Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 26H (Probability and Measure)

Put $\mu =\mathbb{E}[X]$ and ${\sigma }^2=\mathrm{Var}(X)$. Independence of $X$ and the uniform distribution variable $U$ givesAlso $\mathbb{E}[U^2]=1/3$, soTherefore

The map $(x,u)\mapsto (x,y)=(x,xu)$ has inverse $u=y/x$ and Jacobian magnitude $1/x$. Hence the density of $(X,Y)$ is

For $(y,z)=(xu,x(1-u))$, the inverse isand the Jacobian magnitude of $(y,z)\mapsto (x,u)$ is $1/(y+z)$. Thus the uniform split of a random total has joint densityIntegrating out the other piece yieldsThe transformation $U\mapsto 1-U$ preserves the uniform law, so $Z$ has the same density:

Because $h^{\prime }(x)=-g(x)/x$, the joint density from part (b) can be writtenThe pieces are therefore independent if and only ifThis condition is also sufficient because $h$ is the common marginal density.

Set $c=h(0)$. Taking $z=0$ gives the differential equationSince $h$ is a probability density, $c$ must be positive, and normalization givesConversely this function satisfies the displayed factorization. Thus the exponential characterization by a uniform random split shows thatEquivalently, $g(x)=-x{h}^{\prime }(x)=c^{2}x{e}^{-cx}$, so $X$ has the gamma distribution with shape two and rate $c$.