0.13 Maximum likelihood estimation

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In the last lecture we derived a risk (MSE) bound for regression problems; i.e., select an $f\in \mathcal{F}$ so that $E\left[{\left(f\left(X\right)-Y\right)}^{2}\right]-E\left[{\left({f}^{*}\left(X\right)-Y\right)}^{2}\right]$ is small, where ${f}^{*}\left(x\right)=E\left[Y|X=x\right]$ . The result is summarized below.

Theorem

Complexity regularization with squared error loss

Let $\mathcal{X}={\mathbb{R}}^{d}$ , $\mathcal{Y}=\left[-b/2,b/2\right]$ , ${\left\{{X}_{i},{Y}_{i}\right\}}_{i=1}^{n}$ iid, ${P}_{XY}$ unknown, $\mathcal{F}$ = {collection of candidate functions},

$f:{\mathbb{R}}^{d}\to \mathcal{Y},\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}R\left(f\right)=E\left[{\left(f\left(X\right)-Y\right)}^{2}\right].$

Let $c\left(f\right)$ , $f\in \mathcal{F}$ , be positive numbers satisfying ${\sum }_{f\in \mathcal{F}}{2}^{-c\left(f\right)}\le 1$ , and select a function from $\mathcal{F}$ according to

${\stackrel{^}{f}}_{n}=argmin\left\{{\stackrel{^}{R}}_{n},\left(f\right),+,\frac{1}{ϵ},\phantom{\rule{0.166667em}{0ex}},\frac{c\left(f\right)log2}{n}\right\},$

with $ϵ\le \frac{3}{5{b}^{2}}$ and ${\stackrel{^}{R}}_{n}\left(f\right)=\frac{1}{n}{\sum }_{i=1}^{n}{\left(f\left({X}_{i}\right)-{Y}_{i}\right)}^{2}$ . Then,

$E\left[R\left({\stackrel{^}{f}}_{n}\right)\right]-R\left({f}^{*}\right)\phantom{\rule{4pt}{0ex}}\le \phantom{\rule{4pt}{0ex}}\left(\frac{1+\alpha }{1-\alpha }\right)\phantom{\rule{0.166667em}{0ex}}\underset{f\in \mathcal{F}}{min}\left\{R,\left(f\right),-,R,\left({f}^{*}\right),+,\frac{1}{ϵ},\phantom{\rule{0.166667em}{0ex}},\frac{c\left(f\right)log2}{n}\right\}+O\left({n}^{-1}\right)$

where $\alpha =\frac{ϵ{b}^{2}}{1-2{b}^{2}ϵ/3}$ .

Maximum likelihood estimation

The focus of this lecture is to consider another approach to learning based on maximum likelihood estimation. Consider the classical signalplus noise model:

${Y}_{i}=f\left(\frac{i}{n}\right)+{W}_{i},i=1,\cdots ,n$

where ${W}_{i}$ are iid zero-mean noises. Furthermore, assume that ${W}_{i}\sim P\left(w\right)$ for some known density $P\left(w\right)$ . Then

${Y}_{i}\sim P\left(y,-,f,\left(\frac{i}{n}\right)\right)\equiv {P}_{{f}_{i}}\left(y\right)$

since ${Y}_{i}-f\left(\frac{i}{n}\right)={W}_{i}$ .

A very common and useful loss function to consider is

${\stackrel{^}{R}}_{n}\left(f\right)=\frac{1}{n}\sum _{i=1}^{n}\left(-log{P}_{{f}_{i}}\left({Y}_{i}\right)\right).$

Minimizing ${\stackrel{^}{R}}_{n}$ with respect to $f$ is equivalent to maximizing

$\frac{1}{n}\sum _{i=1}^{n}log{P}_{{f}_{i}}\left({Y}_{i}\right)$

or

$\prod _{i=1}^{n}{P}_{{f}_{i}}\left({Y}_{i}\right).$

Thus, using the negative log-likelihood as a loss function leads to maximum likelihood estimation. If the ${W}_{i}$ are iid zero-mean Gaussian r.v.s then this is just the squared error loss weconsidered last time. If the ${W}_{i}$ are Laplacian distributed e.g. $P\left(w\right)\propto {e}^{-|w|}$ , then we obtain the absolute error, or ${L}_{1}$ , loss function. We can also handle non-additive models such as thePoisson model

${Y}_{i}\sim P\left(y|f,\left(i,/,n\right)\right)={e}^{-f\left(i/n\right)}\frac{{\left[f\left(i/n\right)\right]}^{y}}{y!}.$

In this case

$-logP\left({Y}_{i},|f,\left(i,/,n\right)\right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}f\left(i,/,n\right)-{Y}_{i}log\left(f,\left(i,/,n\right)\right)+\mathrm{constant}$

which is a very different loss function, but quite appropriate for many imaging problems.

