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$\newcommand{\vZ}{\boldsymbol{\mathbf{Z}}}$I am reading this paper regarding a simple proof of why rejection sampling works. I managed to understand the proof of Lemma 1, but I am struggling with the other two. The second lemma says:

Suppose the $m$-dimensional random variable $\mathbf{Z}$ has a uniform distribution in $A\subset\mathbb{R}^m$, where $0<V(A)<\infty$. Let $B\subset A, V(B) > 0$. Then the conditional distribution of $\mathbf{Z}$, given $\mathbf{Z}\in B$ is uniform in $B$.


The author says the proof is obvious, but I don't see how. Here is my attempt:

Proof

First of all, $\vZ$ has a uniform distribution on $A$ means $$ \newcommand{\vZ}{\boldsymbol{\mathbf{Z}}} \begin{equation} p(\vZ) = \begin{cases} \frac{1}{V(A)} & \text{if } \vZ\in A\\ 0 & \text{otherwise} \end{cases} \end{equation} $$ Now we want to find the conditional distribution of $\vZ$ given that $\vZ\in B$. $$ \begin{align} p(\vZ\mid \vZ\in B) = \frac{p(\vZ, \vZ\in B)}{p(\vZ\in B)} \end{align} $$ We can find the denominator as $$p(\vZ\in B) = \int_B p(\vZ) dA = \int_B \frac{1}{V(A)} dA = \frac{1}{V(A)}\int_B dA = \frac{V(B)}{V(A)}$$

At this point, I only need to find an expression for the numerator, however I since we know $\vZ\in B$, then certainly $$p(\vZ, \vZ\in B) = p(\vZ\in B) = \frac{V(B)}{V(A)}$$ so that we get $$p(\vZ \mid \vZ\in B) = \frac{V(B)}{V(A)} \times \frac{V(A)}{V(B)} = 1$$ However this is not the definition of being uniform in $B$. We should rather obtain something like:

$$ \begin{equation} p(\vZ\mid \vZ\in B) = \begin{cases} \frac{1}{V(B)} & \text{if } \vZ\in B\\ 0 & \text{otherwise} \end{cases} \end{equation} $$

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  • $\begingroup$ the key error is in the line $p(Z, Z\in B) = p(Z\in B)$. The RHS is certainly a probability, but the LHS is not. (Or, as a probability, its value is $0$.) PDFs can be tricky because they are not probabilities, but rather, densities. See if this wikipedia article helps... $\endgroup$ – antkam Apr 10 at 15:01
  • $\begingroup$ Do you mind telling me how I should change it? $\endgroup$ – Euler_Salter Apr 10 at 15:05
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Here is an equivalent definition of being uniformly distributed:

$Z$ is uniform on $A$ if for any (measurable) subset $E\subset A$, $$P(Z\in E)=V(E)/V(A).$$

To verify that $Z$ conditioned on $Z\in B$ is uniform on $B$, it therefore suffices to show that for any measurable $E\subset B$, that $P(Z\in E|Z\in B)=V(E)/V(B)$. Now $$ P(Z\in E|Z\in B)=\frac{P(Z\in E\cap Z\in B)}{P(Z\in B)}=\frac{P(Z\in E)}{P(Z\in B)}=\frac{V(E)/V(A)}{V(B)/V(A)}=\frac{V(E)}{V(B)}. $$

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