Suppose we have a large number of scalar random variables , which each have bounded size on average (e.g. their mean and variance could be ). What can one then say about their sum ? If each individual summand varies in an interval of size , then their sum of course varies in an interval of size . However, a remarkable phenomenon, known as concentration of measure, asserts that assuming a sufficient amount of independence between the component variables , this sum sharply concentrates in a much narrower range, typically in an interval of size . This phenomenon is quantified by a variety of large deviation inequalities that give upper bounds (often exponential in nature) on the probability that such a combined random variable deviates significantly from its mean. The same phenomenon applies not only to linear expressions such as , but more generally to nonlinear combinations of such variables, provided that the nonlinear function is sufficiently regular (in particular, if it is Lipschitz, either separately in each variable, or jointly in all variables).

The basic intuition here is that it is difficult for a large number of independent variables to “work together” to simultaneously pull a sum or a more general combination too far away from its mean. Independence here is the key; concentration of measure results typically fail if the are too highly correlated with each other.

There are many applications of the concentration of measure phenomenon, but we will focus on a specific application which is useful in the random matrix theory topics we will be studying, namely on controlling the behaviour of random -dimensional vectors with independent components, and in particular on the distance between such random vectors and a given subspace.

Once one has a sufficient amount of independence, the concentration of measure tends to be sub-gaussian in nature; thus the probability that one is at least standard deviations from the mean tends to drop off like for some . In particular, one is standard deviations from the mean with high probability, and standard deviations from the mean with overwhelming probability. Indeed, concentration of measure is our primary tool for ensuring that various events hold with overwhelming probability (other moment methods can give high probability, but have difficulty ensuring overwhelming probability).

This is only a brief introduction to the concentration of measure phenomenon. A systematic study of this topic can be found in this book by Ledoux.

— 1. Linear combinations, and the moment method —

We begin with the simple setting of studying a sum of random variables. As we shall see, these linear sums are particularly amenable to the moment method, though to use the more powerful moments, we will require more powerful independence assumptions (and, naturally, we will need more moments to be finite or bounded). As such, we will take the opportunity to use this topic (large deviation inequalities for sums of random variables) to give a tour of the moment method, which we will return to when we consider the analogous questions for the bulk spectral distribution of random matrices.

In this section we shall concern ourselves primarily with bounded random variables; in the next section we describe the basic truncation method that can allow us to extend from the bounded case to the unbounded case (assuming suitable decay hypotheses).

The zeroth moment method gives a crude upper bound when is non-zero,

but in most cases this bound is worse than the trivial bound . This bound, however, will be useful when performing the truncation trick, which we will discuss below.

The first moment method is somewhat better, giving the bound

which when combined with Markov’s inequality gives the rather weak large deviation inequality

As weak as this bound is, this bound is sometimes sharp. For instance, if the are all equal to a single signed Bernoulli variable , then , and so , and so (2) is sharp when . The problem here is a complete lack of independence; the are all simultaneously positive or simultaneously negative, causing huge fluctuations in the value of .

Informally, one can view (2) as the assertion that typically has size .

The first moment method also shows that

and so we can normalise out the means using the identity

Replacing the by (and by ) we may thus assume for simplicity that all the have mean zero.

Now we consider what the second moment method gives us. We square and take expectations to obtain

If we assume that the are pairwise independent (in addition to having mean zero), then vanishes unless , in which case this expectation is equal to . We thus have

which when combined with Chebyshev’s inequality (and the mean zero normalisation) yields the large deviation inequality

Without the normalisation that the have mean zero, we obtain

Informally, this is the assertion that typically has size , if we have pairwise independence. Note also that we do not need the full strength of the pairwise independence assumption; the slightly weaker hypothesis of being pairwise uncorrelated would have sufficed.

