If $N(t)$ denotes the total number of visitors in the interval $[0,t]$. We suppose that $\lbrace N(t),t > 0 \rbrace $ is a Poisson process with rate $\lambda = 10$ per hour, and that we have $2$ types of visitors such that $80%$ of the visitors are of type $1$ (and $20%$ of type $2$).
I want to compute the mean and the variance of the number of visitors of type $1$, from a given time instant, before a second type $2$ visitor arrives.
I know that in this case:
- The number of visitors of type $1$ $\lbrace N_{1}(t),t > 0 \rbrace$ is a Poisson process with rate $\lambda = 8$ per hour
- The number of visitors of type $1$ $\lbrace N_{2}(t),t > 0 \rbrace$is a Poisson process with rate $\lambda = 2$ per hour
I tried to compute $ E(N_{1}(t)| N_{2}(t) = 0)$ but I'm not quite sure that this is what's required in this question.
Any help is appreciated, thanks.
Each visitor is of type 1 with probability $p=80\%$ and of type 2 with probability $1-p$ hence the number of visitors of type 1, from a given time instant, before a type 2 visitor arrives, is the number $N$ of heads before the first tail in a heads-or-tails game of probabilities $p$ and $1-p$.
This is the geometric distribution such that $P(N=n)=p^n(1-p)$, for every $n\geqslant0$, hence $E(N)$ and $\mathrm{var}(N)$ follow.
Thanks to the independence properties of Poisson processes, the number of visitors of type 1, from a given time instant, before the second type 2 visitor arrives, is the sum $M$ of two independent copies of $N$. In particular, one deduces $E(M)=2E(N)$ and $\mathrm{var}(M)=2\mathrm{var}(N)$.