Past exam of the probability, statistics and options I course of the University of Oxford / 2010 / M / 1 (M1)

For an exponential distribution with rate $\lambda >0$, direct integration gives the moment-generating functionIt is finite in a neighborhood of zero, so its derivatives there give moments. Differentiating three times gives $M_X^{(3)}(t)=6\lambda /(\lambda -t{)}^4$, hence

Take $Y=a{X}_1+b$. From the definition of a moment-generating function, $\mathbb{E}{e}^{tY}=e^{bt}\mathbb{E}{e}^{at{X}_1}=e^{bt}{M}_1(at)$, for $t$ where the expectation is finite.

Take $Y=X_1+X_2$. Independence makes the joint exponential expectation factor:This is the product rule for the moment-generating function of an independent sum.

Take $Y=X_1+X_1^{\prime }$, where $X_1^{\prime }$ is an independent copy of $X_1$. The preceding factorization gives $\mathbb{E}{e}^{tY}=M_1(t{)}^2$. Using $2X_1$ would instead give $M_1(2t)$, so an independent copy is required.

Take $Y=U{X}_1$, where $U$ is uniform on $(0,1)$ and independent of $X_1$. Conditioning on $U$ givesFor $t$ near zero this is finite because the moment-generating function of $X_1$ is finite on a neighborhood of zero; Tonelli's theorem justifies the conditioning integral for the nonnegative exponential.

Recognize the mixture distributionIt is a half-and-half mixture of exponential distributions with rates one and two. Uniqueness of a moment-generating function finite near zero identifies the law. Therefore