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Suppose that there are two brands of replacement components, Brand X and Brand Y, and that for political reasons a company buys a replacements of both types. When a Brand X component fails it is replaced with a new Brand Y component and vice-versa. The lifetimes (measured in thousands of hours) of Brand X components are uniform on [1,2] and the Brand Y components have lifetimes that are uniform on [1,3]. Answer the following questions for large time t.

a) What is the probability that the current component is Brand X?

b) What is the distribution of the age of the current component?

c) What is the distribution of the total lifetime of the current component?

d) Would these answers be different if instead of alternating the brands they used the rule that when a component fails they randomly choose a Brand X or Brand Y component with probability 1/2 for each?

This question is given from the section of Renewal Processes but its one of the challenging question. I'm trying to do these questions as practice for my exam and I would really appreciate it if someone could help me out.

Thanks in advance.

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  • $\begingroup$ Presumably it matters whether the first component is Brand X or Brand Y? It isn't clear which in the question. $\endgroup$ – Frank Apr 1 '14 at 21:37
  • $\begingroup$ @Frank The first component is Brand X. $\endgroup$ – user130344 Apr 1 '14 at 21:50
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    $\begingroup$ @Frank Don't think matters in the limit of large time which the question seems to imply is to be assumed $\endgroup$ – Kai Sikorski Apr 2 '14 at 5:01
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Here is to get you started. Let $N_t$ denote the number of X components used between times $0$ and $t$. The number of Y components used between times $0$ and $t$ is $N_t$ or $N_t-1$ or $N_t+1$. When $t\to\infty$, $N_t\to\infty$.

By the strong law of large numbers, the total length of the $n$ first X components is $nE[L_X]+o(n)$, where $L_X$ denotes the length of an X component. Likewise, the total length of the $n$ first Y components is $nE[L_Y]+o(n)$, where $L_Y$ denotes the length of a Y component, and the total length of the $2n$ first components is $n(E[L_X]+E[L_Y])+o(n)$.

This implies that, when $t\to\infty$, $N_t\cdot(E[L_X]+E[L_Y])\sim t$. The amount of time in $(0,t)$ when X components were in use is equivalent to $N_t\cdot E[L_X]$ hence the proportion of time when X components were in use converges to $$ \lim_{t\to\infty}\frac1tN_t\cdot E[L_X]=\frac{E[L_X]}{E[L_Y]+E[L_Y]}. $$ Under some aperiodicity conditions that are met here, this is the probability to observe an X component at time $t$ when $t\to\infty$.

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