The inequality (5) is sharp in two ways. Firstly, we cannot expect any significant concentration in any range narrower than the standard deviation , as this would likely contradict (3). Secondly, the quadratic-type decay in in (5) is sharp given the pairwise independence hypothesis. For instance, suppose that , and that , where is drawn uniformly at random from the cube , and are an enumeration of the non-zero elements of . Then a little Fourier analysis shows that each for has mean zero, variance , and are pairwise independent in ; but is equal to , which is equal to with probability ; this is despite the standard deviation of being just . This shows that (5) is essentially (i.e. up to constants) sharp here when .

Now we turn to higher moments. Let us assume that the are normalised to have mean zero and variance at most , and are also almost surely bounded in magnitude by some : . (The interesting regime here is when , otherwise the variance is in fact strictly smaller than .) To simplify the exposition very slightly we will assume that the are real-valued; the complex-valued case is very analogous (and can also be deduced from the real-valued case) and is left to the reader.

Let us also assume that the are -wise independent for some even positive integer . With this assumption, we can now estimate the moment

To compute the expectation of the product, we can use the -wise independence, but we need to divide into cases (analogous to the and cases in the second moment calculation above) depending on how various indices are repeated. If one of the only appear once, then the entire expectation is zero (since has mean zero), so we may assume that each of the appear at least twice. In particular, there are at most distinct which appear. If exactly such terms appear, then from the unit variance assumption we see that the expectation has magnitude at most ; more generally, if terms appear, then from the unit variance assumption and the upper bound by we see that the expectation has magnitude at most . This leads to the upper bound

where is the number of ways one can assign integers in such that each appears at least twice, and such that exactly integers appear.

We are now faced with the purely combinatorial problem of estimating . We will use a somewhat crude bound. There are ways to choose integers from . Each of the integers has to come from one of these integers, leading to the crude bound

which after using a crude form of Stirling’s formula gives

and so

If we make the mild assumption

then from the geometric series formula we conclude that

(say), which leads to the large deviation inequality

This should be compared with (2), (5). As increases, the rate of decay in the parameter improves, but to compensate for this, the range that concentrates in grows slowly, to rather than .

Remark 1 Note how it was important here that was even. Odd moments, such as , can be estimated, but due to the lack of the absolute value sign, these moments do not give much usable control on the distribution of the . One could be more careful in the combinatorial counting than was done here, but the net effect of such care is only to improve the unspecified constant (which can easily be made explicit, but we will not do so here).

Now suppose that the are not just -wise independent for any fixed , but are in fact jointly independent. Then we can apply (7) for any obeying (6). We can optimise in by setting to be a small multiple of , and conclude the gaussian-type bound

for some absolute constants , provided that for some small . (Note that the bound (8) is trivial for , so we may assume that is small compared to this quantity.) Thus we see that while control of each individual moment only gives polynomial decay in , by using all the moments simultaneously one can obtain square-exponential decay (i.e. subgaussian type decay).

By using Stirling’s formula (Exercise 2 from Notes 0a) one can show that the quadratic decay in (8) cannot be improved; see Exercise 2 below.

It was a little complicated to manage such large moments . A slicker way to proceed (but one which exploits the joint independence and commutativity more strongly) is to work instead with the exponential moments , which can be viewed as a sort of generating function for the power moments. A useful lemma in this regard is

Lemma 1 (Hoeffding’s lemma) Let be a scalar variable taking values in an interval . Then for any , In particular

Proof: It suffices to prove the first inequality, as the second then follows using the bound and from various elementary estimates.

By subtracting the mean from we may normalise . By dividing (and multiplying to balance) we may assume that , which implies that . We then have the Taylor expansion

which on taking expectations gives

and the claim follows.

Exercise 1 Show that the factor in (10) can be replaced with , and that this is sharp. (Hint: use Jensen’s inequality.)

We now have the fundamental Chernoff bound:

Theorem 2 (Chernoff inequality) Let be independent scalar random variables with almost surely, with mean and variance . Then for any , one has for some absolute constants , where and .

