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Why is it an exact simulation from $f$, and not only an approximation?

I get $\begin{split} P(X^*\in A) & = \sum_i^n P(X^*\in A , X^* = X_i)=\sum_i^n P(X^*\in A | X^* = X_i)P(X^* = X_i) \\ & = \sum_i^n P(X_i\in A)\frac{f(X_i)}{ng(X_i)} \\ & = \sum_i^n 1_{A}(X_i)\frac{f(X_i)}{ng(X_i)} \rightarrow^P E(1_{A}(X))=P(X \in A) \end{split}=$

An old man in the sea.
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1 Answers1

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Maybe this notation is better? I write $g(a)$ for the density of $X_i$. Also $P(X^* = X_i \mid X_i = a)$ is a kernel. It's measurable and integrable in $a$.

\begin{aligned} P(X^* \in A) &= \sum_i P(X^* \in A \text{ and } X^* = X_i) \\ &= \sum_i P(X_i \in A \text{ and } X^* = X_i) \\ &= \sum_i\int_{ A} g(a_i)P(X^* = X_i \mid X_i = a_i)da_i \quad{*} \\ &= \sum_i\int_{ A} g(a_i)\frac{f(a_i)}{ng(a_i)}da_i\\ &= \int_{ A} f(a)da. \end{aligned}

Also, this is exact because $\sum_i \frac{f(x_i)}{ng(x_i)} = 1$. A lot of times you have to use the self-normalized weights $\frac{f(x_j)}{g(x_j)}\left[\sum_i \frac{f(x_i)}{g(x_i)}\right]^{-1}$.

Taylor
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