# Let $X_t$ and $Y_t$ Poisson Process

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?

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".

For part d, here is an answer that avoids the direct calculus computation.

First, we compute what is the probability that a store 2 customer arrives before a store 1 customer arrives. Let $Z_1$, $Z_2$ be i.i.d. standard exponential. Then we are interested in

$P(\frac{1}{\lambda_1}Z_1 >\frac{1}{\lambda_2}Z_2)=P(\frac{Z_1}{Z_1+Z_2}>\frac{\lambda_1}{\lambda_1+\lambda_2})=\frac{\lambda_2}{\lambda_1+\lambda_2}$

where the last equality follows because $\frac{Z_1}{Z_1+Z_2} \sim Beta(1,1) \sim Unif$ (this is a well-known property of Gamma/Beta distributions; see the response by "guy" over here).

Now, if a store 2 customer arrives before a store 1 customer, the story ends and $X_T=0$. If a store 1 customer arrives before a store 2 customer, use the memoryless property of exponential and we reset the story after incrementing $X_T$ by 1.

Therefore, we see that $X_T$ counts the number of failures before a success, where the probability of success is $\frac{\lambda_2}{\lambda_1+\lambda_2}$. Thus,

$X_T \sim Geom(\frac{\lambda_2}{\lambda_1+\lambda_2})$