I have two independent Poisson processes $A$ and $B$ with arrival rates $\lambda_A$ and $\lambda_B$, respectively. Now, the expected time for the arrival of the next item for the merged process should be $\frac {1}{\lambda_A+\lambda_B}$.

Assuming $T_{A+B}$ to be the arrival time for the next item of the combined process, and $\{X=A\}$ or $\{X=B\}$ as the events that the items are from processes $A$ or $B$, using the law of total expectations, we get

$$ \begin{align} \mathbb{E}(T_{A+B}) &= \mathbb{E}( T_{A+B} \mid X =A )\mathbb{P}[X = A] + \mathbb{E}( T_{A+B}\mid X =B)\mathbb{P}[X = B]\\ &= \frac 1\lambda_A \frac {\lambda_A}{\lambda_A+\lambda_B} + \frac 1\lambda_B\frac {\lambda_B}{\lambda_A+\lambda_B} \\ &= \frac {2}{\lambda_A+\lambda_B} \end{align} $$ What am I doing wrong ? Thanks.

  • 3
    $\begingroup$ The issue seems to be that the conditional expectation ${\rm E}[T \mid X = A]$ is not $1/a$ once you know that the first arrival is from process $A$. $\endgroup$
    – heropup
    May 18, 2014 at 7:55
  • 2
    $\begingroup$ @heropup Thanks for the response. Given the exponential distribution of next arrival time, I am not sure why it shouldn't be $\frac {1}{\lambda_A}$. $\endgroup$
    – user90476
    May 18, 2014 at 9:06

2 Answers 2


heropup is right. The problem is that once you know that $X=A$, $X$ is not merely drawn from the exponential with rate $\lambda_A$ since you also know that the sampled value had to be small enough to win the comparison with the hypothetical sampled value from $B$.

So, the density given that $X=A$ is the renormalized pointwise product of the density of an exponential with rate $\lambda_A$ and the right cdf of an exponential with rate $\lambda_B$. This gives an exponential density with rate $\lambda_A + \lambda_B$. So:

\begin{align} \mathbb{E}(T_{A+B}) &= \mathbb{E}( T_{A+B} \mid X =A )\mathbb{P}[X = A] + \mathbb{E}( T_{A+B}\mid X =B)\mathbb{P}[X = B]\\ &= \frac 1{\lambda_A+\lambda_B} \frac {\lambda_A}{\lambda_A+\lambda_B} + \frac 1{\lambda_A+\lambda_B}\frac {\lambda_B}{\lambda_A+\lambda_B} \\ &= \frac {1}{\lambda_A+\lambda_B} \end{align} as desired.


\begin{align} & \Pr( T_{A+B} > t \mid X=A) \\[10pt] = {} & \frac{\Pr(T_{A+B} > t\ \&\ X=A)}{\Pr(X=A)} \\[10pt] = {} & \frac{\Pr(t < T_A < T_B)}{\Pr(X=A)}, \tag 1 \end{align} ${}$ \begin{align} \text{and } & \Pr(t < T_A < T_B) \\[10pt] = {} & \int_t^\infty \left( \int_u^\infty e^{-\lambda_A u} e^{-\lambda_B v} (\lambda_B\, dv) \right) (\lambda_A \, du) \\[10pt] = {} & \int_t^\infty e^{-\lambda_A u} e^{-\lambda_B u} (\lambda_A\,du) = e^{-(\lambda_A+\lambda_B)t} \cdot \frac{\lambda_A}{\lambda_A + \lambda_B}. \end{align} Therefore the expression on line $(1)$ is equal to $e^{-(\lambda_A + \lambda_B)t},$ which is the same as $\Pr(T_{A+B} > t).$

Thus the events $[T_{A+B} >t]$ and $[X=A]$ are actually independent.


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