Consider the sum of iid real random variables of finite mean and variance for some . Then the sum has mean and variance , and so (by Chebyshev’s inequality) we expect to usually have size . To put it another way, if we consider the *normalised sum*

then has been normalised to have mean zero and variance , and is thus usually of size .

In the previous set of notes, we were able to establish various tail bounds on . For instance, from Chebyshev’s inequality one has

and if the original distribution was bounded or subgaussian, we had the much stronger Chernoff bound

for some absolute constants ; in other words, the are uniformly subgaussian.

Now we look at the distribution of . The fundamental central limit theorem tells us the asymptotic behaviour of this distribution:

Theorem 1 (Central limit theorem)Let be iid real random variables of finite mean and variance for some , and let be the normalised sum (1). Then as , converges in distribution to the standard normal distribution .

Exercise 1Show that does not converge in probability or in the almost sure sense (in the latter case, we think of as an infinite sequence of iid random variables). (Hint:the intuition here is that for two very different values of , the quantities and are almost independent of each other, since the bulk of the sum is determined by those with . Now make this intuition precise.)

Exercise 2Use Stirling’s formula from Notes 0a to verify the central limit theorem in the case when is a Bernoulli distribution, taking the values and only. (This is a variant of Exercise 2 from those notes, or Exercise 2 from Notes 1. It is easy to see that once one does this, one can rescale and handle any other two-valued distribution also.)

Exercise 3Use Exercise 9 from Notes 1 to verify the central limit theorem in the case when is gaussian.

Note we are only discussing the case of real iid random variables. The case of complex random variables (or more generally, vector-valued random variables) is a little bit more complicated, and will be discussed later in this post.

The central limit theorem (and its variants, which we discuss below) are extremely useful tools in random matrix theory, in particular through the control they give on random walks (which arise naturally from linear functionals of random matrices). But the central limit theorem can also be viewed as a “commutative” analogue of various spectral results in random matrix theory (in particular, we shall see in later lectures that the *Wigner semicircle law* can be viewed in some sense as a “noncommutative” or “free” version of the central limit theorem). Because of this, the *techniques* used to prove the central limit theorem can often be adapted to be useful in random matrix theory. Because of this, we shall use these notes to dwell on several different proofs of the central limit theorem, as this provides a convenient way to showcase some of the basic methods that we will encounter again (in a more sophisticated form) when dealing with random matrices.

** — 1. Reductions — **

We first record some simple reductions one can make regarding the proof of the central limit theorem. Firstly, we observe *scale invariance*: if the central limit theorem holds for one random variable , then it is easy to see that it also holds for for any real with . Because of this, one can *normalise* to the case when has mean and variance , in which case simplifies to

The other reduction we can make is *truncation*: to prove the central limit theorem for arbitrary random variables of finite mean and variance, it suffices to verify the theorem for *bounded* random variables. To see this, we first need a basic linearity principle:

Exercise 4 (Linearity of convergence)Let be a finite-dimensional real or complex vector space, be sequences of -valued random variables (not necessarily independent), and let be another pair of -valued random variables. Let be scalars converging to respectively.

- If converges in distribution to , and converges in distribution to , and at least one of is deterministic, show that converges in distribution to .
- If converges in probability to , and converges in probability to , show that converges in probability to .
- If converges almost surely to , and converges almost surely , show that converges almost surely to .
Show that the first part of the exercise can fail if are not deterministic.

Now suppose that we have established the central limit theorem for bounded random variables, and want to extend to the unbounded case. Let be an unbounded random variable, which we can normalise to have mean zero and unit variance. Let be a truncation parameter depending on which, as usual, we shall optimise later, and split in the usual fashion (; ). Thus we have as usual.

Let be the mean and variance of the bounded random variable . As we are assuming that the central limit theorem is already true in the bounded case, we know that if we fix to be independent of , then

converges in distribution to . By a diagonalisation argument, we conclude that there exists a sequence going (slowly) to infinity with , such that still converges in distribution to .

For such a sequence, we see from dominated convergence that converges to . As a consequence of this and Exercise 4, we see that

converges in distribution to .

Meanwhile, from dominated convergence again, converges to . From this and (2) we see that

converges in distribution to . Finally, from linearity of expectation we have . Summing (using Exercise 4), we obtain the claim.

Remark 1The truncation reduction is not needed for some proofs of the central limit theorem (notably the Fourier-analytic proof), but is very convenient for some of the other proofs that we will give here, and will also be used at several places in later notes.

By applying the scaling reduction after the truncation reduction, we observe that to prove the central limit theorem, it suffices to do so for random variables which are bounded *and* which have mean zero and unit variance. (Why is it important to perform the reductions in this order?)

** — 2. The Fourier method — **

Let us now give the standard Fourier-analytic proof of the central limit theorem. Given any real random variable , we introduce the *characteristic function* , defined by the formula

Equivalently, is the Fourier transform of the probability measure .

Example 1The signed Bernoulli distribution has characteristic function .

Exercise 5Show that the normal distribution has characteristic function .

More generally, for a random variable taking values in a real vector space , we define the characteristic function by

where denotes the Euclidean inner product on . One can similarly define the characteristic function on complex vector spaces by using the complex inner product

(or equivalently, by identifying with in the usual manner.)

More generally, one can define the characteristic function on any finite dimensional real or complex vector space , by identifying with or . (Strictly speaking, one either has to select an inner product on to do this, or else make the characteristic function defined on the *dual space* instead of on itself; see for instance my notes on the Fourier transform in general locally compact abelian groups. But we will not need to care about this subtlety in our applications.)

The characteristic function is clearly bounded in magnitude by , and equals at the origin. By the Lebesgue dominated convergence theorem, is continuous in .

Exercise 6 (Riemann-Lebesgue lemma)Show that if is an absolutely continuous random variable taking values in or , then as . Show that the claim can fail when the absolute continuity hypothesis is dropped.

Exercise 7Show that the characteristic function of a random variable taking values in or is in fact uniformly continuous on its domain.

Let be a real random variable. If we Taylor expand and formally interchange the series and expectation, we arrive at the heuristic identity

which thus interprets the characteristic function of a real random variable as a kind of generating function for the moments. One rigorous version of this identity is as follows.

