Past exam of the probability course of the University of Oxford / 2023 / 2 / c

Fix $0<c<1$ and choose $R_c$, independently of $X$, with the normal distributionThe linear combination of independent normal random variables $cX+R_c$ is normal. Its expected value is $c\mu +(1-c)\mu =\mu$ and its variance is $c^2{\sigma }^2+(1-c^2){\sigma }^2={\sigma }^2$. Hence $cX+R_c\stackrel{d}{=}X$ for every $c$, proving that every specified normal distribution is a self-decomposable distribution.

Fix $|t|<t_0$. Since $|ct|<t_0$, $0<M_X(ct)<\infty$. Using independence of random variables and the Tonelli theorem for the positive exponential integrands gives the extended-real identityThe left side is finite and the first factor is finite and strictly positive. Consequently $M_R(t)$ is finite andThe same reasoning applies to $R^{\prime }$, so their moment-generating functions are finite and equal throughout $(-t_0,t_0)$. The uniqueness theorem for moment-generating functions yieldsThis proves uniqueness of a self-decomposition remainder while also checking the required finiteness; no independence between $R$ and $R^{\prime }$ is needed.

For $X\sim \mathrm{Exp}(\lambda )$, integration gives $M_X(t)=\lambda /(\lambda -t)$ for $t<\lambda$. For $0<c<1$, the candidate remainder moment-generating function isThis is the moment-generating function of the mixture distributionConstruct it by taking an exponential random variable $E$ of rate $\lambda$ and a Bernoulli random variable $J$ with $\mathbb{P}(J=1)=1-c$, with $E,J,X$ mutually independent, and setting $R_c=JE$. Then $R_c$ is independent of $X$, andon a neighborhood of zero. By the uniqueness theorem for moment-generating functions, $cX+R_c\stackrel{d}{=}X$, so the exponential distribution is a self-decomposable distribution.