My question is whether I correctly derived the CDF of the steady-state age process. For context on why I' am asking this question is problem 6.1(a) from Introduction to Stochastic Processes by Lawler which I' am reading on my own. Here is the problem:

$\textbf{6.1}$ Suppose the lifetime of a component $T_i$ in hours is uniformly distributed on [100,200]. Components are replaced as soon as one fails and assume that this process has been going on long enough to reach equilibrium.

$\textbf{(a)}$ What is the probability that the current component has been in operation for at least 50 hours?

Here is what I think the answer should be in order to answer the question. The distribution function of $T_i$ is, $$F(t)=\frac{t-100}{100}$$

Then according the Lawler, the distribution function of the steady-state age distribution is

\begin{align} \psi_A(x)&=\lim_{t\rightarrow \infty}P(A_t\leq x) \\ &=\frac{1}{\mu}\int_{0}^{x}[1-F(s)]ds \\ &=\frac{1}{150}\int_{0}^{x}\left[1-\frac{s-100}{100}\right]ds \\ &=\frac{1}{150}[2s\Big|_0^{x}-\frac{s^2}{200}\Big|_0^{x}] \\ &=\frac{1}{150}[2x-\frac{x^2}{200}] \end{align}

The answer to the question would be $P(A\geq 50)=1-P(A\leq 50)=0.416$. But I do not think I derived the correct CDF for the age process. I say this because I would expect $\psi_A(200)=1$ because the maximum age is 200, however, using my derivation from above $\psi_A(200)=1.333$.


1 Answer 1


Since $T_i \sim \mathcal{U}[100,200]$ then the corresponding CDF is $$ F(t) = \begin{cases} 0 & \mbox{ if } t<100 \\ \frac{t-100}{100} & \mbox{ if } t \in [100,200] \\ 1 & \mbox{ if } t >200\end{cases}.$$ Therefore, the steady-state age distribution is $$ \psi_A(x) = \lim_{t \to \infty} P(A_t \leq x) = \frac{1}{\mu}\int_0^x (1-F(s))\mathrm{d}s= \begin{cases}\frac{x}{\mu} & \mbox{if } 0\leq x<100 \\ \frac{1}{\mu}\left(100 + \int_{100}^x\left(1-\frac{t-100}{100}\right) \mathrm{d}t \right) & \mbox{if } 100 \leq x< 200 \\ 1 & \mbox{if } x\geq 200\end{cases}. $$ Now you should be able to find your answer.


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