Exercise 8 (Taylor expansion of characteristic function)Let be a real random variable with finite moment for some . Show that is times continuously differentiable, and one has the partial Taylor expansionwhere is a quantity that goes to zero as , times . In particular, we have

for all .

Exercise 9Establish (7) in the case that is subgaussian, and show that the series converges locally uniformly in .

Note that the characteristic function depends only on the distribution of : if , then . The converse statement is true also: if , then . This follows from a more general (and useful) fact, known as Lévy’s continuity theorem.

Theorem 2 (Lévy continuity theorem, special case)Let be a finite-dimensional real or complex vector space, and let be a sequence of -valued random variables, and let be an additional -valued random variable. Then the following statements are equivalent:

- (i) converges pointwise to .
- (ii) converges in distribution to .

*Proof:* Without loss of generality we may take .

The implication of (i) from (ii) is immediate from (6) and the definition of convergence in distribution (see Definition 10 of Notes 0), since the function is bounded continuous.

Now suppose that (i) holds, and we wish to show that (ii) holds. By Exercise 23(iv) of Notes 0, it suffices to show that

whenever is a continuous, compactly supported function. By approximating uniformly by Schwartz functions (e.g. using the Stone-Weierstrass theorem), it suffices to show this for Schwartz functions . But then we have the Fourier inversion formula

where

is a Schwartz function, and is in particular absolutely integrable (see e.g. these lecture notes of mine). From the Fubini-Tonelli theorem, we thus have

and similarly for . The claim now follows from the Lebesgue dominated convergence theorem.

Remark 2Setting for all , we see in particular the previous claim that if and only if . It is instructive to use the above proof as a guide to prove this claim directly.

Exercise 10 (Lévy’s continuity theorem, full version)Let be a finite-dimensional real or complex vector space, and let be a sequence of -valued random variables. Suppose that converges pointwise to a limit . Show that the following are equivalent:

- (i) is continuous at .
- (ii) is a tight sequence.
- (iii) is the characteristic function of a -valued random variable (possibly after extending the sample space).
- (iv) converges in distribution to some -valued random variable (possibly after extending the sample space).

Hint: To get from (ii) to the other conclusions, use Prokhorov’s theorem and Theorem 2. To get back to (ii) from (i), use (8) for a suitable Schwartz function . The other implications are easy once Theorem 2 is in hand.

Remark 3Lévy’s continuity theorem is very similar in spirit to Weyl’s criterion in equidistribution theory.

Exercise 11 (Esséen concentration inequality)Let be a random variable taking values in . Then for any , , show thatfor some constant depending only on and . (

Hint:Use (8) for a suitable Schwartz function .) The left-hand side of (9) is known as thesmall ball probabilityof at radius .

In Fourier analysis, we learn that the Fourier transform is a particularly well-suited tool for studying convolutions. The probability theory analogue of this fact is that characteristic functions are a particularly well-suited tool for studying sums of independent random variables. More precisely, we have

Exercise 12 (Fourier identities)Let be a finite-dimensional real or complex vector space, and let be independent random variables taking values in . Thenfor all . Also, for any scalar , one has

and more generally, for any linear transformation , one has

Remark 4Note that this identity (10), combined with Exercise 5 and Remark 2, gives a quick alternate proof of Exercise 9 from Notes 1.

In particular, in the normalised setting (4), we have the simple relationship

that describes the characteristic function of in terms of that of .

We now have enough machinery to give a quick proof of the central limit theorem:

*Proof:* (Proof of Theorem 1) We may normalise to have mean zero and variance . By Exercise 8, we thus have

for sufficiently small , or equivalently

for sufficiently small . Applying (11), we conclude that

as for any fixed . But by Exercise 5, is the characteristic function of the normal distribution . The claim now follows from the Lévy continuity theorem.

Exercise 13 (Vector-valued central limit theorem)Let be a random variable taking values in with finite second moment. Define the covariance matrix to be the matrix whose entry is the covariance .

- Show that the covariance matrix is positive semi-definite real symmetric.
- Conversely, given any positive definite real symmetric matrix and , show that the
normal distribution, given by the absolutely continuous measurehas mean and covariance matrix , and has a characteristic function given by

How would one define the normal distribution if degenerated to be merely positive semi-definite instead of positive definite?

- If is the sum of iid copies of , show that converges in distribution to .

Exercise 14 (Complex central limit theorem)Let be a complex random variable of mean , whose real and imaginary parts have variance and covariance . Let be iid copies of . Show that as , the normalised sums (1) converge in distribution to the standard complex gaussian .

Exercise 15 (Lindeberg central limit theorem)Let be a sequence of independent (but not necessarily identically distributed) real random variables, normalised to have mean zero and variance one. Assume the (strong) Lindeberg conditionwhere is the truncation of to large values. Show that as , converges in distribution to . (

Hint: modify the truncation argument.)

A more sophisticated version of the Fourier-analytic method gives a more quantitative form of the central limit theorem, namely the Berry-Esséen theorem.

Theorem 3 (Berry-Esséen theorem)Let have mean zero, unit variance, and finite third moment. Let , where are iid copies of . Then we haveuniformly for all , where , and the implied constant is absolute.

*Proof:* (Optional) Write ; our task is to show that

for all . We may of course assume that , as the claim is trivial otherwise.

Let be a small absolute constant to be chosen later. Let be an non-negative Schwartz function with total mass whose Fourier transform is supported in , and let be the smoothed out version of , defined as

Observe that is decreasing from to .

We claim that it suffices to show that

for every , where the subscript means that the implied constant depends on . Indeed, suppose that (13) held. Define

thus our task is to show that .

Let be arbitrary, and let be a large absolute constant to be chosen later. We write

and thus by (13)

Meanwhile, from (14) and an integration by parts we see that

From the bounded density of and the rapid decrease of we have

Putting all this together, we see that

A similar argument gives a lower bound

and so

Taking suprema over , we obtain

If is large enough (depending on ), we can make and small, and thus absorb the latter two terms on the right-hand side into the left-hand side. This gives the desired bound .

