Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Discrete mathematics, continuous mathematics and probability (resit) / 8 / b

For independent random variables, the probability generating function of an independent sum is the product of the individual probability generating functions. ThusExpanding this last expression as $e^{-(\lambda +\mu )}{\sum }_{r\ge 0}(\lambda +\mu {)}^r{s}^r/r!$ identifies each probability coefficient. Therefore

For $0\le k\le n$, the event $\{X=k,X+Y=n\}$ is $\{X=k,Y=n-k\}$. Use independence in its numerator and the Poisson distribution of the sum in its denominator:Outside $0\le k\le n$ the conditional probability is zero. ConsequentlyThis binomial distribution includes $n=0$, when it is concentrated at zero. Conditioning on the total converts the independent counts into an allocation of that total between the two rates.

For $n\ge 2$, $X_2+\cdots +X_n$ is independent of $X_1$ and has a Poisson distribution with parameter $(n-1)\lambda$. Part (ii) therefore givesIts probability at zero isFor $n=1$ the event being conditioned on is $X_1=1$, so the answer is $0$, also matching the displayed expression. The cancellation of $\lambda$ reflects the equal rates of all $n$ counts, not an approximation or large-$n$ limit.