Similarly for some other results in this post.) I know some people have experimented with trying to work with other sets of divisors here than primes, though as far as I am aware in most applications of these sorts of techniques the best set of divisors to use are primes.

13 November, 2019 at 5:35 pm

Entropy eigenvectorsHi Terry,

Is the entropy method useful in isolating the plausibility of the eigenvalue/vector formulae that you and co-authors recently discovered (mentioned in Quanta today) without explicitly using those formulae? Something like: the entropy of the eigenvalues of a matrix A, conditioned on the eigenvalues of the minors, equals the entropy of the eigenvectors of A?

13 November, 2019 at 7:13 pm

Lior SilbermanThere’s a missing in the middle term of the equation before equation (41).

[Corrected, thanks -T.]14 November, 2019 at 12:22 am

roland5999.. another funny behavior happens just before (19), where the Latex source code shows instead of the intended neatly-formatted expression…

[Corrected, thanks -T.]14 November, 2019 at 7:07 am

Kirill MI guess a verb after “may” is omitted in “With probabilistic intuition, one may the random variables”. Most likely, it is “assume” or alike.

[Corrected, thanks -T.]17 November, 2019 at 10:02 am

entropySorry for the possibly dumb question but I do not understand the submodularity inequality (vi). Let us suppose that and are independent, and the function is injective, so that . Then, the submodularity inequality would imply , always. I am missing something?

17 November, 2019 at 4:35 pm

Terence TaoYes, that is correct. If is a function of a random variable , and is also a function of an independent random variable , then it is deterministic.

18 November, 2019 at 1:27 am

entropyThank you for your kind answer. However, please consider the following example. Let and be two independent random variables. The random variable assumes values in equiprobably, and let be such that , while . Then assumes value with probability and value with probability . Now, let assume value with probability and value with probability . Clearly, , where is a 1-1 function. Therefore, is a function of two independent random variables, but it is not deterministic. I am missing some additional hypothesis?

18 November, 2019 at 8:13 am

Terence TaoIt is true that and both assume the value with probability 2/3 and with probability 1/3, but this does not imply that , because the events and are not the same event (even though they occur with the same probability). What is true (assuming some additional regularity hypotheses on the underlying sample space ) is that there is a probability preserving transformation such that for almost all in the sample space, but this is a different claim.

18 November, 2019 at 8:49 am

entropyOK, got it!. Thanks a lot.

20 November, 2019 at 2:14 pm

MarcNgIn Lemma 24 , we have the claim “…From this and Taylor expansion with remainder we have the inequality” but instead appears an equality there.

[Corrected, thanks – T.]20 November, 2019 at 8:57 pm

n_tropyGreat post, thanks. Possible typo: the direction of the inequalities just before (41) should be reversed. From h_{n+1} – h_n \leq h_n – h_{n-1} it would follow that h_{n+1} – h_n \leq h_i – h_{i-1} for i = 1..n.

[Corrected, thanks – T.]1 December, 2019 at 12:26 pm

anonymousDear professor Tao,

Erring on youtube I got into a 2015 presentation you did at ipam UCLA on this topic (you were using the prisoner + snake/cliff metaphor that time), which got me really excited because it relates to an initial problem that I worked on in material sciences in the past. Take a random material (maybe it’s a biphasic material or something more complicated), you want to compute its averaged (homogenised) properties, which is something we know how to do very well – only it requires tremendous computational power.

So the idea was, instead of exploring all the probability space, to get one carefully chosen configuration on which to do the complicated thing. So we did a “sudoku like” thing alike the one you were talking about in that presentation (with a bit of variation, because the aim was slightly different: the goal was to have that configuration work out all moments… well it’s related but not the same I think?). But I could never understand that problem as the size went larger, why it was complicated or even if it were complicated in the first place.

So, that presentation was super interesting, now I know the initial problem (from which we of course strayed rather quickly because applied maths :) We kept working with {+1, -1} though. The focus was on the PDEs part of it.) was complicated, and I have some ideas why, and congratulations on the proof of it.

And I hope that made you smile to see that this work could have a ramification in the (mostly) unrelated field of material sciences.

[On a side note, I’m always super happy to read your blog posts (the articles that are close enough to my specialties for me to get something out of them), they shine a light on the little hole into which life and depression have buried me, a place sadly away from maths. Maybe you get that positive feedback all the time, but I’ve been meaning to write this down for a long time.]

