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.
 A: 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.
A: \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.
