(Warning: These materials may be subject to lots of typos and errors. We are grateful if you could spot errors and leave suggestions in the comments, or contact the author at yjhan@stanford.edu.) 

In the last two lectures we have seen two specific categories of the Le Cam’s two-point methods, i.e., testing between two single hypotheses, or between one single hypothesis and a mixture of hypotheses. Of course, the most powerful and natural generalization of the two-point methods is to test between two mixtures of distributions, which by the minimax theorem is potentially the best possible approach to test between two composite hypotheses. However, the least favorable prior (mixture) may be hard to find, and it may be theoretically difficult to upper bound the total variation distance between two mixtures of distributions. In this lecture, we show that both problems are closely related to moment matching via examples in Gaussian and Poisson models. 

1. Fuzzy Hypothesis Testing and Moment Matching 

In this section, we present the general tools necessary for this lecture. First, we prove a generalized two-point method also known as the fuzzy hypothesis testing. To upper bound the crucial divergence term between two mixtures, we introduce the orthogonal polynomials under different distributions and show that the upper bound depends on moment differences of the mixtures. 

1.1. Mixture vs. Mixture 

We first state the theorem of the generalized two-point methods with two mixtures. As usual, we require that any two points in respective mixtures be well separated but the mixtures be indistinguishable from samples. However, since many natural choices of mixtures may not satisfy the well-separated property in the worst case, the next theorem will be a bit more flexible to require that the mixtures are separated with a large probability. 

Theorem 1 (Mixture vs. Mixture) Let be any loss function, and there exist and such that

Then for any probability measures supported on , we have

Proof: For , let be the conditional probability measure of conditioned on , i.e., 

Simple algebra gives . By coupling, we also have 

Now the desired result follows from the standard two-point arguments and the triangle inequality of the total variation distance.  

The central quantity in Theorem 1 is , the total variation distance between mixture distributions with priors and . A general upper bound on this quantity is very hard to obtain, but the next two sections will show that it is small when the moments of and are close to each other if the model is Gaussian or Poisson. 

1.2. Hermite and Charlier Polynomials 

This section reviews some preliminary results on orthogonal polynomials under a fixed probability distribution. Let be a probability measure on with and all moments finite. Recall that functions with are called orthogonal under iff 

for and all . By orthogonal polynomials we mean that for each , is a polynomial with degree . 

The simplest way to construct orthogonal polynomials is via the Gram–Schmidt orthogonalization. Specifically, we may choose , and to be the orthogonal component of projected onto with the inner product structure . This approach works for general distributions and can be easily implemented in practice, but it gives little insight on the properties of . We shall apply a new approach to arrive at orthogonal functions, which turn out to be polynomials in Gaussian and Poisson models. 

Let be a family of distributions on with . Assume that admits the following local expansion around : 

The next lemma claims that under specific conditions of , the functions are orthogonal under . 

Lemma 2 Under the above conditions, if for all the quantity

depends only on their product and , then the functions are orthogonal under . 

Remark 1 Recall that the quantity plays an important role in the Ingster-Suslina method in Lecture 6. The upper bounds in the next section can be thought of as a generalization of the Ingster-Suslina method, with the help of proper orthogonality properties. 

Proof: The local expansion of likelihood ratio gives 

The condition of Lemma 2 implies that the coefficient of the monomial on the RHS with must be zero, as desired.  

Exercise 1 Show that under the conditions in Lemma 2, for some scalar and any . In other words, is an unbiased estimator of up to scaling in the location model . 

The condition of Lemma 2 is satisfied by various well-known probability distributions. For example, 

and for any , 

In fact, the functions given in the defining equation (1) in the Gaussian and Poisson models are both polynomials of degree , known as the Hermite polynomial  and the Charlier polynomial , respectively. The proof of Lemma 2 gives the following orthogonal relations: 

Throughout this lecture, we won’t need the specific forms of the Hermite and Charlier polynomials. We shall only need the defining property (1) and the above orthogonal relations. 

1.3. Divergences between Mixtures 

Now we are ready to present the upper bound on the total variation distance between Gaussian mixture and Poisson mixture models. We also provide upper bounds on the divergence due to its nice tensorization property. 

We first deal with the Gaussian location model. Let and be two random variables on , and let be the Gaussian mixture with random mean . 

Theorem 3 (Divergence between Gaussian Mixtures) For any , we have

Moreover, if and , then 

Proof: By translation we may assume that . Let be the pdf of , and . Then 

where step (a) is due to the defining property (1), step (b) follows from the Cauchy–Schwartz inequality, and step (c) uses the orthogonal relation of . Hence the upper bound on the total variation distance is proved. 

For the -divergence, first note that by Jensen’s inequality, 

As a result, 

where again we have used the defining property (1) and the orthogonality in the last two identities.  

Specifically, Theorem 3 shows that when the moments of and are close, then the corresponding Gaussian mixtures are statistically close. Similar results also hold for Poisson mixtures. 

Theorem 4 (Divergence between Poisson Mixtures) For any and random variables supported on , we have

Moreover, if and almost surely, then 

Proof: The proof of both inequalities essentially follow the same lines as those in the proof of Theorem 3, with the Hermite polynomial replaced by the Charlier polynomial . The only difference is that when and almost surely, for all we have 

2. Examples in Gaussian Models 

In this section, we present examples in Gaussian location models where we need to test between two mixtures. In these examples, we match moments up to either some finite degree, or some large and growing degrees such that the information divergences become extremely small. 

2.1. Gaussian Mixture Models 

Consider the following two-component Gaussian mixture model where i.i.d. samples are drawn from . One possible task of proper learning is to estimate the parameters within -distance to the truth up to permutation. In other words, the target is to recover the components of the Gaussian mixture, which in practice helps to perform tasks such as clustering. The target is to determine the optimal sample complexity of this problem, assuming that is unknown but bounded away from and (e.g., ), and the overall variance of the mixture is at most , where is some prespecified parameter. 

To derive a lower bound, the two-point method suggests to find two sets of parameters which are -seperated, while the information divergence between these two mixtures is . However, since Theorem 3 can only deal with mixtures with an identical variance in each component, we cannot simply take to be a discrete random variable supported on two points . To overcome this difficulty, note that 

where denotes convolution of probability measures. Hence, we may treat the overall mixture to be , and choose and similarly for . Now the desired -divergence becomes 

To apply Theorem 3, we should construct random variables and with as many matched moments as possible. Since there are free parameters in the two-component Gaussian mixtures, we expect that and can only have matched moments up to degree . A specific choice can be as follows: 

Clearly both and have overall variance . Replacing by their centered version, Theorem 3 gives 

as long as . Hence, by the additivity of the -divergence, we conclude that is a lower bound on the sample complexity. 

The seemingly strange bound is also tight for this problem, and the idea is to estimate the first moments of the mixture and then show that close moments imply close parameters. We leave the details to the reference in the bibliographic notes. 

2.2. -norm Estimation of Bounded Gaussian Mean 

Consider the Gaussian location model with unit variance, where the mean vector satisfies . The target here is to estimate the norm of the mean vector , and let be the minimax risk under the absolute value loss. The main result here is to prove the following tight lower bound: 

2.2.1. Failure of Point vs. Mixture

Motivated by the point vs. mixture approach in the last lecture, one natural idea is to test between and a composite hypothesis , where is a parameter to be specified later such that and are indistinguishable in the minimax sense. Consequently, this approach gives the lower bound , and the target is to find some as large as possible. We claim that the largest possible is , which is strictly smaller than the desired minimax risk. 

Let , and consider any prior distribution supported on . Then Ingster-Suslina method gives 

Using the Taylor expansion and the inequality for all , we conclude that 

Note that the above inequality holds for any supported on . Hence, when , these hypotheses and become statistically distinguishable, and therefore the best possible lower bound from the point vs. mixture approach is . 

There is also another way to show the desired failure, i.e., we may construct an explicit test which reliably distinguishes between and . The idea is to apply the test, i.e., compute the statistic . Clearly under we have , and after some algebra we may show that under . Since implies , we conclude that comparing with a suitable threshold results in a reliable test. 