It remains to establish (13). Applying (8), it suffices to show that

Now we estimate each of the various expressions. Standard Fourier-analytic computations show that

and that

Since was supported in , it suffices to show that

for any ; taking expectations and using the definition of we have

and in particular

if and is small enough. Applying (11), we conclude that

if . Meanwhile, from Exercise 5 we have . Elementary calculus then gives us

(say) if is small enough. Inserting this bound into (16) we obtain the claim.

Exercise 16Show that the error terms here are sharp (up to constants) when is a signed Bernoulli random variable.

** — 3. The moment method — **

The above Fourier-analytic proof of the central limit theorem is one of the quickest (and slickest) proofs available for this theorem, and is accordingly the “standard” proof given in probability textbooks. However, it relies quite heavily on the Fourier-analytic identities in Exercise 12, which in turn are extremely dependent on both the commutative nature of the situation (as it uses the identity ) and on the independence of the situation (as it uses identities of the form ). When we turn to random matrix theory, we will often lose (or be forced to modify) one or both of these properties, which often causes the Fourier-analytic methods to fail spectacularly. Because of this, it is also important to look for non-Fourier based methods to prove results such as the central limit theorem. These methods often lead to proofs that are lengthier and more technical than the Fourier proofs, but also tend to be more robust, and in particular can often be extended to random matrix theory situations. Thus both the Fourier and non-Fourier proofs will be of importance in this course.

The most elementary (but still remarkably effective) method available in this regard is the *moment method*, which we have already used in the previous notes, which seeks to understand the distribution of a random variable via its moments . In principle, this method is equivalent to the Fourier method, through the identity (7); but in practice, the moment method proofs tend to look somewhat different than the Fourier-analytic ones, and it is often more apparent how to modify them to non-independent or non-commutative settings.

We first need an analogue of the Lévy continuity theorem. Here we encounter a technical issue: whereas the Fourier phases were bounded, the moment functions become unbounded at infinity. However, one can deal with this issue as long as one has sufficient decay:

Theorem 4 (Moment continuity theorem)Let be a sequence of uniformly subgaussian real random variables, and let be another subgaussian random variable. Then the following statements are equivalent:

- (i) For every , converges pointwise to .
- (ii) converges in distribution to .

*Proof:* We first show how (ii) implies (i). Let be a truncation parameter, and let be a smooth function that equals on and vanishes outside of . Then for any , the convergence in distribution implies that converges to . On the other hand, from the uniform subgaussian hypothesis, one can make and arbitrarily small for fixed by making large enough. Summing, and then letting go to infinity, we obtain (i).

Conversely, suppose (i) is true. From the uniform subgaussian hypothesis, the have moment bounded by for all and some independent of (see Exercise 4 from Notes 0). From Taylor’s theorem with remainder (and Stirling’s formula, Notes 0a) we conclude

uniformly in and . Similarly for . Taking limits using (i) we see that

Then letting , keeping fixed, we see that converges pointwise to for each , and the claim now follows from the Lévy continuity theorem.

Remark 5One corollary of Theorem 4 is that the distribution of a subgaussian random variable is uniquely determined by its moments (actually, this could already be deduced from Exercise 9 and Remark 2). The situation can fail for distributions with slower tails, for much the same reason that a smooth function is not determined by its derivatives at one point if that function is not analytic.The Fourier inversion formula provides an easy way to recover the distribution from the characteristic function. Recovering a distribution from its moments is more difficult, and sometimes requires tools such as analytic continuation; this problem is known as the

inverse moment problemand will not be discussed here.

To prove the central limit theorem, we know from the truncation method that we may assume without loss of generality that is bounded (and in particular subgaussian); we may also normalise to have mean zero and unit variance. From the Chernoff bound (3) we know that the are uniformly subgaussian; so by Theorem 4, it suffices to show that

for all , where is a standard gaussian variable.

The moments are easy to compute:

Exercise 17Let be a natural number, and let . Show that vanishes when is odd, and equal to when is even. (This can either be done directly by using the Gamma function, or by using Exercise 5 and Exercise 9.)

So now we need to compute . Using (4) and linearity of expectation, we can expand this as

To understand this expression, let us first look at some small values of .

- For , this expression is trivially .
- For , this expression is trivially , thanks to the mean zero hypothesis on .
- For , we can split this expression into the diagonal and off-diagonal components:
Each summand in the first sum is , as has unit variance. Each summand in the second sum is , as the have mean zero and are independent. So the second moment is .

- For , we have a similar expansion
The summands in the latter two sums vanish because of the (joint) independence and mean zero hypotheses. The summands in the first sum need not vanish, but are , so the first term is , which is asymptotically negligible, so the third moment goes to .

- For , the expansion becomes quite complicated:
Again, most terms vanish, except for the first sum, which is and is asymptotically negligible, and the sum , which by the independence and unit variance assumptions works out to . Thus the fourth moment goes to (as it should).

Now we tackle the general case. Ordering the indices as for some , with each occuring with multiplicity and using elementary enumerative combinatorics, we see that is the sum of all terms of the form

where , are positive integers adding up to , and is the multinomial coefficient

The total number of such terms depends only on (in fact, it is (exercise!), though we will not need this fact).

As we already saw from the small examples, most of the terms vanish, and many of the other terms are negligible in the limit . Indeed, if any of the are equal to , then every summand in (17) vanishes, by joint independence and the mean zero hypothesis. Thus, we may restrict attention to those expressions (17) for which all the are at least . Since the sum up to , we conclude that is at most .

On the other hand, the total number of summands in (17) is clearly at most (in fact it is ), and the summands are bounded (for fixed ) since is bounded. Thus, if is *strictly* less than , then the expression in (17) is and goes to zero as . So, asymptotically, the only terms (17) which are still relevant are those for which is *equal* to . This already shows that goes to zero when is odd. When is even, the only surviving term in the limit is now when and . But then by independence and unit variance, the expectation in (17) is , and so this term is equal to

and the main term is happily equal to the moment as computed in Exercise 17.

** — 4. The Lindeberg swapping trick — **

The moment method proof of the central limit theorem that we just gave consisted of four steps:

- (Truncation and normalisation step) A reduction to the case when was bounded with zero mean and unit variance.
- (Inverse moment step) A reduction to a computation of asymptotic moments .
- (Analysis step) Showing that most terms in the expansion of this asymptotic moment were zero, or went to zero as .
- (Algebra step) Using enumerative combinatorics to compute the remaining terms in the expansion.

