Conditional probability in continuous jointly distributed random variables.

Question

This is a follow up question with respect to the question here. Quickly summarising the question and the answer, if $X \sim U(0, S)$ and $Y \sim U(0, X)$, the joint cumulative probability distribution function in the case when $0 \leq y \leq x \leq S$ is given by $P(X \leq x, Y \leq y) = \frac{x}{S} + \frac{y}{S}(\ln x - \ln y)$.

I need the conditional probability $P(X < x | Y = y)$. I am unable to proceed thinking P(Y = y) is zero since it is point probability. How does one formulate this ?

Thanks in advacne.

Answer 1

First, for $0<y< x< S$ and $\epsilon\in(y,x-y)$, $$ \mathsf{P}(X\le x,Y\in[y-\epsilon,y+\epsilon])=\int_0^x\mathsf{P}(Y\in[y-\epsilon,y+\epsilon]\mid X=z)f_X(z)\,dz\\ =\frac{1}{S}\left(2(1+\ln(x))+\ln(y-\epsilon)(y-\epsilon)-\ln(y+\epsilon)(y+\epsilon)\right), $$ and, similarly, $$ \mathsf{P}(Y\in[y-\epsilon,y+\epsilon])=\int_0^S \mathsf{P}(Y\in[y-\epsilon,y+\epsilon]\mid X=z)f_X(z)\,dz \\ =\frac{1}{S}\left(2(1+\ln(S))+\ln(y-\epsilon)(y-\epsilon)-\ln(y+\epsilon)(y+\epsilon)\right). $$ Then for $0<y<x$, $$ \mathsf{P}(X\le x\mid Y=y)=\lim_{\epsilon\downarrow0}\frac{\mathsf{P}(X\le x,Y\in[y-\epsilon,y+\epsilon])}{\mathsf{P}(Y\in[y-\epsilon,y+\epsilon])}=\frac{\ln(x/y)}{\ln(S/y)}. $$ Note that $\mathsf{P}(X\le x\mid Y=y)=0$ when $y\ge x$.