1
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ 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. $\endgroup$
    – angryavian
    May 12, 2016 at 16:58

2 Answers 2

1
$\begingroup$

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 $$ \mathbb{E} [h(Y)\mid X=p] = \mathbb P (Y>q\mid X=p) =\psi(p). $$

$\endgroup$
0
$\begingroup$

You should use the probability density functions.

$$\begin{align}\mathsf P(Y>q\mid X=p) &= \int_{y>q} f_{\small Y\mid X}(y\mid p)~\mathsf d y\\[2ex]&= \dfrac{\int_{y>q}f_{\small X,Y}(p,y)\,\mathsf d y}{\int_{\Bbb R}f_{\small X,Y}(p,y)\,\mathrm d y}\\[2ex]&=\left(\dfrac{\int_q^1\mathrm d y}{\int_0^1\mathrm dy}\,\mathbf 1_{0\leqslant q\lt 1}+\mathbf 1_{1\leqslant q}\right)\,\mathbf 1_{0\leqslant p\leqslant 1} \\[2ex]&=\left((1-q)\,\mathbf 1_{0\leqslant q\lt 1}+\mathbf 1_{1\leqslant q}\right)\,\mathbf 1_{0\leqslant p\leqslant 1}\end{align}$$

$\endgroup$

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.