Before we investigate maximum likelihood estimation for model selection, let's review some of the basic concepts. Let $\Theta$ denote a parameter space (e.g., $\Theta =R$ ), and assume we have observations

${Y}_{i}\stackrel{iid}{\sim }{P}_{{\theta }^{*}}\left(y\right),\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}i=1,\cdots ,n$

where ${\theta }^{*}\in \Theta$ is a parameter determining the density of the { ${Y}_{i}$ }. The ML estimator of ${\theta }^{*}$ is

$\begin{array}{ccc}\hfill {\stackrel{^}{\theta }}_{n}& =& arg\underset{\theta \in \Theta }{max}\prod _{i=1}^{n}{P}_{\theta }\left({Y}_{i}\right)\hfill \\ & =& arg\underset{\theta \in \Theta }{max}\sum _{i=1}^{n}log{P}_{\theta }\left({Y}_{i}\right)\hfill \\ & =& arg\underset{\theta \in \Theta }{min}\sum _{i=1}^{n}-log{P}_{\theta }\left({Y}_{i}\right).\hfill \end{array}$

$\stackrel{^}{\theta }$ maximizes the expected log-likelihood. To see this, let's compare the expected log-likelihood of ${\theta }^{*}$ with any other $\theta \in \Theta$ .

$\begin{array}{ccc}\hfill E\left[log{P}_{{\theta }^{*}}\left(Y\right)-log{P}_{\theta }\left(Y\right)\right]& =& E\left[log,\frac{{P}_{{\theta }^{*}}\left(Y\right)}{{P}_{\theta }\left(Y\right)}\right]\hfill \\ & =& \int log\frac{{P}_{{\theta }^{*}}\left(y\right)}{{P}_{\theta }\left(y\right)}{P}_{{\theta }^{*}}\left(y\right)dy\hfill \\ & =& K\left({P}_{\theta },{P}_{{\theta }^{*}}\right)\phantom{\rule{1.em}{0ex}}\text{the}\phantom{\rule{4.pt}{0ex}}\text{KL}\phantom{\rule{4.pt}{0ex}}\text{divergence}\hfill \\ & \ge & 0\phantom{\rule{1.em}{0ex}}\text{with}\phantom{\rule{4.pt}{0ex}}\text{equality}\phantom{\rule{4.pt}{0ex}}\text{iff}\phantom{\rule{0.166667em}{0ex}}{P}_{{\theta }^{*}}={P}_{\theta }.\hfill \end{array}$

Why?

$\begin{array}{ccc}\hfill -E\left[log,\frac{{P}_{{\theta }^{*}}\left(y\right)}{{P}_{\theta }\left(y\right)}\right]& =& E\left[log,\frac{{P}_{\theta }\left(y\right)}{{P}_{{\theta }^{*}}\left(y\right)}\right]\hfill \\ & \le & logE\left[\frac{{P}_{\theta }\left(y\right)}{{P}_{{\theta }^{*}}\left(y\right)}\right]\hfill \\ & =& log\int {P}_{\theta }\left(y\right)dy=0\hfill \\ & ⇒& K\left({P}_{\theta },{P}_{{\theta }^{*}}\right)\ge 0\hfill \end{array}$

On the other hand, since ${\stackrel{^}{\theta }}_{n}$ maximizes the likelihood over $\theta \in \Theta$ , we have

$\sum _{i=1}^{n}log\frac{{P}_{{\theta }^{*}}\left({Y}_{i}\right)}{{P}_{{\stackrel{^}{\theta }}_{n}}\left({Y}_{i}\right)}=\sum _{i=1}^{n}log{P}_{{\theta }^{*}}\left({Y}_{i}\right)-log{P}_{{\stackrel{^}{\theta }}_{n}}\left({Y}_{i}\right)\le 0.$

