Let $X_t$ and $Y_t$ be two independent Poisson Process with rate
parameters $\lambda_1$ and $\lambda_2$ respectively, measuring the
number of customers arriving in stores 1 and 2.a)What is the probability that a customer arrives in store 1 before
any customers arrive in store 2?b)What is the probability that in the first your, a total of exactly
four customers have arrived at the two stores?c)Given that exactly four customers arrived at the two stores, what is
the probabity that all four went to store 1?d)Let $T$ denote the time of arrival of the first customer at store 2.
Then $X_T$ is the number of customers in store 1 at the time of first
customer arrival at store 2.Find the probability distribution of $X_T$
What I did
a)Let $T_{1i}$ and $T_{2i}$ the time of the ith customer arrives at the store 1 and 2 respectively, let $T=min(T_1,T_2)$ then we want
$$P(T_1=T)=\int_0^\infty P(T_2>t)dP(T_1=t)=\int_0^\infty e^{-\lambda_2 t}\lambda_1 e^{-\lambda_1 t}=\frac{\lambda_1}{\lambda_1+\lambda_2}$$
b) I know that $Z_t=X_t+Y_t\sim PP(\lambda_1+\lambda_2)$
$$P(Z_t=k)=\frac{e^{-(\lambda_1+\lambda_2)t}(\lambda_1+\lambda_2)^k}{k!}$$
then for $t=1$
$$P(Z=4)=\frac{e^{-(\lambda_1+\lambda_2)}(\lambda_1+\lambda_2)^4}{4!}$$
c)Using the same idea as in a) and given that the time between arrivals is independent
$$P=(\frac{\lambda_1}{\lambda_1+\lambda_2})^4$$
My ideas are right? How can I make the last item?
Best Answer
I think you have the first three answers right. For (d), we have $T\sim Exp(\lambda_2)$ so for $k=0,1,2,\ldots$,
\begin{eqnarray*} P(X_T=k) &=& \int_{t=0}^\infty P(X_t(t)=k)P(T=t)\; dt \\ &=& \int_{t=0}^\infty \left(\dfrac{(\lambda_1 t)^k}{k!} e^{-\lambda_1 t}\right) \left(\lambda_2 e^{-\lambda_2 t}\right)\; dt \\ &=& \dfrac{\lambda_1^k\lambda_2}{k!} \int_{t=0}^\infty t^k e^{-t(\lambda_1+\lambda_2)}\; dt \\ &=& \dfrac{\lambda_1^k\lambda_2}{k!} \left[\dfrac{-e^{-t(\lambda_1+\lambda_2)}}{(\lambda_1+\lambda_2)^{k+1}}\left(\sum_{j=0}^k \dfrac{k!}{j!} t^j(\lambda_1+\lambda_2)^j\right) \right]_{t=0}^\infty \\ &=& \dfrac{\lambda_1^k\lambda_2}{k!} \left[\dfrac{k!}{(\lambda_1+\lambda_2)^{k+1}} \right] \\ &=& \dfrac{\lambda_2}{\lambda_1+\lambda_2} \left(\dfrac{\lambda_1}{\lambda_1+\lambda_2} \right)^k. \end{eqnarray*}
To see the integration result (at line $4$) you might like to use Wolfram Alpha or other automated integration - or else by repeated "integration by parts".