0
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Do you want $\mathsf E(X_1\mid 0<X_1<1\mathrm s, 1\mathrm s< X_2)$ or $\mathsf E(X_1\mid 0<X_1<1\mathrm s, X_1<X_2)$ ? $\endgroup$ – Graham Kemp Oct 24 '17 at 2:36
  • 1
    $\begingroup$ "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? $\endgroup$ – Nap D. Lover Oct 24 '17 at 11:47
  • $\begingroup$ @Graham Yes, $E(X_{1}| 0<X_{1}<1s, X_{1}<X_{2})$ $\endgroup$ – Hallian1990 Oct 24 '17 at 18:10
1
$\begingroup$

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 $\lambda_2$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.