Past exam of the options I course of the University of Oxford / 2007 / M (Probability)

The unit disk has area $\pi$, so its joint density is $f_{X,Y}(x,y)={\pi }^{-1}{1}_{\{x^2+y^2\le 1\}}$. The polar-coordinate Jacobian has absolute determinant $r$. Thereforeand is zero outside this rectangle, up to irrelevant boundary choices. Integrating givesThe joint density equals the product of the two marginal densities, so $R$ and $\Theta$ are independent. The angle at the centre can be chosen arbitrarily since that point has probability zero.

A state $i$ is absorbing if $p_{ii}=1$, so once visited it is never left. Condition on the first step and use the Markov property: for $i\ne 1$, the subsequent probability of ever reaching state one is $h_j$ given $X_1=j$. Hence $h_i={\sum }_j{p}_{ij}{h}_j$, with $h_1=1$.

In this chain state four is a separate absorbing state, so $h_4=0$. The other two equations areSolving givesLet $t_i$ be the mean time to hit the absorbing set. From either transient state the next-step absorption probability is at least $1/3$, so its tail is geometrically bounded and the expectations are finite. First-step analysis gives $t_2=1+\frac{2}{3}{t}_3$ and $t_3=1+\frac{1}{4}{t}_2$. Thus $t_3=5/4+t_3/6$ and