$\newcommand{\vZ}{\boldsymbol{\mathbf{Z}}}$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:


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:

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

  • $\begingroup$ 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

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 $$ P(Z\in E|Z\in B)=\frac{P(Z\in E\cap Z\in B)}{P(Z\in B)}=\frac{P(Z\in E)}{P(Z\in B)}=\frac{V(E)/V(A)}{V(B)/V(A)}=\frac{V(E)}{V(B)}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.