Past exam of the probability course of the University of Oxford / 2015 / 2 / b / iii / Solution

For fixed $v>0$, the support inequalities become $v/2<u<v$. The marginal density is thereforeIt is nonnegative and integrates to $2-1=1$. Another interpretation is that $V=X+2Y$ sums independent exponential random variables of rates $1$ and $1/2$, so it has a hypoexponential distribution. The mean is $1+2=3$, which agrees with ${\int }_0^{\infty }v(e^{-v/2}-e^{-v})dv=4-1=3$.