Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / M (Probability)

The moment generating function is $M_X(t)=\mathbb{E}(e^{tX})$ wherever that expectation is finite. A useful continuity theorem says that if the moment generating functions $M_n$ are finite on a common open interval containing zero and converge pointwise there to the moment generating function of a probability distribution, then the corresponding variables converge in distribution to that law. Equivalently one may assume a finite limiting function on such a neighborhood, with the usual continuity at zero, and obtain the uniquely determined limiting law.

For the gamma density, integrating the exponential against the density and changing variable to $(\lambda -t)x$ givesFor the waiting time to one head, independence of successive tosses gives $\mathrm{Pr}(Y=j)=p(1-p{)}^{j-1}$ for $j\ge 1$. Summing the geometric series givesThe waiting times between consecutive heads are independent copies of $Y$, so their sum $N$ has moment generating function $M_Y(t{)}^k$. ConsequentlyFor fixed $t<1/2$, numerator and denominator have leading terms $p$ and $p(1-2t)$ as $p\downarrow 0$. A fixed neighborhood of zero is inside the convergence domain for all sufficiently small $p$, so the continuity theorem applies. The limiting distribution isequivalently a chi-squared distribution with $2k$ degrees of freedom.

Condition on the intermediate state at time $m$. The Markov property and time homogeneity giveThese nonnegative sums are valid on a countable state space. Starting from $P(0)=I$, $P(1)=P$, induction gives $\boxed{P(n)=P^n}$. Conditioning on the initial state now yields $\boxed{\mathrm{Pr}(X_n=i)=(\mu P^n{)}_i}$. If $\mu P=\mu$, induction gives $\mu P^n=\mu$ for every $n$, so all one-time distributions equal the stationary initial distribution.

A state is recurrent if, starting there, the probability of returning at some strictly positive time is one. It is transient if that return probability is strictly less than one. Suppose $i,j$ communicate. Choose $r,s$ with $p_{ji}(r)>0$ and $p_{ij}(s)>0$. Chapman–Kolmogorov givesIf $i$ is recurrent, summing over $n$ and using the supplied return-series criterion makes the return series for $j$ diverge, so $j$ is recurrent. Interchanging the states proves equivalence; therefore transience is also constant on a communicating class.

For the nearest-neighbor walk, return in an odd number of steps is impossible. A return in $2n$ steps requires exactly $n$ steps each way, soAt $p=1/2$, the return series diverges like $\sum n^{-1/2}$. If $0<p<1$ and $p\ne 1/2$, then $4p(1-p)<1$, so it converges by exponential decay. At $p=0$ or one the walk never returns. Hence $\boxed{\text{recurrence occurs exactly at }p=1/2}$.