Conditional probability for poisson random variables.

Question

I am trying to solve the following: $P(N(14)-N(10)=0|N(60)=50)$ where $N(t)\sim\mbox{Poisson}(\lambda t)$

Using Bayes Rule we get $\frac{P(N(14)-N(10)=0,N(60)=50)}{P(N(60)=50)}$

The denominator is straight forward using the pdf of a Poisson distribution.

I am trying to compute the numerator using the law of total probability, but I can't figure out how to set it up. Can anyone give me a hint or help?

I am thinking of setting $P(N(14)-N(10)=0,N(60)=50)=P(N(10)+N(46)=50) = P(N(10)=n,N(46)=50-n)=P(N(10=n)P(N(46)=50-n)$. Does this work?

Answer 1

Your idea is correct. It can be cleaned up a bit though. When $N(60)=50$, this means $0\leq N(10)\leq 50$. There are three independent Poissons:

• $N(10):=X\sim Poisson(10\lambda)$
• $N(14)-N(10):=Y \sim Poisson(4\lambda)$
• $N(60)-N(14):=Z \sim Poisson(46\lambda)$

Each term is $P(X=n)P(Y=0)P(Z=50-n)$. Add this up from $n=0$ to $n=50$.