Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 8

The probability generating function of a nonnegative integer-valued random variable isusing $s^0=1$ also at $s=0$. It is always defined for $0\le s\le 1$, and indeed the power series converges absolutely for all real or complex $s$ with $|s|\le 1$, becauseIt may be defined beyond the unit disk where this power series converges absolutely; such a larger domain depends on the distribution. In particular $G_X(0)=\mathbb{P}(X=0)$ and $G_X(1)=1$.

Let $S_n={\sum }_{i=1}^n{X}_i$. By independence, the expectation of a product of independent random variables factors, and the identical distributions giveThis is valid for $|s|\le 1$ and wherever the required absolute expectations exist. It is the probability generating function of an independent sum.

For $0\le s\le 1$, condition on $N=n$. Its independence from the $X_i$ preserves their independent common distribution, so part (ii) givesFor $n=0$, this is $1$, agreeing with $Y=0$. By the law of total expectation,This proves the probability generating function of an independent random sum. For nonnegative $s$ outside $[0,1]$, the same argument is valid whenever the right side is finite, by Tonelli's theorem for nonnegative sums. For signed or complex $s$, a sufficient absolute-convergence condition is $G_N(G_X(|s|))<\infty$; it holds automatically on $|s|\le 1$. Outside this domain, mere evaluation of an analytically extended expression is not a substitute for existence of the probability generating function.

Using the two masses of the Bernoulli distribution,This probability generating function is a polynomial, so it exists for every finite $s$.

Let $I_i$ indicate that student $i$ stays awake. These indicators have independent Bernoulli distributions with parameter $p$, and $W={\sum }_{i=1}^M{I}_i$. The Poisson distribution has probability generating functionThe probability generating function of an independent random sum now givesRecognizing the Poisson distribution,This is Poisson thinning.

For each student, the probability of both falling asleep and dreaming about mathematics is $(1-p)q$. These indicators are independent across students, so the same Poisson thinning calculation gives, for the requested count $D$,Explicitly $G_D(s)=\exp (\lambda (1-p)q(s-1))$, the probability generating function of that Poisson distribution.

Given $M=m$, there are exactly $m$ independent opportunities to stay awake, each with probability $p$. HenceThe standard binomial distribution formulas are $\mathbb{E}[W\mid M]=pM$ and $\mathrm{var}(W\mid M)=Mp(1-p)$. The Poisson distribution formulas are $\mathbb{E}[M]=\lambda$, $\mathrm{var}(M)=\lambda$, and therefore $\mathbb{E}[M^2]={\lambda }^2+\lambda$. By the law of total expectation,Using the definition of covariance,Thus $W$ and $M$ are not independent, even though the separate awake and asleep counts produced by Poisson thinning are independent.