Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 8

The two outcomes are not independent. Conditional on the selected die, the rolls are independent, but sharing the unknown die creates statistical dependence. Each nonfavored face has probability $(1-1/2)/5=1/10$.

The unconditional marginal distribution assigns the following probability to a $1$ on either roll:However, by the law of total probability,Thus the joint probability does not factor into the two marginals. Conditional independence given a shared latent choice does not imply unconditional independence.

Let $D$ denote the selected die. Bayes' theorem givesOnly the first roll is conditioned on here; the second outcome has not been observed.

This is a Galton-Watson process with offspring count $0$ with probability $q=1-p$ and $2$ with probability $p$. The descendants of distinct initial individuals evolve independently, sofor every finite initial size $i$. Conditioning on the first individual's outcome and applying the law of total probability givesThe equation factors asIts roots are $1$ and $q/p$. Since $p<1/2$, the second root exceeds one and cannot be a probability. Therefore ${\alpha }_1=1$ and ${\alpha }_i=1$ for every finite initial size. The population reaches the absorbing state zero almost surely; no choice between two admissible fixed-point probabilities remains in this subcritical case.

If the initial individual has no offspring, $N=0$. If it splits, count that one event and the splitting events in its two independent descendant trees. Thuswhere $N_1,N_2$ are independent copies of $N$. Taking the probability generating function givesFor $0<s\le 1$, the quadratic equation has two algebraic branches. A probability generating function must remain finite at $s=0$, where it equals $\mathbb{P}(N=0)=q$, so the admissible branch isThe rationalized formula is also defined directly at zero. At $s=1$, $\sqrt{1-4pq}=q-p$, so $G_N(1)=1$, consistent with almost-sure finiteness. Extinction occurs after a finite number of generations almost surely, and each generation is finite, which justifies that finiteness claim for the number of splitting events in a binary branching process.

Differentiate the probability generating function equation:At $s=1$, $G_N(1)=1$, givingThe derivative is finite because $p<1/2$ and the explicit square root has no singularity at $s=1$. Alternatively, the expected population in generation $t$ is $(2p{)}^t$, so the expected number of splits in that generation is $p(2p{)}^t$. Summing this nonnegative geometric series gives the same answer without initially assuming a finite expected value.