Therefore,

$\frac{1}{n}\sum _{i=1}^{n}log\frac{{P}_{{\theta }^{*}}\left({Y}_{i}\right)}{{P}_{{\stackrel{^}{\theta }}_{n}}\left({Y}_{i}\right)}-K\left({P}_{{\stackrel{^}{\theta }}_{n}},{P}_{{\theta }^{*}}\right)+K\left({P}_{{\stackrel{^}{\theta }}_{n}},{P}_{{\theta }^{*}}\right)\le 0$

or re-arranging

$K\left({P}_{{\stackrel{^}{\theta }}_{n}},{P}_{{\theta }^{*}}\right)\le \left|\frac{1}{n},\sum _{i=1}^{n},log,\frac{{P}_{{\theta }^{*}}\left({Y}_{i}\right)}{{P}_{{\stackrel{^}{\theta }}_{n}}\left({Y}_{i}\right)},-,K,\left({P}_{{\stackrel{^}{\theta }}_{n}},{P}_{{\theta }^{*}}\right)\right|.$

Notice that the quantity

$\frac{1}{n}\sum _{i=1}^{n}log\frac{{P}_{{\theta }^{*}}\left({Y}_{i}\right)}{{P}_{\theta }\left({Y}_{i}\right)}$

is an empirical average whose mean is $K\left({P}_{\theta },{P}_{{\theta }^{*}}\right)$ . By the law of large numbers, for each $\theta \in \Theta$ ,

$\left|\frac{1}{n},\sum _{i=1}^{n},log,\frac{{P}_{{\theta }^{*}}\left({Y}_{i}\right)}{{P}_{\theta }\left({Y}_{i}\right)},-,K,\left({P}_{\theta },{P}_{{\theta }^{*}}\right)\right|\stackrel{a.s.}{\to }0.$

If this also holds for the sequence $\left\{{\stackrel{^}{\theta }}_{n}\right\}$ , then we have

$K\left({P}_{{\stackrel{^}{\theta }}_{n}},{P}_{{\theta }^{*}}\right)\le \left|\frac{1}{n},\sum ,log,\frac{{P}_{{\theta }^{*}}\left({Y}_{i}\right)}{{P}_{{\stackrel{^}{\theta }}_{n}}\left({Y}_{i}\right)},-,K,\left({P}_{{\stackrel{^}{\theta }}_{n}},{P}_{{\theta }^{*}}\right)\right|\to 0\phantom{\rule{0.166667em}{0ex}}\mathrm{as}\phantom{\rule{0.166667em}{0ex}}n\to \infty$

which implies that

${P}_{{\stackrel{^}{\theta }}_{n}}\to {P}_{{\theta }^{*}}$

which often implies that

${\stackrel{^}{\theta }}_{n}\to {\theta }^{*}$

in some appropriate sense (e.g., point-wise or in norm).

Gaussian distributions

${P}_{{\theta }^{*}}\left(y\right)=\frac{1}{\sqrt{\pi }}{e}^{-{\left(y-{\theta }^{*}\right)}^{2}}$
$\Theta =\mathbb{R},\phantom{\rule{1.em}{0ex}}{\left\{{Y}_{i}\right\}}_{i=1}^{n}\stackrel{iid}{\sim }{P}_{{\theta }^{*}}\left(y\right)$
$\begin{array}{ccc}\hfill K\left({P}_{\theta },{P}_{{\theta }^{*}}\right)& =& \int log\frac{{P}_{{\theta }^{*}}\left(y\right)}{{P}_{\theta }\left(y\right)}{P}_{{\theta }^{*}}\left(y\right)dy\hfill \\ & =& \int \left[{\left(y-\theta \right)}^{2}-{\left(y-{\theta }^{*}\right)}^{2}\right]{P}_{{\theta }^{*}}\left(y\right)dy\hfill \\ & =& {E}_{{\theta }^{*}}\left[{\left(y-\theta \right)}^{2}\right]-{E}_{{\theta }^{*}}\left[{\left(y-{\theta }^{*}\right)}^{2}\right]\hfill \\ & =& {E}_{{\theta }^{*}}\left[{Y}^{2}-2Y\theta +{\theta }^{2}\right]-1/2\hfill \\ & =& {\left({\theta }^{*}\right)}^{2}+1/2-2{\theta }^{*}\theta +{\theta }^{2}-1/2\hfill \\ & =& {\left({\theta }^{*}-\theta \right)}^{2}\hfill \end{array}$
$⇒{\theta }^{*}\phantom{\rule{0.166667em}{0ex}}\text{maximizes}\phantom{\rule{0.166667em}{0ex}}E\left[log{P}_{\theta }\left(Y\right)\right]\phantom{\rule{0.166667em}{0ex}}\text{wrt}\phantom{\rule{0.166667em}{0ex}}\theta \in \Theta$
$\begin{array}{ccc}\hfill {\stackrel{^}{\theta }}_{n}& =& arg\underset{\theta }{max}\left\{-\sum {\left({Y}_{i}-\theta \right)}^{2}\right\}\hfill \\ & =& arg\underset{\theta }{min}\left\{\sum {\left({Y}_{i}-\theta \right)}^{2}\right\}\hfill \\ & =& \frac{1}{n}\sum _{i=1}^{n}{Y}_{i}\hfill \end{array}$

Hellinger distance

The KL divergence is not a distance function.