Proof: By taking real and imaginary parts we may assume that the are real. By subtracting off the mean (and adjusting appropriately) we may assume that (and so ); dividing the (and ) through by we may assume that . By symmetry it then suffices to establish the upper tail estimate

(with slightly different constants ).

To do this, we shall first compute the exponential moments

where is a real parameter to be optimised later. Expanding out the exponential and using the independence hypothesis, we conclude that

To compute , we use the hypothesis that and (9) to obtain

Thus we have

and thus by Markov’s inequality

If we optimise this in , subject to the constraint , we obtain the claim.

Informally, the Chernoff inequality asserts that is sharply concentrated in the range . The bounds here are fairly sharp, at least when is not too large:

Exercise 2 Let be fixed independently of , and let be iid copies of a Bernoulli random variable that equals with probability , thus and , and so and . Using Stirling’s formula (Notes 0a), show that for some absolute constants and all . What happens when is much larger than ?

Exercise 3 Show that the term in (11) can be replaced with (which is superior when ). (Hint: Allow to exceed .) Compare this with the results of Exercise 2.

Exercise 4 (Hoeffding’s inequality) Let be independent real variables, with taking values in an interval , and let . Show that one has for some absolute constants , where .

Remark 2 As we can see, the exponential moment method is very slick compared to the power moment method. Unfortunately, due to its reliance on the identity , this method relies very strongly on commutativity of the underlying variables, and as such will not be as useful when dealing with noncommutative random variables, and in particular with random matrices. Nevertheless, we will still be able to apply the Chernoff bound to good effect to various components of random matrices, such as rows or columns of such matrices.

The full assumption of joint independence is not completely necessary for Chernoff-type bounds to be present. It suffices to have a martingale difference sequence, in which each can depend on the preceding variables , but which always has mean zero even when the preceding variables are conditioned out. More precisely, we have Azuma’s inequality:

Theorem 3 (Azuma’s inequality) Let be a sequence of scalar random variables with almost surely. Assume also that we have the martingale difference property almost surely for all (here we assume the existence of a suitable disintegration in order to define the conditional expectation, though in fact it is possible to state and prove Azuma’s inequality without this disintegration). Then for any , the sum obeys the large deviation inequality for some absolute constants .

A typical example of here is a dependent random walk, in which the magnitude and probabilities of the step are allowed to depend on the outcome of the preceding steps, but where the mean of each step is always fixed to be zero.

Proof: Again, we can reduce to the case when the are real, and it suffices to establish the upper tail estimate

Note that almost surely, so we may assume without loss of generality that .

Once again, we consider the exponential moment for some parameter . We write , so that

We do not have independence between and , so cannot split the expectation as in the proof of Chernoff’s inequality. Nevertheless we can use conditional expectation as a substitute. We can rewrite the above expression as

The quantity is deterministic once we condition on , and so we can pull it out of the conditional expectation:

Applying (10) to the conditional expectation, we have

and

Iterating this argument gives

and thus by Markov’s inequality

Optimising in gives the claim.

Exercise 5 Suppose we replace the hypothesis in Azuma’s inequality with the more general hypothesis for some scalars . Show that we still have (13), but with replaced by .

Remark 3 The exponential moment method is also used frequently in harmonic analysis to deal with lacunary exponential sums, or sums involving Radamacher functions (which are the analogue of lacunary exponential sums for characteristic ). Examples here include Khintchine’s inequality (and the closely related Kahane’s inequality). The exponential moment method also combines very well with log-Sobolev inequalities, as we shall see below (basically because the logarithm inverts the exponential), as well as with the closely related hypercontractivity inequalities.

— 2. The truncation method —

To summarise the discussion so far, we have identified a number of large deviation inequalities to control a sum :

The zeroth moment method bound (1), which requires no moment assumptions on the but is only useful when is usually zero, and has no decay in .

but is only useful when is usually zero, and has no decay in . The first moment method bound (2), which only requires absolute integrability on the , but has only a linear decay in .

