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

For an integrable discrete random variable, the conditional expectation given an event $B$ of positive probability isThe sum ranges over the possible values of $X$; integrability ensures absolute convergence.

Using the indicator random variable, sum over the values of the discrete random variable $X$:Dividing by the positive probability $\mathbb{P}(B)$ proves $\mathbb{E}[X\mid B]=\mathbb{E}[X{1}_B]/\mathbb{P}(B)$. On $B^c$ the product vanishes, so no additional contribution is omitted.

The finite-partition form of the law of total expectation statesBecause the events form a partition of a sample space of the sample space, ${\sum }_j{1}_{B_j}=1$ pointwise. Multiply by $X$ and use linearity of expectation to obtainAn event of zero probability contributes zero to the middle expression and is omitted from the final expression, so no undefined conditional expectation is multiplied by zero.

Complete the square in the joint probability density function:The density therefore factors as the standard normal distribution density for $Y$ times a normal distribution density in $x$ with mean $\rho y$ and variance $1-{\rho }^2$. It follows that $\mathbb{E}[Y]=0$, $\mathbb{E}[Y^2]=1$, and $\mathbb{E}[X\mid Y]=\rho Y$. Symmetry gives $\mathbb{E}[X]=0$. By the law of total expectation,Consequently the covariance of this bivariate normal distribution is $\mathrm{Cov}(X,Y)=\rho$. This calculation derives the result from the given density.

The two captures form a uniform ordered sample of distinct birds, that is, sampling without replacement. By exchangeability, each of the ten birds has the same chance of occupying the second position, so the probability is $4/10=2/5$. Equivalently, conditioning on the first capture gives $(4/10)(3/9)+(6/10)(4/9)=2/5$.

Let $R_1,R_2$ be the events that the first and second birds are ringed. Under sampling without replacement,Thus the conditional probability is $\mathbb{P}(R_1\mid R_2)=\frac{2/15}{2/5}=\frac{1}{3}$. Alternatively, once a particular ringed second bird is specified, three of the nine possible first birds are ringed.

Let $M$ count the ringed birds among the two captures, and let $D$ be the event that the detector records at least one ring. The hypergeometric distribution gives the prior probabilitiesUnder the usual additional assumption of independent per-ring detection, with no false records, the likelihoods $\mathbb{P}(D\mid M=m)$ are $0$, $9/10$, and $1-(1/10{)}^2=99/100$. Therefore $\mathbb{P}(D)=12/25+33/250=153/250$. Applying Bayes' theorem yields $\mathbb{P}(M=0\mid D)=0,\mathbb{P}(M=1\mid D)=\frac{40}{51},\mathbb{P}(M=2\mid D)=\frac{11}{51}$. Storing only the last detected number means that a stored record does not identify how many rings were present. By label symmetry, the identity of that stored ring does not further alter this distribution.

The text specifies each ring's marginal detection probability, but does not explicitly specify independence of the detection errors. To keep this modelling qualification visible, let $\ell =\mathbb{P}(D\mid M=2)$. The marginal detection rate only implies $9/10\le \ell \le 1$. Without the extra independence assumption, the full family of possible posterior distributions isIn particular $1/5\le \mathbb{P}(M=2\mid D)\le 5/23$. The displayed numerical answer corresponds to $\ell =99/100$; the general answer exposes exactly what the detector's marginal specification alone establishes.