$K\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)\ne K\left({P}_{{\theta }_{2}},{P}_{{\theta }_{1}}\right)$

Therefore, it is often more convenient to work with the Hellinger metric,

$H\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)={\left(\int ,{\left({P}_{{\theta }_{1}}^{\frac{1}{2}},-,{P}_{{\theta }_{2}}^{\frac{1}{2}}\right)}^{2},d,y\right)}^{\frac{1}{2}}.$

The Hellinger metric is symmetric, non-negative and

$H\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)=H\left({P}_{{\theta }_{2}},{P}_{{\theta }_{1}}\right)$

and therefore it is a distance measure. Furthermore, the squared Hellinger distance lower bounds the KL divergence, so convergence in KL divergence implies convergence of the Hellinger distance.

Proposition 1

${H}^{2}\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)\le K\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)$

Proof:

$\begin{array}{ccc}\hfill H\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)& =& \int {\left(\sqrt{{P}_{{\theta }_{1}}\left(y\right)},-,\sqrt{{P}_{{\theta }_{2}}\left(y\right)}\right)}^{2}dy\hfill \\ & =& \int {P}_{{\theta }_{1}}\left(y\right)dy+\int {P}_{{\theta }_{2}}\left(y\right)dy-2\int \sqrt{{P}_{{\theta }_{1}}\left(y\right)}\sqrt{{P}_{{\theta }_{2}}\left(y\right)}dy\hfill \\ & =& 2-2\int \sqrt{{P}_{{\theta }_{1}}\left(y\right)}\sqrt{{P}_{{\theta }_{2}}\left(y\right)}dy,\phantom{\rule{1.em}{0ex}}\text{since}\phantom{\rule{0.166667em}{0ex}}\int {P}_{\theta }\left(y\right)dy=1\phantom{\rule{0.166667em}{0ex}}\forall \theta \hfill \\ & =& 2\left(1,-,{E}_{{\theta }_{2}},\left[\sqrt{{P}_{{\theta }_{1}}\left(Y\right)/{P}_{{\theta }_{2}}\left(Y\right)}\right]\right)\hfill \\ & \le & 2log\left({E}_{{\theta }_{2}},\left[\sqrt{{P}_{{\theta }_{2}}\left(Y\right)/{P}_{{\theta }_{1}}\left(Y\right)}\right]\right),\phantom{\rule{1.em}{0ex}}\text{since}\phantom{\rule{0.166667em}{0ex}}1-x\le -logx\hfill \\ & \le & 2{E}_{{\theta }_{2}}\left[log,\sqrt{{P}_{{\theta }_{2}}\left(Y\right)/{P}_{{\theta }_{1}}\left(Y\right)}\right],\phantom{\rule{1.em}{0ex}}\text{by}\phantom{\rule{4.pt}{0ex}}\text{Jensen's}\phantom{\rule{4.pt}{0ex}}\text{inequality}\hfill \\ & =& {E}_{{\theta }_{2}}\left[log,\left(,{P}_{{\theta }_{2}},\left(Y\right),/,{P}_{{\theta }_{1}},\left(Y\right),\right)\right]\phantom{\rule{4pt}{0ex}}\equiv \phantom{\rule{4pt}{0ex}}K\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)\hfill \end{array}$

Note that in the proof we also showed that

$H\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)=2\left(1,-,\int ,\sqrt{{P}_{{\theta }_{1}}\left(y\right)},\sqrt{{P}_{{\theta }_{2}}\left(y\right)},d,y\right)$

and using the fact $logx\le x-1$ again, we have

$H\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)\le -2log\left(\int ,\sqrt{{P}_{{\theta }_{1}}\left(y\right)},\sqrt{{P}_{{\theta }_{2}}\left(y\right)},d,y\right).$

The quantity inside the log is called the affinity between ${P}_{{\theta }_{1}}$ and ${P}_{{\theta }_{2}}$ :

$A\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)=\int \sqrt{{P}_{{\theta }_{1}}\left(y\right)}\sqrt{{P}_{{\theta }_{2}}\left(y\right)}dy.$

This is another measure of closeness between ${P}_{{\theta }_{1}}$ and ${P}_{{\theta }_{2}}$ .

