# Total expectation theorem for Poisson processes

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.

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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 $A$. – heropup May 18 '14 at 7:55
@heropup Thanks for the response. Given the exponential distribution of next arrival time, I am not sure why it shouldn't be $\frac {1}{\lambda_A}$. – user90476 May 18 '14 at 9:06

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: