Another question from Introducing Monte Carlo Methods with R by Robert and Casella.

Exercise 3.6 basically says the following. Suppose $f$ and $g$ are densities. Draw a random sample $X_1, \ldots, X_n \sim g$, then draw $(w_i \mid X_i) \sim Poisson( (f(X_i)/g(X_i) )$ for $i = 1, 2, \ldots, n$. Also define $X^*_i \mid w_i$ as iid draws from the categorical distribution $(X^*_i \mid w_i) \sim Categorical(\frac{w_1}{\sum_j w_j}, \ldots, \frac{w_n}{\sum_j w_j})$ for all $i$.

Based on the fact (which I've managed to show) that

$$ E_g\Big(w_i h(X_i)\Big)= E_f\Big(h(X_i)\Big),$$

we're supposed to "deduce that this sampling mechanism is marginally distributed from $f$". In other words, $X^{*}_{i} \sim f$.

I'm stuck. What I've tried is (pretend there is an $i$ subscript everywhere):

$$ \begin{aligned} \pi(x^*) &= \int_{-\infty}^{\infty} \sum_{w=0}^{\infty} \pi(x^* , w, x) dx\\ &= \int_{-\infty}^{\infty} \sum_{w=0}^{\infty} \pi(x)\pi(w \mid x)\pi(x^* \mid w , x) dx \\ &= \int_{-\infty}^{\infty} \sum_{w=0}^{\infty} g(x)\frac{\exp\Big(-\frac{f(x)}{g(x)}\Big) \Big(\frac{f(x)}{g(x)}\Big)^w}{w!} \frac{w}{\sum_j w_j} dx \end{aligned} $$

The sum is zero for $w=0$, so we change the bounds on the sum. After some manipulation, we get

$$ \begin{aligned} \pi(x^*) &= \int_{-\infty}^{\infty} \sum_{w=1}^{\infty} f(x)\frac{\exp\Big(-\frac{f(x)}{g(x)}\Big) \Big(\frac{f(x)}{g(x)}\Big)^{(w-1)}}{(w-1)!} \frac{1}{\sum_j w_j} dx \\ &= \int_{-\infty}^{\infty} \frac{f(x)}{\sum_j w_j} \sum_{w=1}^{\infty} \frac{\exp\Big(-\frac{f(x)}{g(x)}\Big) \Big(\frac{f(x)}{g(x)}\Big)^{(w-1)}}{(w-1)!}dx \\ &= \int_{-\infty}^{\infty} \frac{f(x)}{\sum_j w_j}dx \end{aligned} $$

which seems wrong to me. But I can't tell where I'm going wrong. Any help would be appreciated!

  • $\begingroup$ Given that this question is about one exercise you want help with, I would suggest you add the self-study tag. $\endgroup$
    – Xi'an
    Jun 2, 2017 at 19:48

1 Answer 1


Thank you for the question and sorry for being vague! The way the exercise is set is slightly different though:

Exercise 3.6. Given an importance sample $(X_i,f(X_i)/g(X_i))$, show that if $\omega_i$ has a Poisson distribution $\omega_i\sim\mathcal{P}(f(X_i)/g(X_i))$, the estimator $$ \frac{1}{n}\sum_{i=1}^n \omega_i h(x_i) $$ is unbiased. Deduce that the sample derived by this sampling mechanism is marginally distributed from $f$.

The first part of the question is about $$\mathbb{E}_g\Big(w_i h(X_i)\Big)= \mathbb{E}_f\Big(h(X_i)\Big)$$ as you gather. The second and vaguer part means that, if one takes the sample $$(\underbrace{X_1,\ldots,X_1}_\text{$\omega_1$ times},\ldots,\underbrace{X_n,\ldots,X_n}_\text{$\omega_n$ times})$$ then it behaves like a sample from $f$. Almost because $$\sum_{i=1}^n \omega_i \ne n$$ for most realisations,


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