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 / b

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.