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

The inverse linear transformation isThe positive support $x>0,y>0$ is exactly $u>0$ and $u<v<2u$. The independence and rate-one exponential distributions give $f_{X,Y}(x,y)=e^{-x-y}$ there. Since $x+y=u$, the change of variables for probability densities yieldsAlthough the density's value depends only on $u$, its support does not factor as a product of fixed intervals. Thus this expression does not imply independence of $U$ and $V$.

Integrating the joint probability density across $u<v<2u$ gives the marginal densityFor every $u>0$, the conditional probability density is consequentlyHenceThis also follows from two independent exponential variables conditioned on their sum: given $X+Y=u$, the variable $Y$ is uniform on $(0,u)$, and $V=u+Y$. Conditioning on $U=u$ here is defined through the conditional probability density, rather than division by $\mathbb{P}(U=u)$, which is zero. The endpoint $u=0$ lies outside the positive marginal support and needs no specified conditional law.

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 ${\int }_0^{\infty }v(e^{-v/2}-e^{-v})dv=4-1=3$.