Past exam of the probability course of the University of Oxford / 2014 / 2

Integrating $e^{tx}$ against the Gamma density gives the moment-generating function of a gamma distribution:The change of variables $y=(\lambda -t)x$ reduces the integral to $\Gamma (r)=(r-1)!$ using the gamma function. For $t\ge \lambda$ the defining integral is infinite. In particular the moment-generating function is finite on a neighbourhood of zero.

Independence makes the moment-generating function of the sum the product of the two moment-generating functions. For $t<\lambda$,The expectations are finite on, for example, $(-\lambda /2,\lambda /2)$, as is the moment-generating function of the candidate Gamma distribution. The uniqueness theorem for moment-generating functions therefore applies, provingThis is additivity of independent gamma distributions with a common rate; a common rate is essential to this specific formula.

An exponential distribution of rate $\lambda$ is a Gamma distribution with shape one. Repeated additivity of independent gamma distributions with a common rate therefore givesWith integer shape this is also called an Erlang distribution.

The Poisson interarrival times of a homogeneous Poisson process are independent and identically distributed with the exponential distribution of rate $\lambda$. Consequently $Y_1,Y_2$ are independent random variables and their joint probability density isHere $Y_2$ is the second interarrival duration, rather than the time of the second arrival.

The inverse change of variables is $y_1=uv$, $y_2=(1-u)v$, on $0<u<1$, $v>0$. Its absolute Jacobian determinant isThe change of variables for probability densities therefore givesThis factorizes on a product domain into a unit uniform density in $u$ and a shape-two Gamma density in $v$. Thus $U$ and $V$ are independent, with $U\sim \mathrm{Uniform}(0,1)$ and $V\sim \mathrm{Gamma}(2,\lambda )$. This is the shape-$(1,1)$ case of beta-gamma independence.

By part (c) with $\lambda =1$, let $T_n=E_1+\cdots +E_n\sim \mathrm{Gamma}(n,1)$ for independent and identically distributed exponential random variables. The requested integral is $\mathbb{P}(T_n\le n)$. Each summand has expectation one and variance one, so the central limit theorem givesZero is a continuity point of the standard normal distribution, and consequently