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

The moment generating function is $M_X(\theta )=\mathbb{E}(e^{\theta X})$ where finite. The continuity theorem states that, if moment generating functions are finite on a common open interval containing zero and converge there to the moment generating function of a probability law, the corresponding distributions converge weakly to that law.

Integrating the gamma density and using the gamma integral givesFor a geometric waiting time, on $1,2,\dots$, summing the geometric series giveswhen $0<p<1$. For $p=1$, $Y=1$ deterministically. The waiting times between successive heads are independent and identically distributed, so $N=Y_1+\cdots +Y_k$ and $M_N=M_Y^k$. ThereforeIndeed the denominator is $p(1-2\theta )+O(p^2)$ and the numerator is $p+O(p^2)$. On any sufficiently small fixed neighbourhood of zero these moment generating functions exist for all sufficiently small $p$. The limit is the moment generating function of a gamma law with shape $k$ and rate $1/2$, so

Condition on $X_m=k$ and apply the Markov property and time homogeneity:Thus the transition matrices satisfy $P(m+n)=P(m)P(n)$; since $P(1)=P$ and $P(0)=I$, induction gives $P(n)=P^n$. Averaging over the initial state givesIf $\mu P=\mu$, induction gives $\mu P^n=\mu$, so this probability equals ${\mu }_i$ for every $n$. This is stationarity of the initial distribution.

A state $i$ is recurrent if the probability of returning to $i$ at a strictly positive time, starting at $i$, equals one. It is transient if that probability is less than one. If $i$ and $j$ communicate, choose $r,s$ with $p_{ji}(r)>0$ and $p_{ij}(s)>0$. Chapman-Kolmogorov givesIf $i$ is recurrent, summing over $n$ makes the right-hand side diverge, so the allowed recurrence criterion proves $j$ recurrent. Interchanging $i,j$ proves recurrence is shared throughout a communicating class, and therefore transience is shared too.

For the simple random walk, an odd number of steps cannot return to its starting point, andFor $p=1/2$ the return series diverges like $\sum n^{-1/2}$, proving recurrence. For $0<p<1$, $p\ne 1/2$, its tail is bounded by a convergent geometric series since $4p(1-p)<1$, proving transience. At $p=0$ or $1$ the walk is deterministic and never returns. Translation invariance transfers the conclusion to every state. Thus recurrence holds exactly when $p=1/2$.