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?

Re: Statistics - PDF/Probability

Quote:

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

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

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

Re: Statistics - PDF/Probability

Re: Statistics - PDF/Probability

Quote:

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.

Re: Statistics - PDF/Probability

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

Re: Statistics - PDF/Probability

Quote:

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?

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

Re: Statistics - PDF/Probability

Quote:

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?

Re: Statistics - PDF/Probability

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

Re: Statistics - PDF/Probability

Quote:

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 ?

Re: Statistics - PDF/Probability

Re: Statistics - PDF/Probability

Quote:

Originally Posted by

**AwesomeHedgehog** Um, independent?

you call it uniform[0,1]

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.

Re: Statistics - PDF/Probability

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

Re: Statistics - PDF/Probability

Quote:

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

this is correct