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Suppose I have a probability distribution $p_\theta$ on $\Bbb R^n$ dependant on some parameters $\theta$. A natural problem is to evaluate the derivative of some expectation by the parameters: $$d_\theta\Bbb E_{p_\theta}[f]=d_\theta\int d^nx\, f(x)p_\theta(x)=\int d^nx\, f(x)\frac{d_\theta p_\theta(x)}{p_\theta(x)}p_\theta(x) =\Bbb E_{p_\theta}\left[f(x)\,d_\theta\ln(p_\theta(x)\right]$$ It is to be expected that the integrals are intractable, simply because integrals are generally hard. For example if $f$ is a complicated function this will not be easier than integrating $f$.

However if we are in a situation where we can sample from $p_\theta$ and evaluate the derivative of it, are there any barriers in getting an estimator for $d_\theta \Bbb E_{p_\theta}[f]$ by sampling the right hand side? What would be some standard literature about this ansatz?

I would expect yes, in part because this is then a way too simple solution for a problem that seems hard and in part because the logarithm and its derivative explodes for small arguments, so those places where $p_\theta$ is small and hard to access via sampling are given large weighting, necessitating a more samples to approximate the derivative than is possible.

s.harp
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  • FYI: it is possible to sample given the log-probabilities without leaving the log space, see: http://stats.stackexchange.com/questions/64081/how-do-i-sample-from-a-discrete-categorical-distribution-in-log-space/260248#260248 – Tim Apr 04 '17 at 12:45
  • @Tim What you have linked is interesting and highlights something I personally did not know. Its relevance to the question would be that normally a sampler samples from a "rounded down" distribution, where really small probabilities are zero, the log weighting makes these important however and here it is described how we can sample with these really small guys still being in the picture. But I think it does not address the question itself. Am I seeing this correctly? – s.harp Apr 04 '17 at 12:57

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