Poisson process time of failure

Question

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?

Answer 1

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: $\mathbb{E}\left[T\mid K>k\right]= \mathbb{E}T_k + \mathbb{E}T.$

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

Qb Alternate Solution. Since $$\Phi_{T_{k}}(t)=\mathbb{P}(N_{t}\geq k)=1-\mathbb{P}(N_{t}<k)=1-\sum_{j=0}^{k-1}\mathbb{P}(N_{t}=j)=1-e^{-\lambda t}\sum_{j=0}^{k-1}\frac{\left(\lambda t\right)^{j}}{j!},$$ 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.