, but has only a linear decay in . The second moment method bound (5), which requires second moment and pairwise independence bounds on , and gives a quadratic decay in .

, and gives a quadratic decay in . Higher moment bounds (7), which require boundedness and -wise independence, and give a power decay in (or quadratic-exponential decay, after optimising in ).

-wise independence, and give a power decay in (or quadratic-exponential decay, after optimising in ). Exponential moment bounds such as (11) or (13), which require boundedness and joint independence (or martingale behaviour), and give quadratic-exponential decay in .

We thus see that the bounds with the strongest decay in require strong boundedness and independence hypotheses. However, one can often partially extend these strong results from the case of bounded random variables to that of unbounded random variables (provided one still has sufficient control on the decay of these variables) by a simple but fundamental trick, known as the truncation method. The basic idea here is to take each random variable and split it as , where is a truncation parameter to be optimised later (possibly in manner depending on ),

is the restriction of to the event that (thus vanishes when is too large), and

is the complementary event. One can similarly split where

and

The idea is then to estimate the tail of and by two different means. With , the point is that the variables have been made bounded by fiat, and so the more powerful large deviation inequalities can now be put into play. With , in contrast, the underlying variables are certainly not bounded, but they tend to have small zeroth and first moments, and so the bounds based on those moment methods tend to be powerful here. (Readers who are familiar with harmonic analysis may recognise this type of divide and conquer argument as an interpolation argument.)

Let us begin with a simple application of this method.

Theorem 4 (Weak law of large numbers) Let be iid scalar random variables with for all , where is absolutely integrable. Then converges in probability to .

Proof: By subtracting from we may assume without loss of generality that has mean zero. Our task is then to show that for all fixed .

If has finite variance, then the claim follows from (5). If has infinite variance, we cannot apply (5) directly, but we may perform the truncation method as follows. Let be a large parameter to be chosen later, and split , (and ) as discussed above. The variable is bounded and thus has bounded variance; also, from the dominated convergence theorem we see that (say) if is large enough. From (5) we conclude that

(where the rate of decay here depends on and ). Meanwhile, to deal with the tail we use (2) to conclude that

But by the dominated convergence theorem (or monotone convergence theorem), we may make as small as we please (say, smaller than ) by taking large enough. Summing, we conclude that

since is arbitrary, we obtain the claim.

A more sophisticated variant of this argument (which I gave in this earlier blog post, which also has some further discussion and details) gives

Theorem 5 (Strong law of large numbers) Let be iid scalar random variables with for all , where is absolutely integrable. Then converges almost surely to .

Proof: We may assume without loss of generality that is real, since the complex case then follows by splitting into real and imaginary parts. By splitting into positive and negative parts, we may furthermore assume that is non-negative. (Of course, by doing so, we can no longer normalise to have mean zero, but for us the non-negativity will be more convenient than the zero mean property.) In particular, is now non-decreasing in .

Next, we apply a sparsification trick. Let . Suppose that we knew that, almost surely, converged to for of the form for some integer . Then, for all other values of , we see that asymptotically, can only fluctuate by a multiplicative factor of , thanks to the monotone nature of . Because of this and countable additivity, we see that it suffices to show that converges to . Actually, it will be enough to show that almost surely, one has for all but finitely many .

Fix . As before, we split and , but with the twist that we now allow to depend on . Then for large enough we have (say), by dominated convergence. Applying (5) as before, we see that

for some depending only on (the exact value is not important here). To handle the tail, we will not use the first moment bound (2) as done previously, but now turn to the zeroth-moment bound (1) to obtain

summing, we conclude

Applying the Borel-Cantelli lemma (Exercise 1 from Notes 0), we see that we will be done as long as we can choose such that

and

are both finite. But this can be accomplished by setting and interchanging the sum and expectations (writing as ) and using the lacunary nature of the .

