Let $N(T_1)$ is Poisson counting mixture process and $M(T_2)$ is another independent Poisson counting mixture process, such that
$$ N(T_1) \sim (1-p) \cdot \operatorname{P}(\lambda_0T_1)+p \cdot \operatorname{P}(\lambda_1T_1),$$
$$ M(T_2) \sim (1-p) \cdot \operatorname{P}(\lambda_0T_2)+p \cdot \operatorname{P}(\lambda_1T_2),$$
where $0 <p<1$ is a fixed constant, and $T_1$ and $T_2$ are the lengths of intervals in which counting is done.
Question: Let $ U(T_3) \equiv N(T_3)+M(T_3), $
Processes $N$ and $M$ are added first and then counted for time $T_3$. $T_3$ is non over-lapping to $T_1$ and $T_2$.
How to evaluate mass function of conditional random process $U(T_3)\mid N(T_1)$?
Note that time interval $T_3$ is non overlapping to $T_1$ and $T_2.$
The probability that the parameter selected for $N(T)$, which i will call $\Lambda_{N}$, is equal to $\lambda_{0}$ is
$P(\Lambda_{N} = \lambda_{0}|N(T_{1}) = n) = \frac{P(N(T_{1})=n|\Lambda_{N} = \lambda_{0})(1 - p)}{P(N(T_{1})=n|\Lambda_{N} = \lambda_{0})(1 - p) + P(N(T_{1})=n|\Lambda_{N} = \lambda_{1})p}$
This is just Bayes Theorem.
From there we can get:
$$U(T_3)|N(T_1) \sim (1-p) \cdot P(\Lambda_{N} = \lambda_{0}|N(T_{1}) = n) \cdot \operatorname{Poiss}(2\lambda_0T_3)+ \\ (1-p) \cdot (1 - P(\Lambda_{N} = \lambda_{0}|N(T_{1}) = n)) \cdot \operatorname{Poiss}((\lambda_0 + \lambda_1) T_3)+ \\ \> p \cdot P(\Lambda_{N} = \lambda_{0}|N(T_{1}) = n) \cdot \operatorname{Poiss}((\lambda_0 + \lambda_1) T_3)+ p \cdot (1 - P(\Lambda_{N} = \lambda_{0}|N(T_{1}) = n)) \cdot \operatorname{Poiss}(2\lambda_1T_3)$$