Past exam of the options II course of the University of Oxford / 2012 / M / M5

A homogeneous Poisson process of rate $\lambda >0$ is an integer-valued counting process with $N_0=0$ almost surely, nondecreasing right-continuous paths with unit jumps, and independent stationary increments satisfyingIncrements over disjoint time intervals are independent. In particular,The finite Poisson counts ensure that there is no explosion in finite time. The increment formula also gives stochastic continuity, since the probability of any arrival in an interval of length $h$ is $1-e^{-\lambda h}\to 0$.

Since $N_3\sim \mathrm{Poisson}(3\lambda )$,

Stationary increments give $N_5-N_3\sim \mathrm{Poisson}(2\lambda )$. ThereforeThere is zero probability of an arrival at either deterministic endpoint, so endpoint inclusion does not alter the result.

Successive interarrival times $W_1=T_1$ and $W_2=T_2-T_1$ are independent exponential variables of rate $\lambda$. Their joint density is ${\lambda }^2{e}^{-\lambda (w_1+w_2)}$ for $w_1,w_2>0$. The transformation $t_1=w_1$, $t_2=w_1+w_2$ has Jacobian determinant one, givingIntegrating over $0<t_1<t_2<\infty$ gives one, as a joint density should.

Write $N_3=N_2+(N_3-N_2)$. Independent increments make $N_2$ independent of the second term. HenceThe overlapping time interval contributes the covariance; the extra unit of time contributes none.

After the third arrival, the future interarrival times are independent of the first three interarrival times and again generate a rate-$\lambda$ Poisson process. Conditional on $T_3=t<2$, exactly three arrivals have occurred by time $t$, and the number of further arrivals before two is Poisson with mean $\lambda (2-t)$. Consequently, in the regular conditional-distribution sense at $t=1$,Explicitly its conditional mass is zero for $k<3$ and isThis conditions on the third arrival time itself, rather than merely on the event that the third arrival has happened by time one.