To give another illustration of the truncation method, we extend a version of the Chernoff bound to the subgaussian case.

Proposition 6 Let be iid copies of a subgaussian random variable , thus obeys a bound of the form for all and some . Let . Then for any sufficiently large (independent of ) we have for some constants depending on . Furthermore, grows linearly in as .

Proof: By subtracting the mean from we may normalise . We perform a dyadic decomposition

where and . We similarly split

where . Then by the union bound and the pigeonhole principle we have

(say). Each is clearly bounded in magnitude by ; from the subgaussian hypothesis one can also verify that the mean and variance of are at most for some . If is large enough, an application of the Chernoff bound (11) (or more precisely, the refinement in Exercise 3) then gives (after some computation)

(say) for some , and the claim follows.

Exercise 6 Show that the hypothesis that is sufficiently large can be replaced by the hypothesis that is independent of . Hint: There are several approaches available. One can adapt the above proof; one can modify the proof of the Chernoff inequality directly; or one can figure out a way to deduce the small case from the large case.

Exercise 7 Show that the subgaussian hypothesis can be generalised to a sub-exponential tail hypothesis provided that . Show that the result also extends to the case , except with the exponent replaced by for some . (I do not know if the loss can be removed, but it is easy to see that one cannot hope to do much better than this, just by considering the probability that (say) is already as large as .)

— 3. Lipschitz combinations —

In the preceding discussion, we had only considered the linear combination of independent variables . Now we consider more general combinations , where we write for short. Of course, to get any non-trivial results we must make some regularity hypotheses on . It turns out that a particularly useful class of regularity hypothesis here is a Lipschitz hypothesis – that small variations in lead to small variations in . A simple example of this is McDiarmid’s inequality:

Theorem 7 (McDiarmid’s inequality) Let be independent random variables taking values in ranges , and let be a function with the property that if one freezes all but the coordinate of for some , then only fluctuates by most , thus for all , for . Then for any , one has for some absolute constants , where .

Proof: We may assume that is real. By symmetry, it suffices to show the one-sided estimate

To compute this quantity, we again use the exponential moment method. Let be a parameter to be chosen later, and consider the exponential moment

To compute this, let us condition to be fixed, and look at the conditional expectation

We can simplify this as

where

For fixed, only fluctuates by at most and has mean zero. Applying (10), we conclude that

Integrating out the conditioning, we see that we have upper bounded (16) by

We observe that is a function of , where obeys the same hypotheses as (but for instead of ). We can then iterate the above computation times and eventually upper bound (16) by

which we rearrange as

and thus by Markov’s inequality

Optimising in then gives the claim.

Exercise 8 Show that McDiarmid’s inequality implies Hoeffding’s inequality (Exercise 4).

Remark 4 One can view McDiarmid’s inequality as a tensorisation of Hoeffding’s lemma, as it leverages the latter lemma for a single random variable to establish an analogous result for random variables. It is possible to apply this tensorisation trick to random variables taking values in more sophisticated metric spaces than an interval , leading to a class of concentration of measure inequalities known as transportation cost-information inequalities, which will not be discussed here.

The most powerful concentration of measure results, though, do not just exploit Lipschitz type behaviour in each individual variable, but joint Lipschitz behaviour. Let us first give a classical instance of this, in the special case when the are gaussian variables. A key property of gaussian variables is that any linear combination of independent gaussians is again an independent gaussian:

Exercise 9 Let be independent real gaussian variables with , and let be real constants. Show that is a real gaussian with mean and variance . Show that the same claims also hold with complex gaussians and complex constants .

Exercise 10 (Rotation invariance) Let be an -valued random variable, where are iid real gaussians. Show that for any orthogonal matrix , . Show that the same claim holds for complex gaussians (so is now -valued), and with the orthogonal group replaced by the unitary group .