2.2.1. Moment Matching and Polynomial Approximation

Previous section shows that testing between a single distribution and a mixture does not work, where the knowledge of the single distribution can be used for the test and may make the problem significantly easier. Hence, an improvement is to consider two composite hypotheses and , where are parameters to be chosen later. For the priors on , Theorem 3 motivates us to use product priors where are probability measures on with matched moments up to degree (to be chosen later). The specific choices of must fulfill the following requirements: 

1. Have matched moments up to degree while with the quantity as large as possible; 
2. For , the prior is supported on ; 
3. The -divergence is upper bounded by , or in other words, . 

We check the above requirements in the reverse order and specify the choices of and gradually. For the last requirement, Theorem 3 with gives 

As a result, to have , it suffices to take . 

The second constraint that be (almost) supported on is also easy. Set 

where is a large eough numerical constant. The idea behind the above choices is that, under the product distribution , the random variable is the sum of i.i.d. random variables taking value in following distribution . Then by the sub-Gaussian concentration, the norm is centered at with fluctuation . Hence, for large both probabilities and are small, which fulfills the conditions in Theorem 1. 

The most non-trivial requirement is the first requirement, which by our choice of essentially aims to maximize the difference subject to the constraint that the probability measures are supported on and have matching first moments. The following lemma shows the duality between moment matching and best polynomial approximation. 

Lemma 5 For any bounded interval and real-valued function on , let be the maximum difference subject to the constraint that the probability measures are supported on and have matching first moments. Then

where denotes the best degree- polynomial approximation error of on the interval : 

Proof: It is an easy exercise to show that . We present two proofs for the hard direction . The first proof is an abstract proof which holds for general basis functions other than monomials, while the construction of the measures is implicit. The second proof gives an explicit construction, while some properties of polynomials are used in the proof. 

(First Proof) Consider the following linear functional , with for and . Equipped with the norm on functions, it is easy to show that the operator norm of is . By the Hahn-Banach theorem, the linear functional can be extended to without increasing the operator norm. Then by the Riesz representation theorem, there exists a signed Radon measure on such that 

The fact implies that the total variation of is one. Write by Jordan decomposition of signed measures, then and satisfy the desired properties. 

(Second Proof) Let be a degree- polynomial with on . Since is a Haar basis on , Chebyshev’s alternation theorem shows that there exist points such that with or . Consider the signed measure supported on with 

where is a normalizing constant such that . Then by simple algebra, and has total variation . Moreover, the following identity is given by Lagrange interpolation

and comparing the coefficient of on both sides gives for . Now another Jordan decomposition of gives the desired result.  

By Lemma 5, it boils down to the best degree- polynomial approximation error of on . This error is analyzed in approximation theory and summarized in the following lemma. 

Lemma 6 There is a numerical constant (known as the Bernstein’s constant) such that

Hence, by Lemma 5 and 6, the condition of Theorem 1 is satisfied with . As a result, we finally arrive at . 

3. Examples in Poisson Models 

In this section, we present examples in i.i.d. sampling models from a discrete distribution with a large support. We show that a general Poissonization technique allows us to operate in the simpler Poisson models, and use moment matching to either finite or growing degrees to establish tight lower bounds. 

3.1. Poissonization and Approximate Distribution 

Throughout this section the statistical model is i.i.d. sampling from a discrete distribution , where denotes the sample size and denotes the support size. It is well-known that the histogram  with 

constitutes a sufficient statistic. Moreover, for all , and are negatively dependent. To remove the dependence among different bins, recall the following Poissonized model: 

Definition 7 (Poissonization) In the Poissonized model, for all and they are mutually independent. 

In other words, in the Poissonized model we draw a random number of samples from and then compute the histogram. In Lecture 3 we have shown the asymptotic equivalence of the i.i.d. sampling model and the Poissonized model, but the arguments are highly asymptotic. The following lemma establishes a non-asymptotic relationship. 

