3
$\begingroup$

Note: This is a follow-up question inspired by a previous (more difficult) question I asked on MathOverflow.

Let $f:\mathbb{R}\to\mathbb{R}$ be a (sufficiently regular, e.g. smooth) probability density supported on the (compact) interval $[a,b]$. Let $\Vert\cdot\Vert$ be some norm (e.g. $L^2$, total variation). What is the best approximation to $f$ by a single, univariate Gaussian? In other words, solve $$ \text{argmin}_{m\in\mathbb{R},v^2\ge 0}\Vert f-g_{m,v^2}\Vert $$

where $g_{m,v^2}$ is the Gaussian PDF with mean $m$ and variance $v^2$. As I mention above, this question is related to a previous question; I realized while trying to solve that problem that even the simple case of a single Gaussian seems nontrivial.

Hint: I have verified numerically, for both $L^2$ and TV norm, that the answer is not $m=\mathbb{E}_{X\sim f} X$ or $v^2=\text{var}_{X\sim f} (X)$.

Note: The details of which norm (or even metric) or regularity assumptions are made are not important here. I am mostly curious to see if this computation can be carried out for any "reasonable" distance and regularity. In fact, even the case of minimizing over $m$ alone (holding $v^2$ fixed) appears difficult. Note that for the KL-divergence (which is neither a norm nor a metric), this can be solved in closed form quite easily.

$\endgroup$
  • 1
    $\begingroup$ The natural thing to try, for say $L^2$, is to differentiate (under the integral sign) with respect to $m$ and $v$, set them equal to zero, and try to solve. Does anything useful happen when you do that? $\endgroup$ – Nate Eldredge Mar 28 '19 at 16:39
  • $\begingroup$ @NateEldredge Yes, I have tried that and the resulting integral equation is daunting. $\endgroup$ – JohnA Mar 28 '19 at 16:41
  • 1
    $\begingroup$ You certainly shouldn't expect to get a closed-form solution to the equations (say for $L^2$), but numerical methods should work reasonably well. $\endgroup$ – Robert Israel Mar 28 '19 at 17:57
  • $\begingroup$ Among the other possibilities for norms to use: en.wikipedia.org/wiki/Statistical_distance#Examples $\endgroup$ – Matt F. Mar 28 '19 at 18:20
  • $\begingroup$ @MattF Absolutely. The reason I focused the question on norms (and $L^2$ in particular) is that I suspect this should be easier to solve in a Hilbert space. Any probability metric (e.g. Wasserstein, Hellinger) would be of interest as well. $\endgroup$ – JohnA Mar 28 '19 at 20:03
2
$\begingroup$

(i) You can try to minimize the squared Hellinger distance $$H(m,v)^2:=\frac12\,\int(\sqrt f-\sqrt{g_{m,v^2}})^2 =1-\int \sqrt f \sqrt{g_{m,v^2}}, $$ which amounts to the maximization of the comparatively simple expression $$J(m,v):=\int \sqrt{f(x)}\frac1{\sqrt v}\,\exp\Big\{-\frac{(x-m)^2}{4v^2}\Big\}dx $$ in $m,v$.

In particular, if $f$ is log concave, then the integrand in the latter integral is log concave in $(m,x)$. So, by the Prékopa--Leindler theorem (see e.g. Corollary 3.5), $J(m,v)$ is log concave in $m$ and hence will usually have a unique maximum in $m$. If, moreover, $f$ is symmetric about the midpoint $(a+b)/2$ of the interval $[a,b]$, then the maximum of $J(m,v)$ in $m$ is at $m=(a+b)/2$, and then it remains to maximize $J((a+b)/2,v)$ in $v>0$, which should usually be easy to do.

E.g., if $f$ is the density of the beta distribution Beta$(p,p)$ with parameters $\alpha=p$ and $\beta=p$ with $p\ge1$, then $J(m,v)$ is log concave in $m$ and its maximum in $m$ is at $m=1/2$. Here is a graph of the ratio of the maximizer of $J(1/2,v)$ in $v>0$ to the true standard deviation of the Beta$(p,p)$ distribution as the function of $p\ge1$:

enter image description here

We see that this ratio is pretty close to $1$ already for $p=1$.

(ii) Alternatively, you can try to minimize $$\int(\ln f-\ln{g_{m,v^2}})^2d\mu =\int\Big(\ln f(x)+\ln\sqrt{2\pi}+\ln v+\frac{(x-m)^2}{2v^2}\Big)^2\,\mu(dx) $$ for some measure $\mu$. Expanding the latter integrand and then integrating term-wise, we see that, for a given measure $\mu$, this reduces to the minimization in $v>0$ and $m$ of a linear combination of the form $$\sum_{j=1}^{15}c_j w_j(m,v), $$ where, for each $j$, $c_j$ is a known real number and $w_j(m,v)$ is a rather simple elementary function of $m,v$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Why the downvote? This approach seems to lead to a tractable case. Also, is the normalization in your maximization expression correct? $\endgroup$ – lcv Mar 28 '19 at 20:14
  • $\begingroup$ @lcv : Thank you for your comment. I have now corrected that normalization factor. $\endgroup$ – Iosif Pinelis Mar 28 '19 at 20:32
  • $\begingroup$ For i), would you suggest integration over $[a,b]$ or over $\mathbb{R}$? For ii), the only sensible bounds are $[a,b]$. $\endgroup$ – Matt F. Mar 30 '19 at 3:43
  • $\begingroup$ @MattF. : All the integrals are over $\mathbb R$. Note that in (ii) the integral is with respect to a measure $\mu$, to be appropriately chosen. $\endgroup$ – Iosif Pinelis Mar 31 '19 at 0:55
  • $\begingroup$ I have added some details. $\endgroup$ – Iosif Pinelis Apr 1 '19 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.