Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics III / 5

For a nonnegative integer-valued random variable, its probability generating function isThe convention $s^0=1$, including at $s=0$, gives $G(0)=p(0)$. The power series has absolute convergence for every real $|s|\le 1$, since ${\sum }_{j}p(j)|s{|}^j\le 1$. More generally its real domain as a finite expected value is $\{s\in \mathbb{R}:{\sum }_{j}p(j)|s{|}^j<\infty \}$, which may extend beyond $[-1,1]$. No extension beyond that guaranteed interval is automatic.

Under the extra hypothesis that $G(s)$ exists for every real $s$, its power series has infinite radius of convergence. Termwise differentiation of a power series is therefore valid to every finite order. For $k\ge 1$,At $s=1$ this is the expected value of the falling factorial of $X$:In particular $\boxed{G^{\prime }(1)=\mathbb{E}[X]}$. The terms with $j<k$ vanish because their falling factorials contain a zero factor. These factorial moments are finite under the stated hypothesis.

Since $X^2=X(X-1)+X$, the second moment is $\mathbb{E}[X^2]=G^{\prime \prime }(1)+G^{\prime }(1)$. Subtracting the squared mean gives

For the specified Poisson distribution, the exponential series gives the probability generating functionEvery derivative of $G$ is again $e^{s-1}$, so the factorial moments and variance areAlso $\mathbb{E}[X]=1$.

Each of the $n$ distinct elements can receive any of the $x$ labelled colours. The multiplication principle therefore givescolour assignments; colours are allowed to remain unused.

Every assignment determines a unique set partition into its nonempty colour classes. If there are $k$ such blocks of a set partition, there are $A(n,k)$ choices for that set partition. For each fixed choice, assign distinct colours to the $k$ blocks of a set partition inways. Although the blocks of a set partition are unlabelled, they are distinguishable by their elements: ordering them temporarily, for instance by their least elements, counts each assignment exactly once. The falling factorial is zero when $k>x$, as it should be. Summing over the possible numbers of used colours provesHere $A(n,k)$ is the Stirling number of the second kind. Both sides are polynomials in $x$ and agree at infinitely many integers $x>2$, so the identity also holds for every real $x$, including $0,1,2$. This extension will justify substituting the random variable $X$ in part (iii).

Substitute $X$ into the polynomial identity and apply linearity of expectation to its finite sum. The factorial moments of the Poisson distribution in part (a)(iv) are all one, givingThus its $n$th moment counts set partitions: it equals the $n$th Bell number.

Computing the same expected value directly from the Poisson distribution instead givesFor each fixed positive integer $n$, the ratio test shows that this sum converges. This identity is Dobiński's formula. The case $n=0$ also gives $B(0)=1$ if $0^0$ is interpreted as one in this particular moment formula.