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N.B.: Sorry for cross-posting from https://stats.stackexchange.com/posts/347804/edit (I realized it was the wrong venue for the question, but couldn't find an easy way to transfer the question here).

So, consider a set of infintiely-differentiable convex functions $f_i: \mathcal X \rightarrow \mathbb R$, where $i$ varies from $1$ to $m$, and suppose we know all the moments of $f_i(x)$ (and of all the derivatives of $f_i$) for all $i$, as $x$ is sampled from some distribution $P$ on a Hilbert space $\mathcal X$.

Question:

What is a low-variance estimate for the quantities

$$\mathbb E_{x \sim P}\left[\frac{\exp(f_i(x))}{\sum_{j=1}^m\exp(f_j(x))}\right] ?$$

I mean to replace the integrands $\exp(...)/\sum_j \exp(...)$ with other random quantities of same expectation (or approx the same), but with smaller variance (the smaller the better).

Particular case: Affine functons. For simplicity, take $f_i(x) \equiv \langle a_i,x\rangle + b_i$, for some vectors $a_1,\ldots,a_m \in \mathcal X$ and scalars $b_1,\ldots,b_m \in \mathbb R$. Note that in this case, the above sought-for quantity can be rewritten in the form

$$\mathbb \nabla_{b_i} R(b),$$ where $R(b) := E_{x \sim P}\left[\log\left(\sum_{j=1}^m\exp(f_j(x))\right)\right]$. People in finance refer to $R$ as "logarithmic log-returns".

Important note: I should precise that i don't want Monte Carlo (or other black-box simulation technique). I need something more principled which exploits the structure of the problem..

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  • $\begingroup$ What is "a low-variance estimate"? $\endgroup$
    – fedja
    Commented May 28, 2018 at 16:33
  • $\begingroup$ I mean to replace the integrands $\exp(...)/\sum_j \exp(...)$ with another random quantity of same expectation (or approx the same), but with smaller variance. $\endgroup$
    – dohmatob
    Commented May 28, 2018 at 17:13
  • $\begingroup$ I'm still in the dark about what exactly is asked. If anybody can enlighten me, that would be great. $\endgroup$
    – fedja
    Commented Jun 4, 2018 at 1:24
  • $\begingroup$ @fedja I think the question is pretty clear as it stands. If you can say explictly where you're lost, I should be able to clarify... $\endgroup$
    – dohmatob
    Commented Jun 4, 2018 at 5:49
  • $\begingroup$ At the same sentence as before: "What is the low-variance estimate...?" Or, if you prefer, at "I seek to replace the integrands...". What are the rules of the game exactly? I doubt you want the answer "replace them with constants exactly equal to the corresponding expectations". $\endgroup$
    – fedja
    Commented Jun 4, 2018 at 6:19

1 Answer 1

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A way to approach your problem could be to consider the calculation of the expectation value as a Monte Carlo integration. Then you can use established techniques of variance reduction, as described for example in these lecture notes (1) or (2). There is a great variety of techniques, and for many there are ready-to-use software packages that implement the algorithm (for example in MatLab).

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  • $\begingroup$ I should precise in the question that i don't want Monte Carlo (or other black-box simulation technique). I need something more principled which exploits the structure of the problem... $\endgroup$
    – dohmatob
    Commented May 28, 2018 at 18:29
  • $\begingroup$ Well, you could use the "antithetic" variance reduction technique described in the linked references, which combines points $x$ where the integrand is small with those where it is large to achieve a variance reduction; this exploits the structure of the problem, doesn't it? $\endgroup$ Commented May 28, 2018 at 18:36
  • $\begingroup$ Not quite, as it doesn't use any information given in the problem. My best guess might be to resort to "control variates" cooked using the transformed variables $f_i(x)$). BTW, thanks to one of pdfs your referred has helped me identify that he problem is linked to something financial economists refer to as "logarithmic returns". So basically my problem is to produce a low variance estimate of the derivative of logarithmic returns. Rather unexpected connection! $\endgroup$
    – dohmatob
    Commented May 29, 2018 at 7:34

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