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{\bf Question}

If $N_1(t)$ and $N_2(t)$, $t \geq 0$, are two independent Poisson
processes with rates $\lambda_1$ and $\lambda_2$ respectively,
obtain an expression for $$P\left( N_1(t) = n_1 | N_1(t) + N_2(t)
= n\right),$$ where $n_1 \leq n,$ and comment on your result.



\vspace{.25in}

{\bf Answer}

$\ds P(N_1(t) = n_1 | N_1(t) + N_2(t) = n)$
\begin{eqnarray*} & = & \frac{P(N_1(t) = n_1 \, {\rm and} \, N_2(t) = n
- n_1)}{P(N_1(t) + N_2(t) = n)} \\&=& \frac{P(N_1(t) = n_1)\times
P(N_2(t) = n - n_1)}{P(N_1(t) + N_2(t) = n)} \\ & & (N_1,\ N_2\
{\rm are\ independent})\\ & = & \frac{\frac{e^{-\lambda_1 t
}(\lambda_1t)^{n_1}}{n_1!}\cdot \frac{e^{-\lambda_2 t
}(\lambda_2t)^{n-n_1}}{(n-n_1)!}}{\frac{e^{-(\lambda_1 +
\lambda_2)t}(\lambda _1 + \lambda_2)^n t^n}{n!}}\\ &&(N_1+N_2\
{\rm is\ Poisson\ with\ rate\ }\lambda_1+\lambda_2)\\
 & = & \frac{n!}{n_1!(n-n_1)!} \left( \frac{\lambda_1}{\lambda_1
+ \lambda_2} \right)^{n_1} \left( \frac{\lambda_2}{\lambda_1 +
\lambda_2} \right)^{n - n_1} \end{eqnarray*}

This is the binomial probability of $n_1$ successes in $n$ trials,
the probability of success being
$\ds\frac{\lambda_1}{\lambda_1+\lambda_2}$.

A Bernoulli trial is an event in the combined process, and a
success occurs if the event is from the first process $N_1.$



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