In this particular case, the enumerative combinatorics was very classical and easy – it was basically asking for the number of ways one can place balls in boxes, so that the box contains balls, and the answer is well known to be given by the multinomial . By a small algebraic miracle, this result matched up nicely with the computation of the moments of the gaussian .

However, when we apply the moment method to more advanced problems, the enumerative combinatorics can become more non-trivial, requiring a fair amount of combinatorial and algebraic computation. The algebraic miracle that occurs at the end of the argument can then seem like a very fortunate but inexplicable coincidence, making the argument somehow unsatisfying despite being rigorous.

In a 1922 paper, Lindeberg observed that there was a very simple way to *decouple* the algebraic miracle from the analytic computations, so that all relevant algebraic identities only need to be verified in the special case of *gaussian* random variables, in which everything is much easier to compute. This *Lindeberg swapping trick* (or *Lindeberg replacement trick*) will be very useful in the later theory of random matrices, so we pause to give it here in the simple context of the central limit theorem.

The basic idea is follows. We repeat the truncation-and-normalisation and inverse moment steps in the preceding argument. Thus, are iid copies of a boudned real random variable of mean zero and unit variance, and we wish to show that , where , where is fixed.

Now let be iid copies of the gaussian itself: . Because the sum of independent gaussians is again a gaussian (Exercise 9 from Notes 1, we see that the random variable

already has the same distribution as : . Thus, it suffices to show that

Now we perform the analysis part of the moment method argument again. We can expand into terms (17) as before, and discard all terms except for the term as being . Similarly, we can expand into very similar terms (but with the replaced by ) and again discard all but the term.

But by hypothesis, the second moments of and match: . Thus, by joint independence, the term (17) for is exactly equal to that of . And the claim follows.

This is almost exactly the same proof as in the previous section, but note that we did *not* need to compute the multinomial coefficient , nor did we need to verify the miracle that this coefficient matched (up to normalising factors) to the moments of the gaussian. Instead, we used the much more mundane “miracle” that the sum of independent gaussians was again a gaussian.

To put it another way, the Lindeberg replacement trick factors a universal limit theorem, such as the central limit theorem, into two components:

- A
*universality*or*invariance*result, which shows that the distribution (or other statistics, such as moments) of some random variable is asymptotically unchanged in the limit if each of the input variables are replaced by a gaussian substitute ; and - The
*gaussian case*, which computes the asymptotic distribution (or other statistic) of in the case when are all gaussians.

The former type of result tends to be entirely analytic in nature (basically, one just needs to show that all error terms that show up when swapping with are ), while the latter type of result tends to be entirely algebraic in nature (basically, one just needs to exploit the many pleasant algebraic properties of gaussians). This decoupling of the analysis and algebra steps tends to simplify both, both at a technical level and at a conceptual level.

** — 5. Individual swapping — **

In the above argument, we swapped all the original input variables with gaussians *en masse*. There is also a variant of the Lindeberg trick in which the swapping is done *individually*. To illustrate the individual swapping method, let us use it to show the following weak version of the Berry-Esséen theorem:

Theorem 5 (Berry-Esséen theorem, weak form)Let have mean zero, unit variance, and finite third moment, and let be smooth with uniformly bounded derivatives up to third order. Let , where are iid copies of . Then we have

*Proof:* Let and be in the previous section. As , it suffices to show that

We telescope this (using linearity of expectation) as

where

is a partially swapped version of . So it will suffice to show that

uniformly for .

To exploit this, we use Taylor expansion with remainder to write

and

where the implied constants depend on but not on . Now, by construction, the moments of and match to second order, thus

and the claim follows. (Note from Hölder’s inequality that .)

Remark 6The above argument relied on Taylor expansion, and the hypothesis that the moments of and matched to second order. It is not hard to see that if we assume more moments matching (e.g. ), and more smoothness on , we see that we can improve the factor on the right-hand side. Thus we see that we expect swapping methods to become more powerful when more moments are matching. We will see this when we discuss thefour moment theoremof Van Vu and myself in later lectures, which (very) roughly speaking asserts that the spectral statistics of two random matrices are asymptotically indistinguishable if their coefficients have matching moments to fourth order.

Theorem 5 is easily implied by Theorem 3 and an integration by parts. In the reverse direction, let us see what Theorem 5 tells us about the cumulative distribution function

of . For any , one can upper bound this expression by

where is a smooth function equal to on that vanishes outside of , and has third derivative . By Theorem 5, we thus have

On the other hand, as has a bounded probability density function, we have

and so

A very similar argument gives the matching lower bound, thus

Optimising in we conclude that

Comparing this with Theorem 3 we see that we have lost an exponent of . In our applications to random matrices, this type of loss is acceptable, and so the swapping argument is a reasonable substitute for the Fourier-analytic one in this case. Also, this method is quite robust, and in particular extends well to higher dimensions; we will return to this point in later lectures, but see for instance Appendix D of this paper of myself and Van Vu for an example of a multidimensional Berry-Esséen theorem proven by this method.

On the other hand there is another method that can recover this loss while still avoiding Fourier-analytic techniques; we turn to this topic next.

** — 6. Stein’s method — **

Stein’s method, introduced by Charles Stein in 1970 (who should not be confused with a number of other eminent mathematicians with this surname, including my advisor), is a powerful method to show convergence in distribution to a special distribution, such as the gaussian. In several recent papers, this method has been used to control several expressions of interest in random matrix theory (e.g. the distribution of moments, or of the Stieltjes transform.) We will not use it much in this course, but this method is of independent interest, so I will briefly discuss (a very special case of) it here.

The probability density function of the standard normal distribution can be viewed as a solution to the ordinary differential equation

One can take adjoints of this, and conclude (after an integration by parts) that obeys the integral identity

for any continuously differentiable with both and bounded (one can relax these assumptions somewhat). To put it another way, if , then we have

whenever is continuously differentiable with both bounded.

