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Machine $1$ is currently working. Machine $2$ will be put in use at time $t$ from now. If the lifetime of machine $i$ is exponential with rate $\lambda_i=1,2$, what is the probability that machine $1$ is the first machine to fail?

I tried a few things but I can not get the answer

Let $X_1\sim exp(\lambda_1)$ and $X_2\sim exp(\lambda_2)$ maybe I'm wrong playing, but I think the exercise asks $$P(X_1<X_2|X_2=t)=\frac{P(X_1<X_2,X_2=t)}{P(X_2=t)}$$

I believe that times are independent of each machine, but I think I can not take it.

Can anyone give a hand?

EDIT: The corret answer is $$1-e^{-\lambda_1t}+e^{-\lambda_1t}\frac{\lambda_1}{\lambda_1+\lambda_2}$$

Roland
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  • what do you mean :"I believe that times are ind ...." ? – Cardinal Jul 19 '15 at 16:32
  • t is the life time or spent time ? – Cardinal Jul 19 '15 at 16:33
  • x representing spent time of RV X ? am I right ? – Cardinal Jul 19 '15 at 16:34
  • @Cardinal I do not know man, I typed the same as it was in the book, I think t is the time the machine 2 will operate, but not sure. – Roland Jul 19 '15 at 16:35
  • I think since the starting point for machain 2 is independent from machine 1, thus, they will be independent – Cardinal Jul 19 '15 at 16:36
  • @Cardinal I tried using it but did not get the correct answer. You can write your answer? – Roland Jul 19 '15 at 16:41
  • look it depend on definition , what about this: $$ P_r(X_1<t)$$ – Cardinal Jul 19 '15 at 16:42
  • Suppose (if we can) that the first machine starts just now. The desired probability could be $$P( X_1 - t \leq X_2 ) = P(X_1 \leq t) + P(X_1 > t, X_1 - t \leq X_2) $$the first probability is easy. For the second, using the independence between $X_1$ and $X_2$, $$P(X_1 > t, X_1 - t \leq X_2) = \mathbb{E}\Big{ \mathbb{E}\big[ 1_{(X_1 > t, X_1 - t \leq X_2) } \vert X_1 \big]\Big} $$

    $$= \mathbb{E}\Big{ 1_{(X_1 > t)} \cdot \big[ 1- F_2(X_1 - t) \big] \Big} $$ where $F_i$ is the c.d.f. of $X_i$, so $$= \int_t^\infty\big[ 1- F_2( \ell - t) \big] dF_1(\ell) $$

     – Chival Jul 19 '15 at 16:47
  • @Chival I did not really understand his reasoning, what is $l$? If you can write your idea in more detail, thank you. – Roland Jul 19 '15 at 16:54

1 Answers1

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I don't believe what you wrote is exactly what you want. Since $X_2$ starts at time $t$, it gets a total of time $t$ without failing for free. It seems like you should be looking for $\mathbb{P}(X_1 < X_2 + t) = \mathbb{P}(X_1 - X_2 < t)$. From this page, we know the pdf of $X_1 - X_2$ is $$ f(x) = \frac{\lambda_1\lambda_2}{\lambda_1 + \lambda_2}\left\{ \begin{array}{ll}e^{-\lambda_1x} &\text{ if }x > 0 \\ e^{\lambda_2x} &\text{ if }x < 0\end{array}\right. .$$

Integrating this from $-\infty$ to $t$ gives \begin{align} \mathbb{P}(X_1 - X_2 < t) &=\frac{\lambda_1\lambda_2}{\lambda_1 + \lambda_2} \left(\int\limits_{-\infty}^0 e^{\lambda_2 x}dx + \int\limits_{0}^t e^{-\lambda_1 x} dx\right) \\ &= \frac{\lambda_1\lambda_2}{\lambda_1 + \lambda_2}\left(\frac{1}{\lambda_2} + \frac{1}{\lambda_1} - \frac{1}{\lambda_1} e^{-\lambda_1 t} \right) \\ &= 1 - e^{-\lambda_1 t} + e^{-\lambda_1 t}\frac{\lambda_1}{\lambda_1 + \lambda_2}. \end{align}

Community
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Marcus M
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  • Sounds good, actually I did not understand well what is asked. – Roland Jul 19 '15 at 16:59
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    Just to explain a bit more: In words, the probability that you wrote down $\mathbb{P}(X_1 < X_2 | X_2 = t)$ is "the probability that machine $1$ fails before machine $2$ given that machine $2$ failed at time $t$. The problem was definitely awkwardly worded, though, so I understand the confusion. – Marcus M Jul 19 '15 at 17:04