Past exam of the options II course of the University of Oxford / 2009 / M (Probability)

The moment-generating function is $M_X(\theta )=\mathbb{E}(e^{\theta X})$ for those real $\theta$ where this expectation is finite. When it exists on a neighbourhood of zero it determines the distribution, and its derivatives at zero give the corresponding moments.

A Poisson process of rate $\lambda >0$ is a counting process with $N_0=0$, independent stationary increments, and $N_t-N_s\sim \mathrm{Poisson}(\lambda (t-s))$ for $0\le s<t$, with the usual right-continuous unit-jump sample paths.

The interarrival times are independent and identically distributed with exponential distribution of rate $\lambda$: for any finite $k$ their joint density is ${\lambda }^k\exp (-\lambda {\sum }_{j=1}^k{x}_j)$ on $x_j>0$. In particular

Use independent colour marks, also independent of arrival times, as in the standard marked-process model. Let $K$ be the index of the first white arrival. Then $\mathbb{P}(K=k)=p(1-p{)}^{k-1}$ and $K$ is independent of the interarrival times. Thus $T={\sum }_{j=1}^K{X}_j$. Conditional on $K$, independence makes its moment-generating function a power of $M_X$, and summing the geometric series givesThis is the exponential distribution of rate $p\lambda$, so $T\sim \mathrm{Exp}(p\lambda )$. The calculation assumes $p>0$; if $p=0$, there is no white arrival and $T=+\infty$ almost surely.

Write $Z_t={\sum }_{j=1}^{N_t}{Y}_j$, with an empty sum equal to zero. Conditioning on the independent Poisson random variable $N_t$, the moment-generating function isThis is the compound Poisson distribution. Differentiating its log moment-generating function at zero, when the first two moments are finite, givesThe variance includes variability in the number of coaches, and so uses the second moment, not merely $\mathrm{Var}(Y)$.

Since $\mathbb{E}{N}_t=\mathrm{Var}(N_t)=\lambda t$, use the Poisson moment-generating function to get, for $t>0$,This is finite for every real $\theta$.

For each fixed $\theta$, the Taylor expansion of the exponential yieldsThe limiting moment-generating function $e^{{\theta }^2/2}$ is that of the standard normal distribution and is finite around zero. The moment-generating function continuity theorem therefore gives

For $i\notin A$, the event of ever reaching $A$ can be conditioned on the first step. Given $X_1=j$, its remaining hitting probability is $h_j$ by the Markov property, including $h_j=1$ if $j\in A$. The law of total probability thus givesFor $i\in A$ the definition uses time zero and gives $h_i=1$, so the harmonic equation is needed only outside $A$.

First assume $0<q<p<1$. Stop the nearest-neighbour walk on reaching zero or a finite upper level $L>i$. Its probability $h_i^{(L)}$ of reaching zero first solves $h_i^{(L)}=p{h}_{i+1}^{(L)}+q{h}_{i-1}^{(L)}$, with boundary values one and zero. Solving this second-order recurrence givesAbsorption in the finite interval occurs almost surely: from every interior state there is a uniformly positive probability of an absorbing run within $L$ steps. As $L\to \infty$, the events of hitting zero before $L$ increase to the event of ever hitting zero, since a path reaching zero at a finite time has a finite maximum before then. ConsequentlyThis limiting argument selects the actual hitting probability, rather than just an arbitrary bounded solution of the infinite recurrence. If $q=0$, the walk moves deterministically upwards from positive states and the probability is zero there; from state zero it is one. These agree with the formula interpreted separately at $i=0$.

After a first step to $+1$, the walk moves away from zero with probability $3/4$ and towards it with probability $1/4$, so the previous result gives return probability $1/3$. After a first step to $-1$, reflect the line: movement away from zero has probability $2/3$ and towards zero $1/3$, giving return probability $1/2$. Weighting by the first-step probabilities gives

All entries must lie between zero and one. The tightest restriction comes from $p_{01}=\lambda$, so together with the given positivity the parameter range is

Every finite upward path has positive probability, since each upward transition probability is positive. Every state $i\ge 1$ has a positive transition probability to zero because $\lambda /(i+1)<1$. From zero one can therefore reach every state, and from any state return to zero and then reach any other. Hence the chain is irreducible.

If $\lambda <1$, state zero has a positive self-loop, so its period is one. If $\lambda =1$, that self-loop disappears, but the paths $0\to 1\to 0$ and $0\to 1\to 2\to 0$ have positive probabilities and lengths two and three. Their greatest common divisor is one. Thus zero has period one in this boundary case too; irreducibility makes the whole chain aperiodic.

For $j\ge 1$, the only incoming transition is from $j-1$, so a stationary distribution must satisfyNormalization gives ${\pi }_0{e}^{\lambda }=1$, henceThe zero-state balance also holds: ${\sum }_i{\pi }_i(1-\lambda /(i+1))=1-{\sum }_{j\ge 1}{\pi }_j={\pi }_0$. Thus the Poisson distribution is indeed stationary, including when $\lambda =1$. Irreducibility gives uniqueness.