17 December, 2019 at 8:27 am

254A, Notes 10 – mean values of nonpretentious multiplicative functions | What's new[…] for instance, in order to combine this type of result with the entropy decrement methods from Notes 9, it is essential that be allowed to grow more slowly than . See also this survey of Soundararajan […]

15 April, 2020 at 5:37 am

AlexanderIn the proof Proposition 18 in the first sum probably schould be ..p\le e^{\epsilon^{-1}+k_0\epsilon}. On the other hand is not clear the sentence

contradicts Mertens’ theorem if {\varepsilon} is large enough because the left-hand …

We need \epsilon \to 0 ?

[Corrected, thanks – T.]15 April, 2020 at 9:56 am

AlexanderOnce again some question related proof of Proposition 18.

By Mertens Theorem we have

From this not follows the contradiction with inequality

. I am right ?

P.S. In (27) instead p_1 should be p.

[Yes, that is correct. The typo is now corrected, thanks -T.]1 May, 2020 at 1:51 am

AlexanderInProbably In Exercise 17 instead \log^2 x should be \log^2 p, and in (26) instead \frac{e^{\epsilon k}}{k} should be \frac{1}{k}.

12 October, 2020 at 1:28 pm

254A announcement: Analytic prime number theory | What's new[…] The entropy decrement argument […]

13 November, 2020 at 4:33 pm

WBIn exercise 19, it should be “use the {\it concavity} of $t \to t\log\left(\frac{1}{t}\right)$”.

[Corrected, thanks – T.]21 January, 2022 at 2:05 pm

AnonymousIn the equation immediately before Exercise 3, there should be an instead of an .

[Corrected, thanks -T.]23 January, 2022 at 6:17 pm

AnonymousIn the proof of Lemma 6, you say “[…] normalised by the left-hand side of (14)”. I believe it should be equation (13).

[Corrected, thanks – T.]24 January, 2022 at 9:42 am

AnonymousIn the proof of Proposition 11, in the sentence, “[i]f we furthermore restrict to primes larger than , then the contribution of those that are divisible by is also “, should it be larger than ?

[Corrected, thanks – T.]30 January, 2022 at 4:18 pm

AnonymousIn the proof of Lemma 14, you state that “… for all outside of an exceptional set of primes of logarithmic size .” I believe that this should be .

[Corrected, thanks – T.]31 January, 2022 at 8:27 am

AnonymousSorry if I’m missing something obvious, but in the proof of Proposition 16, the integral in the equation immediately below the statement, “We rescale (20) by to conclude that…” has the bounds , when it seems like it should be .

Also, in that same proof, you say that “since , we have “. I want to say that there should be a there.

[Corrected, thanks – T.]7 February, 2022 at 2:09 pm

AnonymousIn the first equation in the proof of Proposition 18, the upper bound on your sum is , when I think you mean . Also, immediately following this, you say that this “contradicts Mertens’ theorem if is large enough”. Do you mean small enough?

[Corrected, thanks – T.]7 February, 2022 at 2:47 pm

AnonymousI think he meant if is large enough.

8 February, 2022 at 11:37 pm

AnonymousThe right-hand side of the main equation in Exercise 31 seems to be cut off.

[Corrected, thanks – T.]9 February, 2022 at 1:55 am

AnonymousAlso, out of curiosity, what notes are you referring to when you mention a “later series of notes” at the very end of this blog post? For instance, in the sentence “It turns out that these double sums on the right-hand side can be estimated by methods which we will cover in later series of notes.”

[

[See https://terrytao.wordpress.com/2019/12/17/254a-notes-10-mean-values-of-nonpretentious-multiplicative-functions/ -T.]9 February, 2022 at 1:52 pm

AnonymousThank you!

25 March, 2022 at 8:55 pm

James LengPerhaps this is a naive question, but could iterating this entropy decrement argument lead to a regularity lemma like its “cousin” the energy decrement argument does?

28 April, 2022 at 6:24 am

Rory McClenaganIn Exercise 30, I think you might should be dividing the denominator of the summands by . Also, in the proofs of Propositions 28 and 29, you say “[m]eanwhile, the left-hand side is equal to”, and I think it should be “[m]eanwhile, the right-hand side is equal to”.

[Corrected, thanks – T.]