Sum of independently distributed exponential random variables

Question

I got two independently distributed exponential random variables $X$, $Y$ with distribution $e^{\lambda}$.

How do we prove that the conditional distribution $X \vert U$ is a uniform distribution $U(0,u)$, i.e. $$f_{X \vert U = u} \sim U(0,u) $$

Answer 1

Use the Bayes Theorem, we have $$f(x|u)=\frac{f_U(u|x)\,f_X(x)}{f_U(u)}=\frac{e^{-\lambda(u-x) } e^{-\lambda x}}{u e^{(-\lambda u)}}$$ Note that $U = X + Y $ is an Erlang random variable with parameters $(2,\lambda)$. This gives us $$f(x \vert u) = \frac{1}{u} \delta(x \leq u)$$ where $\delta(x < u) = 1$ if $x < u$ else it is zero. This is so because the exponential distribution $f_U(u|x)$ is defined for $ u > x$, else it is zero. Consequently, we will have $$X \vert U \sim U(0,u)$$ So you must be asking for $X \vert U$ as @Robert Israel indicates. Honestly, I wouldn't have figured it out without that comment.