Lemma 8 For a given statistical task, let and be the minimax risk under the i.i.d. sampling model and the Poissonized model with design sample size , respectively. Then

Proof: Let be the Bayes risks under prior in respective models. The desired inequality for Bayes risks follows from the identity 

the Poisson tail bounds and the monotonicity of Bayes risks . Now the minimax theorem gives the desired lemma. 

To establish the first identity, simply note that under any prior distribution , the Bayes estimator under the Poissonized model given the realization is exactly the Bayes estimator under the i.i.d. sampling model with samples.  

Lemma 8 shows that the minimax risk essentially does not change after Poissonization. We sometimes also consider approximate distributions in the Poissonized model, where may not necessarily sum into one (note that the distribution of the histogram is still well-defined). The approximate distribution is typically used in lower bounds where a product prior is assigned to and cannot preserve the distribution property. For statistical problems where the objective function or hypothesis depends on the vector in a nice way that changing into with will not change the objective much, in the lower bound it typically suffices to consider the approximate distributions in the Poissonized model. The key idea is that, conditioning on , the histogram is exactly distributed as that in the i.i.d. sampling model from with samples. Then we may construct the estimator in the Poissonized model from the hypothetical optimal estimators (with different sample sizes) in the i.i.d. sampling model, and applying the same tail bounds as in the proof of Lemma 8 suffices. The details of the arguments may vary from example to example, but in most scenarios it will not hurt the lower bound. 

3.2. Generalized Uniformity Testing 

Consider the following generalized uniformity testing problem: given i.i.d. observations from some discrete distribution supported on at most elements, one would like to test whether the underlying distribution is uniform on its support. Note the difference from the traditional uniformity testing problem: the distribution may be supported on a subset of while still be uniform on this subset. Specifically, the task is to determine the sample complexity of distinguishing from the hypothesis uniform on its support and is -away from any uniform distribution with support under distance. We will show that the desired sample complexity is lower bounded by 

Note that the first term trivially follows from the Paninski’s construction in the traditional uniformity testing problem, the goal is to prove the second term. It is expected that the second term captures the difficulty of recovering the support of the distribution, and therefore we should consider a mixture of uniform distributions with random support in . Specifically, we consider the following mixture: let be two random variables with 

Assign the -fold product distribution of (or ) to the probability vector , then forms an approximate probability distribution since . Moveover, under prior the (normalized) distribution is always uniform, and under prior the (normalized) distribution is -far from any uniform distribution supported on a subset of with high probability. Hence, neglecting the additional details for the approximate distribution, it suffices to show that 

To establish the above bound for the -divergence, note that the random variables are chosen carefully with for . Moreover, 

Consequently, Theorem 4 gives that 

Since is a stronger lower bound than if and only if , under this condition and we will have . Then the above inequality gives 

which is the claimed upper bound on the -divergence, establishing the lower bound.

We provide some discussions on the choice of and . Theorem 4 suggests that if and could match more moments, the -divergence could even be smaller. However, the number of matched moments is in fact limited by the problem structure. In the traditional uniformity testing problem, in the last lecture we essentially choose and as above. In this case, and only match the first moment, which is the best possible for must be a constant. In generalized uniformity testing, must only be supported on two points, one of which is zero. Meanwhile, the support of can potentially be arbitrarily large. The next lemma shows that no matter how we choose , we can match at most the first two moments. 

Lemma 9 Let be a probability measure supported on elements of , one of which is zero, and be any probability measure supported on . Then if and match the first moments, we must have . 

Proof: Let . Let the support of be . Consider the polynomial of degree , the assumption gives . Finally, since is always non-negative, we have .  

Lemma 9 applied to shows that moment matching up to degree is the best we can hope for. In fact, the above lower bound is also tight (see bibliographic notes). 

3.3. Shannon Entropy Estimation 

Finally we revisit the Shannon entropy estimation problem where the target is to estimate the Shannon entropy . Let be the minimax risk of estimating under the mean squared error, our target is to show that 

The lower bound has already been shown via the two-point method in Lecture 5, and therefore the remaining target is to establish . 

