# Poisson process time of failure

Background

A typical Poisson process denoted by $$N_t$$ refers to the number of arrivals at time $$t$$, with rate $$\lambda$$, where

$$P(N_t = k) = \frac{e^{-\lambda t} (\lambda t)^k}{k!}$$

Question

A device is subject to shocks which occur according to a Poisson process $$N$$ with rate $$\lambda$$. The device can fail only due to a shock, and the probability that a given shock causes failure is $$p$$ independent of the number and times of previous shocks. Let $$K$$ be the total number of shocks the device takes before failure, and let $$T = T_K$$ be the time of failure.

a) Compute $$E[T]$$ and $$\operatorname{Var}(T)$$.

Usually $$E[T] = k/\lambda$$ and $$Var(T) = k/\lambda ^2$$, but I don't know if you need to factor in $$p$$ here.

b) Compute $$E[T\mid K]$$.

Not sure how to approach this. It seems that $$T$$ implies that $$K$$ occurred since it is given that $$T = T_K$$, so the conditional seems redundant.

c) Compute $$E[T\mid K>9]$$.

Again not sure how to approach this. Is $$K$$ Bernoulli?

Assumption. The failure probability is not $$p=0$$ or $$p=1$$; these cases can be handled separately.

Qa Hint. Let $$X_t\sim\operatorname{Poisson}(\lambda tp)$$. Then, $$\Phi_T(t)\equiv\mathbb{P}(T\leq t)=\mathbb{P}(X_t\geq1)=1-\mathbb{P}(X_t=0)$$.

Since $$\Phi_{T}(t)=1-e^{-\lambda tp}$$ for $$t \geq 0$$, $$T$$ is exponentially distributed with parameter $$\lambda p$$. Therefore, $$\mathbb{E}T=1/(\lambda p)$$ and $$\operatorname{Var}T=1/(\lambda p)^2$$.

Qb Hint. Use the memoryless property.

Let $$T_k$$ be time of the $$k$$-th shock. Then, $$\mathbb{E}[T\mid K=k]=\mathbb{E}[T\boldsymbol{1}_{\{K=k\}}]\,/\,\mathbb{P}(K=k)$$ so that $$\mathbb{E}[T\boldsymbol{1}_{\{K=k\}}]=\mathbb{E}[T_{k}\boldsymbol{1}_{\{K=k\}}] = \mathbb{E}T_k \mathbb{P}(K = k)$$ and $$\mathbb{E}[T\mid K=k]=\mathbb{E}T_k$$. Now, split the expectation into $$k$$ terms, each of which is the time it takes to experience a single shock: $$\mathbb{E}T_k = k\mathbb{E}T_1 = k/\lambda$$.

Qc Hint. Use the memoryless property.

Split the expectation up into two terms: the time it takes to experience $$k$$ non-failing shocks and the time it takes to fail after having experienced $$k$$ shocks: $$\mathbb{E}\left[T\mid K>k\right]= \mathbb{E}T_k + \mathbb{E}T.$$

It is also possible to solve the question without directly using the memoryless property.

Qb Alternate Solution. Since $$\Phi_{T_{k}}(t)=\mathbb{P}(N_{t}\geq k)=1-\mathbb{P}(N_{t} it follows that $$\Phi_{T_{k}}^{\prime}(t)=\lambda e^{-\lambda t}\sum_{j=0}^{k-1}\frac{\left(\lambda t\right)^{j}}{j!}-\lambda e^{-\lambda t}\sum_{j=0}^{k-2}\frac{\left(\lambda t\right)^{j}}{j!}=\frac{1}{t}\frac{e^{-\lambda t}\left(\lambda t\right)^{k}}{\left(k-1\right)!}.$$ Therefore, $$\mathbb{E}T_{k}=\int_{0}^{\infty}t\Phi_{T_{k}}^{\prime}(t)dt=\frac{k!}{\lambda\left(k-1\right)!}\int_{0}^{\infty}\frac{\lambda e^{-\lambda t}\left(\lambda t\right)^{k}}{k!}dt=\frac{k!}{\lambda\left(k-1\right)!}=\frac{k}{\lambda}.$$ Qc Alternate Solution. Use the fact that $$\mathbb{E}T=\mathbb{E}\left[T\boldsymbol{1}_{\{K\leq k\}}+T\boldsymbol{1}_{\{K>k\}}\right]=\mathbb{E}\left[T\boldsymbol{1}_{\{K\leq k\}}\right]+\mathbb{E}\left[T\boldsymbol{1}_{\{K>k\}}\right]$$ which implies $$\mathbb{E}\left[T \mid K > k\right] \mathbb{P}(K > k)=\mathbb{E}\left[T\boldsymbol{1}_{\{K>k\}}\right]=\mathbb{E}T-\mathbb{E}\left[T\boldsymbol{1}_{\{K\leq k\}}\right]=\mathbb{E}T-\sum_{j=1}^{k}\mathbb{E}T_{j}\mathbb{P}(K=j).$$ Now, do a bunch of algebra to arrive at the same solution as above.