Past exam of the probability course of the University of Oxford / 2015 / 1 / c

The Poisson process gives $N_k\sim \mathrm{Pois}(k)$, with expected value and variance both equal to $k$. Consequently $Y_k=(N_k-k)/\sqrt{k}$. Scaling and translating the argument in the moment-generating function of a Poisson distribution gives, for every real $t$,The centering term must remain inside the exponent: omitting it would leave a divergent first-order contribution.

For fixed real $t$, Taylor's theorem for the exponential function givesSubstituting into the previous moment-generating function,The continuity theorem for moment-generating functions states that, if the moment-generating functions of $Z_n$ and $Z$ are finite on a common open interval containing zero and $M_{Z_n}(t)\to M_Z(t)$ throughout that interval, then $Z_n\stackrel{\mathrm{d}}{\to }Z$. Here finiteness holds for all real $t$, and the limit is the moment-generating function of a standard normal variable. ThereforeThis proves the required instance of the Poisson central limit theorem directly from moment-generating functions; no appeal to a general central limit theorem is needed.

Write $A_k=N_{2k}-N_k$ and $B_k=N_k$. The stationary increments and independent increments of the Poisson process make $A_k$ and $B_k$ independent random variables, each with Poisson distribution of parameter $k$. ThusMultiplication of the two moment-generating functions givesThe odd terms cancel in Taylor's theorem, so $\log M_k(t)=t^2+O(k^{-1})$. The limit $e^{t^2}$ is the moment-generating function of a normal distribution with mean zero and variance two. The same continuity theorem for moment-generating functions therefore yieldsIn particular, the two equal-length increments must be separated before using independence: $N_{2k}$ and $N_k$ themselves are dependent.