Theorem 8 (Gaussian concentration inequality for Lipschitz functions) Let be iid real gaussian variables, and let be a -Lipschitz function (i.e. for all , where we use the Euclidean metric on ). Then for any one has for some absolute constants .

Proof: We use the following elegant argument of Maurey and Pisier. By subtracting a constant from , we may normalise . By symmetry it then suffices to show the upper tail estimate

By smoothing slightly we may assume that is smooth, since the general case then follows from a limiting argument. In particular, the Lipschitz bound on now implies the gradient estimate

for all .

Once again, we use the exponential moment method. It will suffice to show that

for some constant and all , as the claim follows from Markov’s inequality and optimisation in as in previous arguments.

To exploit the Lipschitz nature of , we will need to introduce a second copy of . Let be an independent copy of . Since , we see from Jensen’s inequality that

and thus (by independence of and )

It is tempting to use the fundamental theorem of calculus along a line segment,

to estimate , but it turns out for technical reasons to be better to use a circular arc instead,

The reason for this is that is another gaussian random variable equivalent to , as is its derivative (by Exercise 9); furthermore, and crucially, these two random variables are independent (by Exercise 10).

To exploit this, we first use Jensen’s inequality to bound

Applying the chain rule and taking expectations, we have

Let us condition to be fixed, then ; applying Exercise 9 and (17), we conclude that is normally distributed with standard deviation at most . As such we have

for some absolute constant ; integrating out the conditioning on we obtain the claim.

Exercise 11 Show that Theorem 8 is equivalent to the inequality holding for all and all measurable sets , where is an -valued random variable with iid gaussian components , and is the -neighbourhood of .

Now we give a powerful concentration inequality of Talagrand, which we will rely heavily on later in this course.

Theorem 9 (Talagrand concentration inequality) Let , and let be independent complex variables with for all . Let be a -Lipschitz convex function (where we identify with for the purposes of defining “Lipschitz” and “convex”). Then for any one has and for some absolute constants , where is a median of .

We now prove the theorem, following the remarkable argument of Talagrand.

By dividing through by we may normalise . now takes values in the convex set , where is the unit disk in . It will suffice to establish the inequality

for any convex set in and some absolute constant , where is the Euclidean distance between and . Indeed, if one obtains this estimate, then one has

for any (as can be seen by applying (20) to the convex set ). Applying this inequality of one of equal to the median of yields (18), which in turn implies that

which then gives (19).

We would like to establish (20) by induction on dimension . In the case when are Bernoulli variables, this can be done; see this previous blog post. In the general case, it turns out that in order to close the induction properly, one must strengthen (20) by replacing the Euclidean distance by an essentially larger quantity, which I will call the combinatorial distance from to . For each vector and , we say that supports if is non-zero only when is non-zero. Define the combinatorial support of relative to to be all the vectors in that support at least one vector in . Define the combinatorial hull of relative to to be the convex hull of , and then define the combinatorial distance to be the distance between and the origin.

Lemma 10 (Combinatorial distance controls Euclidean distance) Let be a convex subset of . .

Proof: Suppose . Then there exists a convex combination of elements which has magnitude at most . For each such , we can find a vector supported by . As both lie in , every coefficient of has magnitude at most , and is thus bounded in magnitude by twice the corresponding coefficent of . If we then let be the convex combination of the indicated by , then the magnitude of each coefficient of is bounded by twice the corresponding coefficient of , and so . On the other hand, as is convex, lies in , and so . The claim follows.

Thus to show (20) it suffices (after a modification of the constant ) to show that

We first verify the one-dimensional case. In this case, equals when , and otherwise, and the claim follows from elementary calculus (for small enough).

Now suppose that and the claim has already been proven for . We write , and let be a slice of . We also let . We have the following basic inequality:

Lemma 11 For any , we have

Proof: Observe that contains both and , and so by convexity, contains one of or whenever and . The claim then follows from Pythagoras’ theorem and the Cauchy-Schwarz inequality.

