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Fred’s beloved computer will last an $Expo(λ)$ amount of time until it has a malfunction. When that happens, Fred will try to get it fixed. With probability $p$, he will be able to get it fixed. If he is able to get it fixed, the computer is good as new again and will last an additional, independent $Expo(λ)$ amount of time until the next malfunction (when again he is able to get it fixed with probability p, and so on). If after any malfunction Fred is unable to get it fixed, he will buy a new computer. Find the expected amount of time until Fred buys a new computer. (Assume that the time spent on computer diagnosis, repair, and shopping is negligible.)

$T$~$Expo(λ)$; Let $X$ be the time untill he buys a new computer:

$E[X]=E[X|I_p=1]p+E[X|I_p=0]q$, where the first term in the right by meaning says that with prob. $p$ computer on average lasted $E[T]$ time untill it get broken +$E[X]$ after being repaired till the moment of being replaced.

But this logic is wrong. Can you give me a hint?

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Let $N$ be the number of malfunctions until Fred buys a new computer and $\{T_1,T_2,\ldots,T_N\}$ be the time intervals between malfunctions. Given $\mathbb{E}[T_i] = \dfrac{1}{\lambda}$ and $\mathbb{E}[N] = \dfrac{1}{1-p}$. The time to buy a new computer is $ X = \sum_{i=1}^N T_i$. Using Wald's lemma, the expected time is $\mathbb{E}[X] = \mathbb{E}[T]\mathbb{E}[N] = \dfrac{1}{\lambda(1-p)}$.

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Slick approach using Poisson process splitting/superposition:

The problem can be viewed as a Poisson process with rate $\lambda$, with each malfunction (arrival) being independently labeled "fixable" or "unfixable" with probability $p$ or $1-p$ respectively. By Poisson splitting, the fixable malfunctions form a Poisson process of rate $p\lambda$, and and the unfixable malfunctions form a Poisson process of rate $(1-p)\lambda$, and these two processes are independent. You are then asking for the expected time until the first unfixable malfunction, which is the expectation of an $\text{Expo}((1-p)\lambda)$ random variable, i.e. $\frac{1}{(1-p) \lambda}$.


Direct approach:

Let $T_1$ be the time until the first malfunction, and let $X$ be the time until the first unfixable malfunction. Let $I_1$ be the indicator for the event that the first malfunction is fixable.

$$E[X] = (1-p) E[X \mid I_1=0] + p E[X \mid I_1 = 1].$$ Note $E[X \mid I_1 = 0] = E[T_1]$. Note also that $E[X \mid I_1 = 1] = E[X] + E[T_1]$, since after this malfunction the process is identical to the original process. Thus $$E[X] = (1-p) E[T_1] + p(E[X] + E[T_1]).$$ Solving for $E[X]$ yields the same answer as above.

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Using your defined variables, we have $$E(X)=(1−p)E(T)+p(E(X)+E(T))$$

What this means is that, if the computer malfunctions and is not successfully fixed, the lifetime is simply the value of $T$. If the computer is fixed, time elapsed is that of $T$ and further time is reset so it's back to $E(X)$. Solving this is surely not an issue.

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