Distribution of the product of an exponential and a uniform distribution I'm trying to show that $U(X+Y) = X$ in distribution, where X and Y are independent $\exp(\lambda)$ distributed and $U$ is uniformly distributed on (0,1) independent of $X+Y$.
I've been able to show that $X+Y$ has a $\Gamma(2, \lambda)$ distribution, but how do I calculate the distribution of this product? 
To clarify: The answer by Sasha works, but I was looking for a solution that does something like integrate over a suitable area.

One way to show equality in distribution is to establish point-wise equality of characteristic functions:
$$
    \phi_X(t) = \mathbb{E}\left(\mathrm{e}^{i t X} \right) = \frac{\lambda}{\lambda- i t}
$$
For the left-hand-side:
$$
   \mathbb{E}\left(\mathrm{e}^{i t U (X+Y)} \right) \stackrel{\text{tower law}}{=} \mathbb{E}\left(  \mathbb{E}\left( \mathrm{e}^{i t U (X+Y)} | U\right)  \right) \stackrel{\text{cond. indep.}}{=} \mathbb{E}\left(  \mathbb{E}\left( \mathrm{e}^{i t U X} | U\right)  \mathbb{E}\left( \mathrm{e}^{i t U Y} | U\right) \right) = \mathbb{E}\left( \left(\frac{\lambda}{\lambda - i U t}\right)^2\right) = \int_0^1 \left(\frac{\lambda}{\lambda - i u t}\right)^2 \mathrm{d} u = \left. \frac{\lambda}{i t} \frac{\lambda}{\lambda - i t u} \right|_0^1 = \frac{\lambda}{\lambda - i t}
$$
N.B.: Incidentally, your conclusion that $X+Y$ follows $\exp(2\lambda)$ is not correct. $X+Y$ follows gamma distribution $\Gamma(2,\lambda)$.