It turns out that the converse is true: if is a real random variable with the property that

whenever is continuously differentiable with both bounded, then is Gaussian. In fact, more is true, in the spirit of Theorem 2 and Theorem 4:

Theorem 6 (Stein continuity theorem)Let be a sequence of real random variables with uniformly bounded second moment, and let . Then the following are equivalent:

- (i) converges to zero whenever is continuously differentiable with both bounded.
- (ii) converges in distribution to .

*Proof:* To show that (ii) implies (i), it is not difficult to use the uniform bounded second moment hypothesis and a truncation argument to show that converges to when is continuously differentiable with both bounded, and the claim then follows from (22).

Now we establish the converse. It suffices to show that

whenever is a bounded Lipschitz function. We may normalise to be bounded in magnitude by .

Trivially, the function has zero expectation when one substitutes for the argument , thus

Comparing this with (22), one may thus hope to find a representation of the form

for some continuously differentiable with both bounded. This is a simple ODE and can be easily solved (by the method of integrating factors) to give a solution , namely

(One could dub the *Stein transform* of , although this term does not seem to be in widespread use.) By the fundamental theorem of calculus, is continuously differentiable and solves (24). Using (23), we may also write as

By completing the square, we see that . Inserting this into (25) and using the bounded nature of , we conclude that for ; inserting it instead into (26), we have for . Finally, easy estimates give for . Thus for all we have

which when inserted back into (24) gives the boundedness of (and also of course gives the boundedness of ). In fact, if we rewrite (26) as

we see on differentiation under the integral sign (and using the Lipschitz nature of ) that for ; a similar manipulation (starting from (25)) applies for , and we in fact conclude that for all .

Applying (24) with and taking expectations, we have

By the hypothesis (i), the right-hand side goes to zero, hence the left-hand side does also, and the claim follows.

The above theorem gave only a qualitative result (convergence in distribution), but the proof is quite quantitative, and can be used to in particular to give Berry-Esséen type results. To illustrate this, we begin with a strengthening of Theorem 5 that reduces the number of derivatives of that need to be controlled:

Theorem 7 (Berry-Esséen theorem, less weak form)Let have mean zero, unit variance, and finite third moment, and let be smooth, bounded in magnitude by , and Lipschitz. Let , where are iid copies of . Then we have

*Proof:* Set .

Let be the Stein transform (25) of , then by (24) we have

We expand . For each , we then split , where (cf. (19)). By the fundamental theorem of calculus, we have

where is uniformly distributed in and independent of all of the . Now observe that and are independent, and has mean zero, so the first term on the right-hand side vanishes. Thus

Another application of independendence gives

so we may rewrite (28) as

Recall from the proof of Theorem 6 that and . By the product rule, this implies that has a Lipschitz constant of . Applying (24) and the definition of , we conclude that has a Lipschitz constant of . Thus we can bound the previous expression as

and the claim follows from Hölder’s inequality.

This improvement already reduces the loss in (20) to . But one can do better still by pushing the arguments further. Let us illustrate this in the model case when the not only have bounded third moment, but are in fact bounded:

Theorem 8 (Berry-Esséen theorem, bounded case)Let have mean zero, unit variance, and be bounded by . Let , where are iid copies of . Then we have

*Proof:* Write , thus we seek to show that

Let be the Stein transform (25) of . is not continuous, but it is not difficult to see (e.g. by a limiting argument) that we still have the estimates and (in a weak sense), and that has a Lipschitz norm of (here we use the hypothesis ). A similar limiting argument gives

and by arguing as in the proof of Theorem 7, we can write the right-hand side as

From (24), is equal to , plus a function with Lipschitz norm . Thus, we can write the above expression as

The terms cancel (due to the independence of and , and the normalised mean and variance of ), so we can simplify this as

and so we conclude that

Since and are bounded, and is non-increasing, we have

applying the second inequality and using independence to once again eliminate the factor, we see that

which implies (by another appeal to the non-increasing nature of and the bounded nature of ) that

or in other words that

Similarly, using the lower bound inequalities, one has

Moving up and down by , and using the bounded density of , we obtain the claim.

Actually, one can use Stein’s method to obtain the full Berry-Esséen theorem, but the computations get somewhat technical, requiring an induction on to deal with the contribution of the exceptionally large values of : see this paper of Barbour and Hall.

** — 7. Predecessor comparison — **

Suppose one had never heard of the normal distribution, but one still suspected the existence of the central limit theorem – thus, one thought that the sequence of normalised distributions was converging in distribution to something, but was unsure what the limit was. Could one still work out what that limit was?

Certainly in the case of Bernoulli distributions, one could work explicitly using Stirling’s formula (see Exercise 2), and the Fourier-analytic method would also eventually work. Let us now give a third way to (heuristically) derive the normal distribution as the limit of the central limit theorem. The idea is to compare with its predecessor , using the recursive formula

(normalising to have mean zero and unit variance as usual; let us also truncate to be bounded, for simplicity). Let us hypothesise that and are approximately the same distribution; let us also conjecture that this distribution is absolutely continuous, given as for some smooth . (If we secretly knew the central limit theorem, we would know that is in fact , but let us pretend that we did not yet know this fact.) Thus, for any test function , we expect

Now let us try to combine this with (30). We assume to be smooth, and Taylor expand to third order:

Taking expectations, and using the independence of and , together with the normalisations on , we obtain

Up to errors of , one can approximate the second term here by . We then insert (31) and are led to the heuristic equation

Changing variables for the first term on the right hand side, and integrating by parts for the second term, we have

Since was an arbitrary test function, this suggests the heuristic equation

Taylor expansion gives

which leads us to the heuristic ODE

where is the Ornstein-Uhlenbeck operator

Observe that is the total derivative of ; integrating from infinity, we thus get

which is (21), and can be solved by standard ODE methods as for some ; the requirement that probability density functions have total mass then gives the constant as , as we knew it must.

