In a process where $\lambda$ = 3 with probability .3, and $\lambda$ = 5 with probability .7, what is the probability that $\lambda = 3$, given that at time $t=1, N(1) = 3$.
I have tried to use a combined $\lambda = 4.4$, but I have found that
$P(N(1)= 3, \lambda = 4.4)$ != $.3*P(N(1)=3,\lambda = 3) + .7*P(N(1) = 3, \lambda = 5)$
Which means I cannot use the typical definition of conditional probability.
We know that for a Poisson process with parameter $\lambda,$ $$\Pr(N(1)=3)=\frac{\lambda^3}6e^{-\lambda}$$ By Bayes' rule, $$\begin{align} \Pr(\lambda=3|N(1)=3)&= \frac{\Pr(N(1)=3|\lambda=3)\Pr(\lambda=3)}{\Pr(N(1)=3)}\\ &=\frac{.3\cdot\frac{3^3e^{-3}}6}{.3\cdot\frac{3^3e^{-3}}6+.7\cdot\frac{5^3e^{-5}}6}\\ &=\frac{81e^{-3}}{81e^{-3}+875e^{-5}}\\ &=\frac{81}{81+875e^{-2}}\\ &\approx.406181 \end{align}$$
When you took $\lambda=4.4$, I think you were thinking of the theorem that if $X_1$ and $X_2$ are independent Poisson random variables with parameters $\lambda_1$ and $\lambda_2$ respectively, then $X_1+X_2$ is a Poisson random variable with parameter $\lambda_1+\lambda_2$, but that's not the situation we have here. We have only one random variable, and there's no sum.