Finding an expression for the probability that one random variable is less than another, given a condition.
Let $X$ and $Y$ be two independent random variables, who's supports are $[0,\infty]$. We can express $\mathbb{P}[X<Y]$ as:
$$\mathbb{P}[X < Y] = \int_{y=0}^{\infty}\int_{x=0}^{y}P_{X}(x)P_{Y}(y)dxdy.$$
Can we find a similar expression for:
$$\mathbb{P}[X < Y|Y<k],\; \mathrm{given}\;k\in[0,\infty)?$$
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$\begingroup$You can find an expression for the desired conditional probability. Assume for simplicity that $X$ and $Y$ have respectively density functions $f_X(x)$ and $f_Y(y)$. Then $$\Pr(X\lt Y|Y\lt k)=\frac{\Pr((X\lt Y)\cap (Y\lt k)}{\Pr(Y\lt k))}.$$ Both numerator and denominator can be expressed as integrals. For the numerator, we want $\int_{y=0}^k\int_{x=0}^y f_X(x)f_Y(y)\,dx\,dy$. For the denominator, it is much the same, except that $x$ goes from $0$ to $\infty$.
Remarks: $1.$ The independence does not play a large role here, apart from (in concrete cases) making the integrations easier. For joint density functions $f_{X,Y}(x,y)$ the expression for the conditional probability is $$\frac{\int_{-\infty}^k\int_{-\infty}^y f_{X,Y}\,dx\,dy}{\int_{-\infty}^k\int_{-\infty}^\infty f_{X,Y}\,dx\,dy}.$$
$2.$ Note that as pointed out by @Jon Claus, the denominator is just the probability that $Y\lt k$, so it can be expressed in the simple form $\int_0^k f_Y(y)\,dy$.
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