The above argument was not rigorous, but one can make it so with a significant amount of PDE machinery. If we view (or more precisely, ) as a time parameter, and view as depending on time, the above computations heuristically lead us eventually to the Fokker-Planck equation for the Ornstein-Uhlenbeck process,

which is a linear parabolic equation that is fortunate enough that it can be solved exactly (indeed, it is not difficult to transform this equation to the linear heat equation by some straightforward changes of variable). Using the spectral theory of the Ornstein-Uhlenbeck operator , one can show that solutions to this equation starting from an arbitrary probability distribution, are attracted to the gaussian density function , which as we saw is the steady state for this equation. The stable nature of this attraction can eventually be used to make the above heuristic analysis rigorous. However, this requires a substantial amount of technical effort (e.g. developing the theory of Sobolev spaces associated to ) and will not be attempted here. One can also proceed by relating the Fokker-Planck equation to the associated stochastic process, namely the Ornstein-Uhlenbeck process, but this requires one to first set up stochastic calculus, which we will not do here. (The various Taylor expansion calculations we have performed in these notes, though, are closely related to stochastic calculus tools such as Ito’s lemma.) Stein’s method, discussed above, can also be interpreted as a way of making the above computations rigorous (by not working with the density function directly, but instead testing the random variable against various test functions ).

This argument does, though highlight two ideas which we will see again in later notes when studying random matrices. Firstly, that it is profitable to study the distribution of some random object by comparing it with its predecessor , which one presumes to have almost the same distribution. Secondly, we see that it may potentially be helpful to approximate (in some weak sense) a discrete process (such as the iteration of the scheme (30)) with a continuous evolution (in this case, a Fokker-Planck equation) which can then be controlled using PDE methods.

## 54 comments

Comments feed for this article

6 January, 2010 at 2:02 am

JeanVery nice notes! Thank you!

Just a remark: in http://arxiv.org/abs/0705.1224 and http://arxiv.org/abs/0908.0391, one can find some applications of Stein’s method in random matrix theory.

6 January, 2010 at 7:27 am

Mark MeckesMore applications of Stein’s method in random matrix theory appear here, here, and here.

6 January, 2010 at 2:53 am

karabasovA couple of very minor typos:

1) in the second-last paragraph of section 3, it should be “to those terms in (17)”

2) Beginning of section 4: “Lindeberg observed that” – the comma between observed and that is not correct.

6 January, 2010 at 3:25 am

karabasovI think there might be a parsing error around “where L is the rhox” in the last section… or maybe I just don’t get it.

6 January, 2010 at 9:12 am

Giovanni PeccatiHi Terry,

thank you for these inspiring lecture notes !

Small typo, in the brackets starting 5 lines after formula (30) (If we secretely…) there is a square root missing in the normalizing factor of the Gaussian density.

Best, G

6 January, 2010 at 12:04 pm

Terence TaoThanks for the corrections!

Somewhat embarrassingly, I was only dimly aware of the work on applying Stein’s method to random matrices; it seems that the fraction of literature on the subject that I am familiar with is still not fully representative. Thanks for the references!

7 January, 2010 at 5:45 am

Tim vBDear Terry,

thanks for this marvelous lecture notes!

I hope the following two remarks are not too pedantic, but both points made me stop for a short while:

1. The part “now we get the Fokker-Planck-Equation and that can be solved exactly” confused me a bit, because the Fokker-Planck-Equation can “usually” not be solved excatly – the one for the Ornstein-Uhlenbeck process can. It’s of course clear from the context that you mean exactly this, but I think it would help me if you redundantly write ” the above computations heuristically lead us eventually to the Fokker-Planck equation of the Ornstein-Uhlenbeck process”.

2. Wouldn’t most people say the Ornstein-Uhlenbeck process is the solution of a stochastic ordinary differential equation (stochastic ODE) rather than a stochastic partial differential equation?

[Fair enough; I’ve adjusted the notes accordingly.]15 January, 2010 at 12:31 pm

Steven HeilmanMore tedious [potential] corrections:

1. Two Eqs. after Eq. (14): P(…)\leq E(…)

2. Eq (16): square in denominator correct?

3. Theorem 4,5: F should be phi (or vice versa)

4(a). Proof of Theorem 5: several missing parentheses? (or E symbols), e.g.

\displaystyle {\bf E} (\varphi(Z_n) – \varphi(W_n)) = o(1).

This is done quite a lot, so I will assume it was done on purpose.

4(b). telescoping sum, multiply by -1

5. After Eq. (26): tex error for bound on f. I assume you want

{f(x) = O_\varphi( 1/|x|)}

[Corrected, thanks – T.]17 January, 2010 at 7:28 pm

salazarSecond 1 and 3. For 4, since W_n is a RV, there should be no ambiguity. For 2, I haven’t been able to see how to replace \hat{eta}.

17 January, 2010 at 8:55 pm

Terence Taocan be bounded by an absolute constant. (The denominator of was a relic from an earlier version of the argument, but is now redundant due to the restriction to the range .)

18 January, 2010 at 6:29 pm

254A, Notes 3b: Brownian motion and Dyson Brownian motion « What’s new[…] the central limit theorem from the fundamental solution of the heat equation (cf. Section 7 of Notes 2), although the derivation is only heuristic because one first needs to know that some limiting […]

29 January, 2010 at 7:16 am

Andreas NaiveExcellent notes!

Just a small fix:

In (30), should be , which slightly simplify the following argument.

[Corrected, thanks – T.]2 February, 2010 at 1:34 pm

254A, Notes 4: The semi-circular law « What’s new[…] vein, we may apply the truncation argument (much as was done for the central limit theorem in Notes 2) to reduce the semi-circular law to the bounded case: Exercise 5 Show that in order to prove the […]

10 February, 2010 at 10:56 pm

245A, Notes 5: Free probability « What’s new[…] an interesting analogue in the freely independent setting. For instance, the central limit theorem (Notes 2) for averages of classically independent random variables, which roughly speaking asserts that such […]

25 February, 2010 at 10:19 am

PDEbeginnerDear Prof. Tao,

I have some problems on Ex 12 (Esseen concentration inequaity) and Thm 3:

Ex 12. Let , the value on LHS of (9) should be 1, while that on RHS should be 0. Then the inequality breaks down.

Thm 3. In the paragraph immediately below (14), I do not understand how to apply an integration by parts argument, since one does not know the exact distribution of .

Thanks in advance!

25 February, 2010 at 11:09 am

Terence TaoAh, there was a factor of r^d missing in the concentration inequality; it is fixed now.

As for Theorem 3, the point is that the distribution of X is the Stieltjes derivative of the cumulative distribution function of X; . If we integrate the latter integral by parts, we see that small perturbations of F in the uniform norm lead to small perturbations of when f has bounded total variation.