Gaussian distributions

${P}_{\theta }\left(y\right)=\frac{1}{\pi }{e}^{-{\left(y-\theta \right)}^{2}}$
$\begin{array}{ccc}& & -2log\int \sqrt{{P}_{{\theta }_{1}}\left(y\right)}\sqrt{{P}_{{\theta }_{2}}\left(y\right)}dy\hfill \\ & =& -2log\int {\left(\frac{1}{\sqrt{\pi }},{e}^{-{\left(y-{\theta }_{1}\right)}^{2}}\right)}^{\frac{1}{2}}{\left(\frac{1}{\sqrt{\pi }},{e}^{-{\left(y-{\theta }_{2}\right)}^{2}}\right)}^{\frac{1}{2}}dy\hfill \\ & =& -2log\left(\int ,\frac{1}{\sqrt{\pi }},{e}^{-\left[\frac{{\left(y-{\theta }_{1}\right)}^{2}}{2},+,\frac{{\left(y-{\theta }_{2}\right)}^{2}}{2}\right]},d,y\right)\hfill \\ & =& -2log\left(\int ,\frac{1}{\sqrt{\pi }},{e}^{-\left[{\left(y,-,\left(,\frac{{\theta }_{1}+{\theta }_{2}}{2},\right)\right)}^{2},+,{\left(\frac{{\theta }_{1}-{\theta }_{2}}{2}\right)}^{2}\right]},d,y\right)\hfill \\ & =& -2log{e}^{-{\left(\frac{{\theta }_{1}-{\theta }_{2}}{2}\right)}^{2}}\hfill \\ & =& \frac{1}{2}{\left({\theta }_{1}-{\theta }_{2}\right)}^{2}\hfill \end{array}$
$\begin{array}{ccc}& ⇒& -2logA\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)=\frac{1}{2}{\left({\theta }_{1}-{\theta }_{2}\right)}^{2}\phantom{\rule{1.em}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}\text{Gaussian}\phantom{\rule{4.pt}{0ex}}\text{distributions}\hfill \\ & ⇒& H\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)\le \frac{1}{2}{\left({\theta }_{1}-{\theta }_{2}\right)}^{2}\phantom{\rule{1.em}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}\text{Gaussian}.\hfill \end{array}$

Poisson distributions

If ${P}_{\theta }\left(y\right)={e}^{-\theta }\frac{{\theta }^{y}}{y!},\theta \ge 0$ , then

$-2logA\left({P}_{{\theta }_{1}},{P}_{{\theta }_{2}}\right)={\left(\sqrt{{\theta }_{1}}-\sqrt{{\theta }_{2}}\right)}^{2}.$

Summary

${Y}_{i}\stackrel{iid}{\sim }{P}_{{\theta }^{*}}$
1. Maximum likelihood estimator maximizes the empirical average
$\frac{1}{n}\sum _{i=1}^{n}log{P}_{\theta }\left({Y}_{i}\right)$
(our empirical risk is negative log-likelihood)
2. ${\theta }^{*}$ maximizes the expectation
$E\left[\frac{1}{n},\sum _{i=1}^{n},log,{P}_{\theta },\left({Y}_{i}\right)\right]$
(the risk is the expected negative log-likelihood)
3. $\frac{1}{n}\sum _{i=1}^{n}log{P}_{\theta }\left({Y}_{i}\right)\stackrel{a.s.}{\to }E\left[\frac{1}{n},\sum _{i=1}^{n},log,{P}_{\theta },\left({Y}_{i}\right)\right]$
so we expect some sort of concentration of measure.
4. In particular, since
$\frac{1}{n}\sum _{i=1}^{n}log\frac{{P}_{{\theta }^{*}}\left({Y}_{i}\right)}{{P}_{\theta }\left({Y}_{i}\right)}\stackrel{a.s.}{\to }K\left({P}_{\theta },{P}_{{\theta }^{*}}\right)$
we might expect that $K\left({P}_{{\stackrel{^}{\theta }}_{n}},{P}_{{\theta }^{*}}\right)\to 0$ for the sequence of estimates ${\left\{{P}_{{\stackrel{^}{\theta }}_{n}}\right\}}_{n=1}^{\infty }$ .

So, the point is that maximum likelihood estimator is just a special case of a loss function in learning. Due to its special structure, weare naturally led to consider KL divergences, Hellinger distances, and Affinities.

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