Similar to the norm example above, the Shannon entropy is a symmetric sum of individual functions of , where the individual function is non-differentiable at zero. It suggests us to apply similar ideas based on moment matching and best polynomial approximation to establish the lower bound. Specifically, the target is to construct two priors on the interval (with parameter to be chosen later) such that: 

1. The priors have matched moments up to degree (to be chosen later); 
2. The difference between and is large; 
3. With high probability, the Shannon entropy under and is well-separated; 
4. The common mean is at most . 

Note that the first requirement (moment matching) ensures a small TV distance between the mixtures, the second and third requirements ensure the separation property (i.e., lower bound the mean difference and upper bound the fluctuations), and the last requirement ensures that sums into a constant smaller than one (recall that by assumption) and therefore setting gives a valid probability vector . We check these requirements one by one to find proper parameters . 

For the first requirement, recall that Theorem 4 gives 

Since the individual TV distance should be at most (for the future triangle inequality), we should choose , where is due to the assumption and that to make the first term become dominate in the minimax risk. 

For the second requirement, by the duality result in Lemma 5 it is essentially the best degree- polynomial approximation error of on . The next lemma gives the best approximation error. 

Lemma 10

By Lemma 10, the target is to maximize subject to the previous condition . Simple algebra shows that the maximum is , with the maximizer . 

To resolve the third requirement, note that the mean difference of is by the above choice of and . Moreover, since for some constant if , the sub-Gaussian concentration shows that the fluctuation of under both is at most . Since , the fluctuation is indeed negligible compared with the mean difference. Careful analysis also shows that the contribution of to the entropy difference is negligible. 

The last requirement requires proper modifications on the priors constructed in Lemma 5 to satisfy the mean constraint. This can be done via the following trick of change of measures: let be the priors constructed in Lemma 5which attains , whose value is summarized in the following lemma. 

Lemma 11

Next we construct the priors as follows: for , set 

Then it is easy to show that both and are probability measures, have matched moments up to degree , and have mean . Moreover, 

Hence, the fourth requirement is fulfilled without hurting the previous ones. 

In summary, Theorem 1 holds with , and we arrive at the desired lower bound . 

4. Bibliographic Notes 

The method of two fuzzy hypotheses (Theorem 1) are systematically used in Ingster and Suslina (2012) on nonparametric testing, and the current form of the theorem is taken from Theorem 2.14 of Tsybakov (2009). Statistical closeness of Gaussian mixture models via moment matching (Theorem 3) was established in Cai and Low (2011), Hardt and Price (2015), Wu and Yang (2018) for the -divergence, where the result is new for the TV distance. For Theorem 4, the TV version was established in part by Jiao, Kartik, Han and Weissman (2015) and Jiao, Han and Weissman (2018), where the version is new here. When , a stronger bound of the TV distance without the squared root is also available in Wu and Yang (2016). For more properties of Hermite and Charlier polynomials, we refer to Labelle and Yeh (1989). 

For Gaussian examples, the proper learning of two-component Gaussian mixture was established in Hardt and Price (2015). The norm estimation problem was taken from Cai and Low (2011), which was further motivated by Lepski, Nemirovski and Spokoiny (1999). The Gaussian mean testing example under metric was taken from Ingster and Suslina (2012). For technical lemmas, proofs of Lemma 5 is taken from Lepski, Nemirovski and Spokoiny (1999) and Wu and Yang (2016), respectively, and Lemma 6 was due to Bernstein (1912). 

For Poisson examples, the non-asymptotic equivalence between i.i.d. sampling model and the Poissonized model was due to Jiao, Kartik, Han and Weissman (2015). The tight bounds of the generalized uniformity testing problem were due to Batu and Canonne (2017) for constant , and Diakonikolas, Kane and Stewart (2018) for general , where their proof was greatly simplified here thanks to Theorem 4. For Shannon entropy estimation, the optimal sample complexity was obtained in Valiant and Valiant (2011), and the minimax risk was obtained independently in Jiao, Kartik, Han and Weissman (2015) and Wu and Yang (2016). For tools in approximation theory to establish Lemma 10 and 11, we refer to books Devore and Lorentz (1993), Ditzian and Totik (2012) for wonderful toolsets. 

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