25 February, 2010 at 10:20 am

PDEbeginnerSorry, it is Ex 11.

26 February, 2010 at 2:41 pm

YPThe derivation of Gaussian in Section 7 is amazing! (To me it seems very much in the spirit of physics). Isn’t it strange that to our brain it seems very natural to make this argument as it is, whereas for scientific community you actually need to add days of hard work (in Sobolev spaces etc)?

27 February, 2010 at 6:44 pm

vedadiDear Prof. Tao,

We know that for i.i.d mean zero, variance one random variables, converges in distribution to the point mass at zero, and standard normal r.v. for respectively. Do we have convergence in distribution for other values

Thanks

6 March, 2010 at 7:14 pm

254A, Notes 7: The least singular value « What’s new[…] understand this walk, we apply (a slight variant) of the Berry-Esséen theorem from Notes 2: Exercise 1 Show […]

10 March, 2010 at 10:35 am

PDEbeginnerDear Prof. Tao,

I finished reading this note. As usual, I have some problems :-)

1. For the moment method, we first apply Chernoff bound (3), and then we can assume that are uniformly guassian. But when proving Chernoff bound in Note 1, we have used the identity . When we are dealing random matrices, if we also use Chernoff bounds, then it seems we shall have some trouble.

2. I still do not know how to prove the conclusion in Ex 11. Suppose one is trying to bound the easy probability (with and ):

.

By Fourier transform, we obtain

Clearly, to show has the same bound as in exercise, we need to show that has that bound. I don’t know to show this.

3. A small typo: the in Theorem 5 seems to be .

10 March, 2010 at 6:38 pm

Terence Tao1. For random matrices, one does not apply the Chernoff bound to the matrix directly, but to other scalar expressions related to that matrix, e.g. linear combinations of the matrix entries.

2. One first needs to replace the sharp cutoff to the ball of radius r by a smoother cutoff. This is the method of smoothing sums:

http://www.tricki.org/article/Smoothing_sums

3. Thanks for the correction!

5 May, 2010 at 8:48 pm

AnonymousThese notes are very helpful, thank you very much!

I do not have a strong background in the Fourier analysis, could you point me to a refernce to the solutions to Exercise 11 and Exercise 15?

A few comments I have:

In Exercise 13, the characteristic function should be $e^{i\mu}t…$ instead of $e^{-i\mu}t$?

In Part 1 of this note, the “N” in the last two singled-out equations should be replaced with “N_n”?

6 May, 2010 at 9:06 am

Terence TaoThanks for the corrections,’

For Ex11, see Lemma 7.17 of my book “Additive combinatorics” with Van Vu. Ex15 can be found in most graduate texts in probability, including those listed at the link given.

27 July, 2010 at 3:50 pm

Ahmet ArivanGreat notes. I have probably a very stupid question. But in the Stein method section, it is claimed that if f(t) =O(1/1+|t|) and f'(t)=O(1) then tf(t) is Lipschitz with constant of O(1). Is it possible to give a hint on why this is true?

Thanks,

31 July, 2010 at 8:35 am

Terence TaoAh yes, that requires an additional argument (which needs the Lipschitz hypothesis on phi). I’ve modified the text accordingly (and in particular weakened the Berry-Esseen type claim, since the Lipschitz property is not present in that case.)

29 October, 2012 at 3:53 pm

Nick CookSmall thing: I think the last part of the proof of Theorem 6, showing that , with implicit constant linear in the bounded Lipschitz norm of , belongs just after this proof since we aren’t assuming is Lipschitz here. – or leave it in with modification of the phrase “using the Lipschitz nature of ” :)

[Corrected by assuming $\phi$ to be Lipschitz. -T.]4 November, 2010 at 8:33 am

Qiaochu YuanRegarding Section 7: I had always vaguely figured that if were going to converge to a particular distribution, it would have to converge to an eigenfunction of the Fourier transform. Do you know of any heuristic (or rigorous!) derivations of the CLT using this idea?

4 November, 2010 at 9:02 am

Terence TaoI think the more accurate statement would be that a stable law should have a Fourier transform whose logarithm is an eigenfunction of the scaling operation, since the convolution of with itself needs to be a rescaling of .

16 November, 2010 at 1:04 pm

karabasovAt the beginning of the proof of theorem 1, it seems that the first “for sufficiently small t” is not needed. Only the second is needed.

15 March, 2011 at 6:57 pm

SujitHi Terry,

I am a bit puzzled about Exercise 1. I am thinking of $Z_n$ as a map from $n$ copies of $\Omega$ to $R$. If this is the case, what does it mean to say $Z_n$ converges almost surely? Isn’t the underlying sample space varying as $n$ varies?

The CLT says something about the push forward of the measures from $\Omega^n$ to $R$ and I can understand it. So even if the sample space changes, it doesn’t matter as we are comparing the push forward measures. In the strong LLN case, it still sounds fine to me if I interpret the almost sure convergence as almost sure convergence to the constant map on $\Omega^n$ given by the expectation. But I am having trouble trying to understand what Exercise 1 means.

Thanks

15 March, 2011 at 7:08 pm

Terence TaoThe intent here is to work in a common extension of all these probability spaces, where we have an infinite sequence of iid random variables (and this is also the correct way to interpret the strong law of large numbers).

16 March, 2011 at 1:03 pm

AnonymousDear Prof. Tao,

Let be the moment generating function of a non degenerate r.v . Then is always strictly increasing and strictly convex on ?

Thanks

28 August, 2011 at 6:12 am

The law of large numbers and the central limit theorem | Controlled Complexity[…] theorem (CLT). There are excellent resources on the net for LLN and CLT. For example, this and this are highly recommended readings. This blog will play a complementary with figures and animations to […]

1 October, 2011 at 10:32 am

AnonymousCan anyone help me to compute the expected waiting time of the first occurrence of the patten, say, TTHT. here

is there a general approach to solve this sort of problems?

