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The following is the problem I am solving for.

Let $X$ and $Y$ be the fail time for a machine with two components. The component $Y$ will start working if and only if component $X$ fails. The machine fails when component $Y$ fails as well. Given the joint density function

$$f_{X,Y}(x,y)=6e^{-x}e^{-2y}, \quad 0<x<y< \infty$$

find the expected time of failure of the machine.

My method was to find

$$\int_0^{\infty} \int_x^{\infty} (x+y)6e^{-x}e^{-2y} dydx$$

which ended up being equal to $ 7 \over 6$.

However, the answer was supposedly approximately 0.83.

I am not quite sure how to solve this problem.

Can someone explain me what is going on?

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hyg17
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    The expected time of failure of the machine is E(Y), not E(X+Y). – Did Sep 09 '14 at 08:09
  • Hello, Did. That's the part I am not understanding. If Component Y starts moving after component X does, then the time until complete failure would be X+Y, would it not? – hyg17 Sep 10 '14 at 18:26
  • No, it would not. Quote: "Let X and Y be the fail time for a machine with two components". Thus Y is the fail time of component Y, hence also the fail time of the machine itself. – Did Sep 10 '14 at 21:17
  • But Y starts after X, right? So if Y=2, for example, under the situation X=1 then the fail time would be 3. But if X=4 then the fail time would be 6. – hyg17 Sep 11 '14 at 03:27
  • @hyg17, Use the fact that$0\lt x\lt y\lt \infty$. There will never be a time when x=4 and y = 2. If Y = y, it will find the marginal distribution of Y by integrating from 0 to y. That is how you are taking care of the fact that fail time of X is factored in. Now for the machine to fail, it is only Y that you will have to take care. Expected time will be $\int y.f(y)$. – Satish Ramanathan Sep 11 '14 at 03:50
  • I guess I get it a bit more. I'm worried that the next time I see this I won't be able to understand it, but I kind of get this one at least. – hyg17 Sep 11 '14 at 09:23

1 Answers1

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Hint:

The answer is not approximately 0.83, but $\frac{5}{6}$. Use the following formulation $$\int_{0}^{\infty} \int_{0}^{y} (y)(6e^{-x}e^{-2y})dxdy$$. A tedious integration by parts and you will land up with the answer $\frac{5}{6}$

Satish Ramanathan
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