Let us now freeze and consider the conditional expectation

Using the above lemma (with some depending on to be chosen later), we may bound the left-hand side of (21) by

applying Hölder’s inequality and the induction hypothesis (21), we can bound this by

which we can rearrange as

where (here we note that the event is independent of ). Note that . We then apply the elementary inequality

which can be verified by elementary calculus if is small enough (in fact one can take ). We conclude that

Taking expectations in we conclude that

Using the inequality with we conclude (21) as desired.

The above argument was elementary, but rather “magical” in nature. Let us now give a somewhat different argument of Ledoux, based on log-Sobolev inequalities, which gives the upper tail bound

but curiously does not give the lower tail bound. (The situation is not symmetric, due to the convexity hypothesis on .)

Once again we can normalise . By regularising we may assume that is smooth. The first step is to establish the following log-Sobolev inequality:

Lemma 12 (Log-Sobolev inequality) Let be a smooth convex function. Then for some absolute constant (independent of ).

Remark 5 If one sets and normalises , this inequality becomes which more closely resembles the classical log-Sobolev inequality of Gross. The constant here can in fact be taken to be ; see Ledoux’s paper.

Proof: We first establish the -dimensional case. If we let be an independent copy of , observe that the left-hand side can be rewritten as

From Jensen’s inequality, , so it will suffice to show that

From convexity of (and hence of ) and the bounded nature of , we have

and

when , which leads to

in this case. Similarly when (swapping and ). The claim follows.

To show the general case, we induct on (keeping care to ensure that the constant does not change in this induction process). Write , where . From induction hypothesis, we have

where is the -dimensional gradient and . Taking expectations, we conclude that

From the convexity of and Hölder’s inequality we see that is also convex, and . By the case already established, we have

Now, by the chain rule

where is the derivative of in the direction. Applying Cauchy-Schwarz, we conclude

Inserting this into (23), (24) we close the induction.

Now let be convex and -Lipschitz. Applying the above lemma to for any , we conclude that

setting , we can rewrite this as a differential inequality

which we can rewrite as

From Taylor expansion we see that

as , and thus

for any . In other words,

By Markov’s inequality, we conclude that

optimising in gives (22).

Remark 6 The same argument, starting with Gross’s log-Sobolev inequality for the gaussian measure, gives the upper tail component of Theorem 8, with no convexity hypothesis on . The situation is now symmetric with respect to reflections , and so one obtains the lower tail component as well. The method of obtaining concentration inequalities from log-Sobolev inequalities (or related inequalities, such as Poincaré-type ienqualities) by combining the latter with the exponential moment method is known as Herbst’s argument, and can be used to establish a number of other functional inequalities of interest.

We now close with a simple corollary of the Talagrand concentration inequality, which will be extremely useful in the sequel.

Corollary 13 (Distance between random vector and a subspace) Let be independent complex-valued random variables with mean zero and variance , and bounded almost surely in magnitude by . Let be a subspace of of dimension . Then for any , one has for some absolute constants .

Informally, this corollary asserts that the distance between a random vector and an arbitrary subspace is typically equal to .

Proof: The function is convex and -Lipschitz. From Theorem 9, one has

To finish the argument, it then suffices to show that

We begin with a second moment calculation. Observe that

where is the orthogonal projection matrix to the complement of , and are the components of . Taking expectations, we obtain

where the latter follows by representing in terms of an orthonormal basis of . This is close to what we need, but to finish the task we need to obtain some concentration of around its mean. For this, we write

where is the Kronecker delta. The summands here are pairwise uncorrelated for , and the cases can be combined with the cases by symmetry. Each summand also has a variance of . We thus have the variance bound

where the latter bound comes from representing in terms of an orthonormal basis of . From this, (25), and Chebyshev’s inequality, we see that the median of is equal to , which implies on taking square roots that the median of is , as desired.