Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 9 / a

For a Poisson distribution with parameter $\lambda >0$, the probability mass function isand is zero at all other values. Its normalization follows from the exponential series ${\sum }_{k\ge 0}{\lambda }^k/k!=e^{\lambda }$.

Subtract the first three values of the probability mass function of the Poisson distribution from one:

Let $X_1,X_2$ be the two independent Poisson random variables. For $k\ge 0$, their discrete convolution iswhere the last equality is the binomial theorem. Thus $X_1+X_2\sim \mathrm{Poisson}(\lambda +\mu )$.

By the preceding convolution, $Z=X_1+X_2$ has a Poisson distribution with parameter $\nu =\lambda +\mu$. Directly from its probability mass function, shifting the index in the absolutely convergent sums gives $\mathbb{E}[Z]=\nu$ and $\mathbb{E}[Z(Z-1)]={\nu }^2$. Since $Z^2=Z(Z-1)+Z$, $\mathbb{E}[(X_1+X_2{)}^2]=(\lambda +\mu {)}^2+(\lambda +\mu )$. This is also the identity $\mathbb{E}[Z^2]=\mathrm{Var}(Z)+(\mathbb{E}[Z]{)}^2$.

This is the ordinary factorial of $X$, rather than a falling-factorial moment. Substituting the probability mass function of the Poisson distribution gives the nonnegative seriesThe geometric series converges exactly when $0<\lambda <1$. Therefore $\mathbb{E}[X!]=\frac{e^{-\lambda }}{1-\lambda }$ for $0<\lambda <1$. For $\lambda \ge 1$, its terms fail to tend to zero and the expectation is infinite. This is the factorial of a Poisson random variable.