1. ## Statistics - PDF/Probability

The length of time that a machine operates without failure is denoted by X and the length of time to repair a failure is denoted by Y.

X and Y are independent each with exponential distribution, mean 1.

a.) Find the pdf of U = X/[X + Y], the proportion of time that the machine is in operation during any one operation-repair cycle.

b.) Use the pdf of U to find the probability that U is greater than 0.75.

For part a, I did:

f(x) = λ*e^(-λx) for 0 < x

F(u) = P( U < u ) = P( x/(x+y) < u) = P ( x < u/(1-u) ) = integral from 0 to u/(1-y) of λ*e^(-λx) dx

I'm wondering if I set that up correctly?

2. ## Re: Statistics - PDF/Probability

Originally Posted by AwesomeHedgehog
The length of time that a machine operates without failure is denoted by X and the length of time to repair a failure is denoted by Y.

X and Y are independent each with exponential distribution, mean 1.

a.) Find the pdf of U = X/[X + Y], the proportion of time that the machine is in operation during any one operation-repair cycle.

b.) Use the pdf of U to find the probability that U is greater than 0.75.

For part a, I did:

f(x) = λ*e^(-λx) for 0 < x

F(u) = P( U < u ) = P( x/(x+y) < u) = P ( x < u/(1-u) ) = integral from 0 to u/(1-y) of λ*e^(-λx) dx

I'm wondering if I set that up correctly?
Nah. You have a 2D pdf here. If you work it through you find

$Pr[\frac{X}{X+Y}<u] = Pr[\frac{1-u}{u} x < y]$

so you have a 2D integral

$\large Pr[U<u]=\displaystyle{\int_0^\infty \int_{\frac{1-u}{u}x}^\infty} f_{XY}(x,y)~dy~dx =$

$\large \displaystyle{\int_0^\infty \int_{\frac{1-u}{u}x}^\infty} e^{-(x+y)}~dy~dx$

4. ## Re: Statistics - PDF/Probability

Originally Posted by AwesomeHedgehog
the link doesn't exist. The site supports enough LaTex to type in any math you might want.

The answer to this one can be stated in 1 line.

5. ## Re: Statistics - PDF/Probability

Basically, I did the integral above and as my answer I got: P[U < u] = u

6. ## Re: Statistics - PDF/Probability

Originally Posted by AwesomeHedgehog
Basically, I did the integral above and as my answer I got: P[U < u] = u
and what does that get you for a pdf? Also note the "support" for U, i.e. the possible values U can take.

what's a common name for this pdf?

7. ## Re: Statistics - PDF/Probability

Wouldn't the pdf be the derivative of the cdf? Which in this case, the cdf is equal to u. So, the pdf would be:

f(u) = 1 for 0 < u and 0 elsewhere

8. ## Re: Statistics - PDF/Probability

Originally Posted by AwesomeHedgehog
Wouldn't the pdf be the derivative of the cdf? Which in this case, the cdf is equal to u. So, the pdf would be:

f(u) = 1 for 0 < u and 0 elsewhere
not quite. What's an upper bound for u? Besides, is what you've written a valid pdf? Does it integrate to 1?

9. ## Re: Statistics - PDF/Probability

Would the upper bound just be 0 <= u <= 1

10. ## Re: Statistics - PDF/Probability

Originally Posted by AwesomeHedgehog
Would the upper bound just be 0 <= u <= 1
yes! and what do they call a pdf that's 1 for 0<=u<=1 ?

11. ## Re: Statistics - PDF/Probability

Um, independent?

12. ## Re: Statistics - PDF/Probability

Originally Posted by AwesomeHedgehog
Um, independent?
you call it uniform[0,1]

13. ## Re: Statistics - PDF/Probability

Of course, the most obvious answer, and once again I'm over thinking things haha.

Then, for the second part of the problem (b) I did:

P[ u > 0.75] = the integral from 0 to 1 of 0.75 dx = 0.75

Which does not seem right.

14. ## Re: Statistics - PDF/Probability

I did the integral wrong above, I meant the integral from 0.75 to 1 of 1 du = 0.25

15. ## Re: Statistics - PDF/Probability

I did the integral wrong above, I meant the integral from 0.75 to 1 of 1 du = 0.25
this is correct