# Conditional Probability on Joint Uniform Distribution

This is a very basic question, but I somehow manage to confuse myself all the time. So any help is greatly appreciated. Suppose we have two random variables $X$ and $Y$ with joint distribution function $f$ where $f$ is uniform. That is

$$f(x,y)= \begin{cases} 1 & \text{if 0\leq x\leq 1 and 0\leq y \leq 1}, \\ 0 & \text{elsewhere}. \end{cases}$$

If I want to determine the conditional probability $\Pr(Y>q\mid X=p)$ I get confused since

$$\Pr(Y>q\mid X=p)=\frac{\Pr((Y>q)\cap (X=p))}{\Pr(X=p)}$$ but then isn't the denominator a zero probability event? In general what is the answer of this conditional probability?

• In this particular case, $X$ and $Y$ are independent (do you see why?) so $P(Y>q \mid X=p) = P(Y > q)$. More generally, you can do a similar computation with densities: $f_{Y \mid X=p}(y) = f(p,y)/\int f(p,y')\mathop{dy'}$ but this requires some justification. Hopefully someone can give you a more concrete explanation below. May 12, 2016 at 16:58

the question on how to condition on a set of the form $\{X=p\}$ which has zero probability used to bug as well. It is defined via the conditional expectation given the random variable $X$.
Recall that for a random variable $Y$ and a measurable function $h$ such that $h(Y)$ is integrable the conditional expectation of $h(Y)$ given $X$, $\mathbb{E} [h(Y)| X]$, is a $\sigma(X)$-measurable random variable (with certain other properties). From the $\sigma(X)$-measurability we can deduce that there is a function $\psi$ (which depends on $h$ and $Y$) such that $$\mathbb{E} [h(Y)\mid X]= \psi(X).$$ Then we can define $$\mathbb{E} [h(Y)\mid X=p]:= \psi(p).$$ In your case, if we choose $h(Y)=1_{\{Y>q\}}$ we get that $$\mathbb{E} [h(Y)\mid X=p] = \mathbb P (Y>q\mid X=p) =\psi(p).$$
\begin{align}\mathsf P(Y>q\mid X=p) &= \int_{y>q} f_{\small Y\mid X}(y\mid p)~\mathsf d y\\[2ex]&= \dfrac{\int_{y>q}f_{\small X,Y}(p,y)\,\mathsf d y}{\int_{\Bbb R}f_{\small X,Y}(p,y)\,\mathrm d y}\\[2ex]&=\left(\dfrac{\int_q^1\mathrm d y}{\int_0^1\mathrm dy}\,\mathbf 1_{0\leqslant q\lt 1}+\mathbf 1_{1\leqslant q}\right)\,\mathbf 1_{0\leqslant p\leqslant 1} \\[2ex]&=\left((1-q)\,\mathbf 1_{0\leqslant q\lt 1}+\mathbf 1_{1\leqslant q}\right)\,\mathbf 1_{0\leqslant p\leqslant 1}\end{align}