# Conditioning uniform distribution on subset of support gives uniform distribution

$$\newcommand{\vZ}{\boldsymbol{\mathbf{Z}}}$$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 $$0. Let $$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 $$\newcommand{\vZ}{\boldsymbol{\mathbf{Z}}} $$p(\vZ) = \begin{cases} \frac{1}{V(A)} & \text{if } \vZ\in A\\ 0 & \text{otherwise} \end{cases}$$$$ Now we want to find the conditional distribution of $$\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:

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

• 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... – antkam Apr 10 at 15:01
• Do you mind telling me how I should change it? – Euler_Salter Apr 10 at 15:05

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