# Poisson Process.

The number of applications forms a Poisson process $$(N_t,t \geq 0)$$ of rate $$\lambda = 6$$. Let $$M_t$$ be the number bad applications in the time interval $$[0,t]$$ with rate equal to $$\frac{2}{3}$$. Arrivals of bad applications are independent. Considering possible values of $$N_t$$, compute the probability that $$P(M_t = m)$$.

• Are you saying each application has a probability of being bad of $\frac23$ or of $\frac{2/3}{6}=\frac19$? Jan 12 '20 at 22:54
• Some versions of this question have appeared here before. It's a standard exercise. Jan 12 '20 at 22:58

## 1 Answer

$$\newcommand{\e}{\operatorname{E}}$$ \begin{align} & \Pr(M_t=m) \\[8pt] = {} & \e(\Pr(M_t = m\mid N_t)) \\[10pt] = {} & \e\left( \binom{N_t} m \left( \frac 2 3 \right)^m \left( \frac 1 3 \right)^{N_t-m} \right) \\[8pt] = {} & 2^m \e\left( \binom {N_t} m \left( \frac 1 3 \right)^{N_t} \right) \\[8pt] = {} & 2^m \sum_{n=m}^\infty \binom n m \left( \frac 1 3 \right)^n \Pr(N_t=n) \\ & \text{(This starts at m, not at 0, because} \\ & \phantom{\text{(}} \text{the probability is 0 when n So the marginal distribution of $$M_t$$ is Poisson with expected value $$4t.$$