# Arrival time expectation value - Merged Poisson Process

Consider a merged Poisson process comprising two Independent Poisson processes, i.e., Process $1$ and Process $2$ with arrival rate $\lambda_{1}$ and $\lambda_{2}$, respectively. Consider the the Merged poisson processes for a time duration of $[0 - 1s]$

IF We assume that $\lambda_{1} << \lambda_{2}$. $\lambda_{1}$ arrival rate is too less than $\lambda_{2}$.

What is the expected value of the arrival time of the $FIRST$ arrival from Process $1$ within the given time period $[0 - 1s]$ before any arrival from Process $2$? It is like a conditional expectation value of the $FIRST$ arrival from Process $1$ before any arrival from Process $2$?

Hint: In general, for any poisson process with rate $\lambda$, If $T$ is the time elapsed until the $FIRST$ arrival, then we can say that $E[X]=1/\lambda$.

But, in this case, I am interested to calculate the expectation value of the arrival time of the $FIRST$ arrival from process $1$ in the given time period $[0-1s]$ and with a condition that no arrival occurred before from process $2$ within $[0-1s]$. If the X is the random variable for the time elapsed until this FIRST arrival from process $1$, then I can say, I need to calculate

$$E[X_{1} | 0<X<1s\ and \ No \ arrival \ from \ process \ 2]$$

• Do you want $\mathsf E(X_1\mid 0<X_1<1\mathrm s, 1\mathrm s< X_2)$ or $\mathsf E(X_1\mid 0<X_1<1\mathrm s, X_1<X_2)$ ? – Graham Kemp Oct 24 '17 at 2:36
• "If $T$ is the time elapsed until the FIRST arrival, then we can say that $E(X)=1/\lambda$"—where did $X$ come from and where did $T$ go? – Nap D. Lover Oct 24 '17 at 11:47
• @Graham Yes, $E(X_{1}| 0<X_{1}<1s, X_{1}<X_{2})$ – Hallian1990 Oct 24 '17 at 18:10

## 1 Answer

Units in seconds.

$\mathsf E(X_1\mid 0\leqslant X_1\leqslant 1, X_1\leqslant X_2) ~{= \dfrac{\int_0^1\int_s^\infty \lambda_1\lambda_2 s~e^{-\lambda_1s-\lambda_2t}\mathrm d t\mathrm d s}{\int_0^1\int_s^\infty \lambda_1\lambda_2 e^{-\lambda_1s-\lambda_2t}\mathrm d t\mathrm d s}\\ = \dfrac{\int_0^1 se^{-(\lambda_1+\lambda_2) s}\mathrm d s}{\int_0^1 e^{-(\lambda_1+\lambda_2) s}\mathrm d s} \\=\dfrac{1-(\lambda_1+\lambda_2+1)e^{\lambda_1+\lambda_2}}{(\lambda_1+\lambda_2)(1-e^{\lambda_1+\lambda_2})}\\= \dfrac{1}{\lambda_1+\lambda_2}-\dfrac{1}{e^{\lambda_1+\lambda_2}-1}}$

Then ${\left.\mathsf E(X_1\mid 0\leqslant X_1\leqslant 1, X_1\leqslant X_2)\right\lvert}_{\lambda_1\ll\lambda_2} \approx\dfrac{1}{\lambda_2}-\dfrac{1}{e^{\lambda_2}-1}$, which approaches zero for larger $\lambda_2$

• Thank You So Much @Graham. May I ask you another question here. – Hallian1990 Oct 30 '17 at 0:59