Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Discrete mathematics, continuous mathematics and probability / 8 / b

Given $X=k$, the score $Y$ has a binomial distribution with success probability $1/2$. Its conditional expected value and variance are $k/2$ and $k/4$. ThusThis is binomial thinning of the first-round count; the second formula uses the identity $\mathbb{E}[Y^2\mid X]=\mathrm{Var}(Y\mid X)+\mathbb{E}[Y\mid X{]}^2$.

The law of total expectation gives $\mathbb{E}Y=\mathbb{E}[X/2]=q/(2p)$. The law of total variance separates the two sources of randomness:Substitution givesThe binomial thinning of a geometric random variable identity supplies a further check: $Y$ itself has a zero-based geometric distribution with success probability $2p/(1+p)$.

Put $S=X+Y$. Given $X$, its conditional expected value is $3X/2$ and its conditional variance is $X/4$, since $X$ is then fixed. The law of total expectation and law of total variance giveThe two scores are dependent: the law of total covariance gives $\mathrm{Cov}(X,Y)=\mathrm{Cov}(X,X/2)=q/(2p^2)$. Adding their two variances without that covariance term would give an incorrect total.