Let $X_1$ and $X_2$ be independent random variables with the exponential distribution and expected value equal to $1$. How to obtain distribution $\operatorname{Pr}\left(\dfrac{X_1}{X_2}|X_2=x_2\right)$?
I have evaluated $X_1/X_2$ distribution myself and it is $f_{X_1/X_2}(t)=\dfrac{1}{(1+t)^2}.$
Primarily I wanted to use the formula $$ f_{X|Y=y}=\dfrac{f_{XY}(x,y)}{f_Y(y)}, \ f_Y(y)\neq 0, $$ but the problem is in fact, that our variables ($X_1/X_2$ and $X_2$) aren't independent.
Any advices are going to be helpful for me. Thanks in advance.
There is a very useful rule in probability theory, which states that the conditional distribution of $f(X_1,X_2)$ conditioned on $X_2=x_2$ is simply the conditional distribution of $f(X_1,x_2)$ conditioned on $X_2=x_2$ for any measurable function $f$.
This means that $$\frac{X_1}{X_2} \: | \: X_2=x_2 \quad \sim \quad\frac{X_1}{x_2} \: | \: X_2=x_2 \quad \sim \quad \frac{X_1}{x_2}$$