Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Discrete mathematics, continuous mathematics, and probability / 9

For a nonnegative integer-valued random variable, its probability generating function is $G_X(s)=\mathbb{E}[s^X]={\sum }_{k\ge 0}\mathbb{P}(X=k)s^k$, at least for $|s|\le 1$. Insert the Poisson distribution masses and sum the exponential series:

Independence gives the probability generating function of the sum as the product:This is the probability generating function of the Poisson distribution with parameter $\mu +\lambda$. Uniqueness of the coefficients of a power series therefore gives $X+Y\sim \mathrm{Poisson}(\mu +\lambda )$. The product rule for these generating functions requires the stated independence.

Condition first on $X$. A Poisson distribution with parameter $k$ has conditional probability generating function $e^{k(s-1)}$, including the degenerate zero case when $k=0$. The law of total expectation givesFor $0\le s\le 1$ the inner argument is in $(0,1]$, so this use of the generating function is justified directly. The result is a compound Poisson distribution with a Poisson distribution of unit parameter for each jump, also called a Neyman Type A distribution in this case. It is generally not an ordinary Poisson distribution.

The conditional mean of a Poisson distribution equals its parameter, so $\mathbb{E}[Z\mid X]=X$. The law of total expectation givesAs a check, differentiating the probability generating function in part (iii) at one also gives $G_Z^{\prime }(1)=\mu$.

Let $R$ count red balls and let $h_r$ be the probability of eventual absorption at $R=0$ starting from $R=r$. The birth-death chain moves to $r-1$ with probability $r/4$ and to $r+1$ with probability $(4-r)/4$. Its absorbing states have $h_0=1$, $h_4=0$. First-step analysis givesSubstitution yields $h_2=1/2$, $h_1=5/8$ and $h_3=3/8$. The initial state has one red ball, so the probability that all balls end white is $5/8$.

Absorption occurs almost surely: from each interior state there is probability at least $1/4$ of absorption within the next two draws. The chance of surviving $2m$ draws is therefore at most $(3/4{)}^m$. This also justifies the finite expected times used in part (ii). The process is the absorbing four-ball Ehrenfest chain.

Let $t_r$ be the expected hitting time of either absorbing state. The boundary values are $t_0=t_4=0$. The colour symmetry gives $t_1=t_3$, and first-step analysis givesSolving gives $t_1=t_3=7$ and $t_2=8$. Thus the expected total number of draws from the stated initial condition is seven. The additive one counts the current draw. Finiteness follows from the uniform two-step absorption bound in part (i), so the first-step equations do not hide an assumption of finite expectation.