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Math Help - Statistics - PDF/Probability

  1. #1
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    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?
    Last edited by AwesomeHedgehog; March 4th 2014 at 03:15 AM.
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  2. #2
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    Re: Statistics - PDF/Probability

    Quote Originally Posted by AwesomeHedgehog View Post
    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

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

    $\large \displaystyle{\int_0^\infty \int_{\frac{1-u}{u}x}^\infty} e^{-(x+y)}~dy~dx$
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    Re: Statistics - PDF/Probability

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    Re: Statistics - PDF/Probability

    Quote Originally Posted by AwesomeHedgehog View Post
    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.
    Last edited by romsek; March 4th 2014 at 11:21 AM.
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  5. #5
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    Re: Statistics - PDF/Probability

    Basically, I did the integral above and as my answer I got: P[U < u] = u
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    Re: Statistics - PDF/Probability

    Quote Originally Posted by AwesomeHedgehog View Post
    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?
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    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
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    Re: Statistics - PDF/Probability

    Quote Originally Posted by AwesomeHedgehog View Post
    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?
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    Re: Statistics - PDF/Probability

    Would the upper bound just be 0 <= u <= 1
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    Re: Statistics - PDF/Probability

    Quote Originally Posted by AwesomeHedgehog View Post
    Would the upper bound just be 0 <= u <= 1
    yes! and what do they call a pdf that's 1 for 0<=u<=1 ?
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    Re: Statistics - PDF/Probability

    Um, independent?
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    Re: Statistics - PDF/Probability

    Quote Originally Posted by AwesomeHedgehog View Post
    Um, independent?
    you call it uniform[0,1]
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    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.
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  14. #14
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    Re: Statistics - PDF/Probability

    I did the integral wrong above, I meant the integral from 0.75 to 1 of 1 du = 0.25
    Last edited by AwesomeHedgehog; March 4th 2014 at 03:55 PM.
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  15. #15
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    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
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