Thanks

18 November, 2011 at 2:38 am

Diffusion in Ehrenfest wind-tree model « Disquisitiones Mathematicae[…] of the word “abnormal” comes by comparison with Brownian motion and/or central limit theorem: once we know that the diffusion is “sublinear” (maybe after removing the […]

17 March, 2012 at 5:48 am

AnonymousDear Prof. Tao,

Let be the simple symmetric random walk in one dimension.

Assume that K is a large constant and I need a lower bound for the following probability which does not depend on

What is the idea behind finding a lower bound for this type of quantities?

Thank you

10 October, 2012 at 3:33 pm

Robert KohnHi, thanks for insightful presentation. My question is, suppose that the are independent with finite third moments, zero mean and unit variance. What can we say about the convergence of the third moment of the normalized sum ?

In addition, what results of the same kind can we get if the are correlated?

Thanks,

Robert Kohn

20 November, 2012 at 8:14 am

JackRegarding Exercise 1: I don’t see how the intuition help. What I thought is to give a lower bound of for some , which is from the definition of convergence in probability. Could you elaborate the intuition? Why would independence help here?

20 November, 2012 at 8:41 am

Terence TaoActually, one needs a lower bound on to contradict convergence in probability to an unspecified limit (lower bounding merely prevents convergence in probability to zero, or to a non-positive random variable). But if one can quantify some approximate independence between and for n,m widely separated, then this, together with the central limit theorem, can supply the desired lower bound.

20 November, 2012 at 2:07 pm

JackAn attempt is that both and $Z_{2n}$ converge to in distribution by CLT. Thus one has . I still don’t see how one can bound .

20 November, 2012 at 2:14 pm

Terence Taoand are still quite correlated, so it is a bit tricky to separate them by . Try and instead for some large M (larger than or so). The point is that these two random variables are almost independent (more precisely, and are independent) which makes it difficult for them to stay so close to each other so often.

20 November, 2012 at 2:46 pm

JackWhat puzzles me is the use of “independence” here. As a beginner, I only think that and are independent if and only if where and are Borel sets in the corresponding Borel sigma algebra. Why does this concept have anything to do with the “distance” between these random variables? It’s quite intuitive though: two things are “independent”, then they should not be too “close” to each other. But I don’t see the connection. (Eventually I come up with something like . Is this the point? )

20 November, 2012 at 6:24 pm

Terence TaoIf X and Y are independent, then (for instance) will be large if and are large, thus giving some separation between X and Y. (One can also in principle work with expectations, but this requires some hypotheses on absolute integrability that one then has to justify separately.)

26 November, 2012 at 3:13 pm

JackFinding a lower bound for is still not “obvious” to me. I tried . And the independence between and gives . Since is chosen such that , implies that this is very close to the desired estimate. But I don’t see the way to bound and . Since I haven’t use CLT so far, I am wondering there must be lack of something here.

26 November, 2012 at 3:42 pm

Terence TaoThe CLT describes the limiting value of in the limit as with fixed, and similarly for (which is basically a rescaled variant of ).

26 November, 2012 at 4:14 pm

JackI found in some book that random variables converging in distribution to $X$ is defined as (and the author also calls it “weak convergence”)

for all bounded, continuous function . Is it “equivalent” to the definition in [254A Note4 Exercise 18](https://terrytao.wordpress.com/2010/10/02/245a-notes-4-modes-of-convergence/)? But in [245B Note 11](https://terrytao.wordpress.com/2009/02/21/245b-notes-11-the-strong-and-weak-topologies/), weak convergence is convergence in weak topology. I am totally confused with the terminologies.

[I don’t know how to give links like people do in MathOverflow.]

26 November, 2012 at 4:39 pm

Terence TaoYes, these notions are all connected to each other; see Exercise 23 of https://terrytao.wordpress.com/2010/01/01/254a-notes-0-a-review-of-probability-theory/ . (Though, strictly speaking, vague convergence of measures – which is equivalent to convergence in distribution – is an example of weak* convergence rather than weak convergence.)

22 March, 2013 at 12:34 am

AnonymousHi professor Tao,

does the CLT hold if we substitute the hypothesis of finite mean and variance with the hypothesis of finite second order moment, ?

22 March, 2013 at 9:10 am

Terence TaoBy Holder’s inequality (or the Cauchy-Schwarz inequality), finite second order moment is equivalent to finite mean and variance.

3 April, 2013 at 3:13 am

AnonymousThanks a lot for the answer! By the way, does { \Z_n := \frac{S_n – n \mu}{\sqrt{n} \sigma} to N(0,1)} implies that {\frac{S_n}{n} to N(\mu,\frac{\sigma^2}{n}) }? This implication seems quite weird to me, since {n to \infty}, but on a book about generalized polynomial chaos I fount the following sentence: “the numerical average of a sequence of i.i.d. random variables will converge, as n is increased, to a Gaussian distribution { N(\mu,\frac{\sigma^2}{n}) } […]”.

3 April, 2013 at 10:30 am

Terence TaoIf one uses a rescaled notion of convergence, then this statement is true, although if one uses unrescaled versions of convergence (e.g. convergence in distribution, vague convergence, total variation convergence, etc.) then the statement is either false or trivially true, depending on exactly which mode of convergence is specified. It is somewhat of an abuse of notation to refer to convergence of random variables without specifying the exact nature of the convergence, although for informal mathematical discussion it is generally permissible to be a bit loose in this regard.

7 May, 2013 at 3:44 am

Anonymous(i) In Remark 1 the word theorem is missing from “central limit theorem.”

(ii) The equation following the sentence “From the bounded density of $G$ and the rapid decrease of $\eta$ we have…” I think the first term should be an expectation and not probability.

(iii) Do you use any convention for capitalizing theorems? For example, you write “central limit theorem” in lower case but “Taylor” expansion in upper case.

(iv) What is the diagonalisation argument in “By a diagonalisation argument, we conclude that there exists a sequence…”? Is there a Wikipedia page? Thanks.

[Corrected, thanks. Taylor is a proper noun and is therefore capitalised in English. The diagonalisation argument, originally due to Cantor, refers to any procedure in which one repeatedly extracts sequences with various properties and then passes to a diagonal subsequence with an even better property. Actually, the Arzela-Ascoli diagonalisation argument is closer in spirit to the one used here than the Cantor diagonalisation argument.]13 May, 2013 at 12:00 pm

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