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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 $P(M_t = m)$.

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  • $\begingroup$ Are you saying each application has a probability of being bad of $\frac23$ or of $\frac{2/3}{6}=\frac19$? $\endgroup$
    – Henry
    Jan 12 '20 at 22:54
  • $\begingroup$ Some versions of this question have appeared here before. It's a standard exercise. $\endgroup$ Jan 12 '20 at 22:58
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$\newcommand{\e}{\operatorname{E}}$ \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 $n<m$.)} \\[8pt] = {} & 2^m \sum_{n=m}^\infty \binom n m \left( \frac 1 3 \right)^n \frac{e^{-6t} (6t)^n}{n!} \\[8pt] = {} & \frac{2^m (2t)^m e^{-6t}}{m!} \sum_{n=m}^\infty \frac{(2t)^{n-m}}{(n-m)!} \\[8pt] = {} & \frac{(4t)^m e^{-6t}}{m!} \sum_{k=0}^\infty \frac{(2t)^k}{k!} \\[8pt] = {} & \frac{(4t)^m e^{-6t}}{m!} \cdot e^{2t} \\[8pt] = {} & \frac{(4t)^m e^{-4t}}{m!} \end{align} So the marginal distribution of $M_t$ is Poisson with expected value $4t.$

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