Average Length of Life of Two Exponentially Distributed Components

[kjartan]

1. Suppose that two electronic components in the guidance system for a missile operate independently and that each has a length of life governed by the exponential distribution with mean 1 (with measurements in hundreds of hours).

(a) Find the probability density function for the average length of life of the two components.

(b) Find the mean and variance.
2. f(y) = e^(-y) for y>=0
For (a), I obtained, f(u) = u*exp(-u) for u>=0. Still an exponential distribution.
The back of the text had f(u) = 4u*exp(-2u), where f has a gamma distribution with α=2, β=1/2.

Thanks for any and all help!

 

[vela]

How'd you come up with your result of f(u) = ue^-u?

 

[kjartan]

Thanks for taking a look!
What I did was this.

f(y1) = e^(-y1), and f(y2) = e^(-y2), since the general form for an exponentially distributed random variable is f(y) = (1/$$\beta$$)*e^(-y/$$\beta$$). Where the mean is beta, and we are given that the mean is 1.

Then, since y1 & y2 are independent, f(y1,y2) = f(y1)*f(y2).

Thus, e^(-y1)*e^(-y2) = e^(-(y1+y2)).

Then, the distribution function, P(Y1+Y2<u) is given by
$$\int^{u}_{0}$$$$\int^{u-y2}_{0}e^(-(y1+y2))$$dy1dy2 = 1-ue^(-u)-e^(-u)

Then, to get the density function, we differentiate with respect to u, obtaining:
f(u) = ue^(-u)

 

[vela]

I think you just need to calculate the average length of life as (y1+y2)/2, not simply (y1+y2).

 

[kjartan]

Ok, almost there . . .

So, since we have (y1+y2)/2, we switch the upper limits of integration as such:
$$\int^{2u}_{0}$$$$\int^{2u-y2}_{0}e^(-(y1+y2)/2)dy1dy2$$= 4(1-e^(-u)-ue^(-u)).

Then take d/du, obtaining:
4ue^(-u)I must be tired, I'll have to take another look at this to figure out what I'm doing wrong . . . need to obtain 4ue^(-2u) . . .

Thanks! I think that's the crux of the problem that I was missing, now I just need to fill in the last detail . . . I appreciate your help.

 

[vela]

The joint probability density doesn't change; only the limits do.

 

[kjartan]

Whoops! Thanks!

Ok, now finally . . .

integrating e^(-(y1+y2)) with our new limits of integration and then taking d/du, to obtain

f(u)=4ue^(-2u)

Presto!

Thanks! Ok, so I should have paid more attention to their use of "average" in the question, I kind of read over that.
I appreciate your help, I stared and stared at that problem and could not figure out what I had done wrong.