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Thread: Exponential decay question

  1. #1
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    Exponential decay question

    What is the memoryless property?
    Last edited by bluebiro; Aug 28th 2009 at 05:11 AM.
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    Quote Originally Posted by bluebiro View Post
    Let X represent the life time of a washing machine with c.d.f. F(X) = 1 - exp (-X^ p) for values of X greater than 0, where p > 0 is a parameter. If I have been using the washing machine for X years without failured, is it now better, worse or the same quality as a new washing machine?

    I've tried to apply the exponential model to this but i find the question quite confusing Any help/explanation would be greatly appreciated
    Read 3.2 of this: Exponential distribution - Wikipedia, the free encyclopedia
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    Super Member Random Variable's Avatar
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    let X be the lifetime of the washing machine

    $\displaystyle f(x) = pe^{-px} $

    $\displaystyle P(X>x+t|X>x) = \frac{P(X>x,X>x+t)}{P(X>x)} $

    EDIT: $\displaystyle = \frac{P(X>x+t)}{P(X>x)} = \frac{e^{-p(x+t)}}{e^{-px}}$ $\displaystyle = \frac{e^{-px}e^{-pt}}{e^{-px}} = e^{-pt}$

    notice that $\displaystyle P(X>t) = \int_{t}^{\infty} pe^{-px}dx = e^{-pt} $

    so the probabilty that the washing machine will continue working for at least another t years is the same probability if the washing machine were new
    Last edited by Random Variable; Jul 25th 2009 at 11:49 AM.
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    Quote Originally Posted by Random Variable View Post
    let X be the lifetime of the washing machine

    $\displaystyle f(x) = pe^{-px} $

    $\displaystyle P(X>x+t|X>x) = \frac{P(X>x,X>x+t)}{P(X>x)} $

    $\displaystyle = \frac{P(X>x+t)}{P(X>x)} = \frac{pe^{-p(x+t)}}{pe^{-px}}$ $\displaystyle = \frac{pe^{-px}e^{-pt}}{pe^{-px}} = e^{-pt}$

    notice that $\displaystyle P(X>t) = \int_{t}^{\infty} pe^{-px}dx = e^{-pt} $

    so the probabilty that the washing machine will continue working for at least another t years is the same probability if the washing machine were new
    You've explained this very clearly. Thank you!
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    Super Member Random Variable's Avatar
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    I made two errors. But it doesn't change the final result.

    $\displaystyle \frac{P(X>x+t)}{P(X>x)} = \frac{\int^{\infty}_{x+t} pe^{-px} \ dx} {\int^{\infty}_{x} pe^{-px} \ dx} $$\displaystyle = \frac{e^{-p(x+t)}}{e^{-px}} = \frac{e^{-px}e^{-pt}}{e^{-px}}= e^{-pt}= P(X>t)$
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    Quote Originally Posted by Random Variable View Post
    I made two errors. But it doesn't change the final result.

    $\displaystyle \frac{P(X>x+t)}{P(X>x)} = \frac{\int^{\infty}_{x+t} pe^{-px} \ dx} {\int^{\infty}_{x} pe^{-px} \ dx} $$\displaystyle = \frac{e^{-p(x+t)}}{e^{-px}} = \frac{e^{-px}e^{-pt}}{e^{-px}}= e^{-pt}= P(X>t)$
    Oh, I see. Thanks for informing me about this!
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    Quote Originally Posted by Random Variable View Post
    let X be the lifetime of the washing machine

    $\displaystyle f(x) = pe^{-px} $

    $\displaystyle P(X>x+t|X>x) = \frac{P(X>x,X>x+t)}{P(X>x)} $

    EDIT: $\displaystyle = \frac{P(X>x+t)}{P(X>x)} = \frac{e^{-p(x+t)}}{e^{-px}}$ $\displaystyle = \frac{e^{-px}e^{-pt}}{e^{-px}} = e^{-pt}$

    notice that $\displaystyle P(X>t) = \int_{t}^{\infty} pe^{-px}dx = e^{-pt} $

    so the probabilty that the washing machine will continue working for at least another t years is the same probability if the washing machine were new
    I just realised that your working is for F(x) = 1 - exp(-X*p), not F(x) = 1 - exp(-X^p). Does the solution still hold?
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    Quote Originally Posted by bluebiro View Post
    I just realised that your working is for F(x) = 1 - exp(-X*p), not F(x) = 1 - exp(-X^p). Does the solution still hold?
    Why don't you try to answer that question yourself?

    Use, what's right above and.....

    $\displaystyle P(X>x+t|X>x) = \frac{P(X>x,X>x+t)}{P(X>x)} =\frac{P(X>x+t)}{P(X>x)} =\frac{1-F(x+t)}{1-F(x)} $

    see what happens.
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    Quote Originally Posted by matheagle View Post
    Why don't you try to answer that question yourself?

    Use, what's right above and.....

    $\displaystyle P(X>x+t|X>x) = \frac{P(X>x,X>x+t)}{P(X>x)} =\frac{P(X>x+t)}{P(X>x)} =\frac{1-F(x+t)}{1-F(x)} $

    see what happens.
    Good point. I'm just worried I'll have to differentiate something.
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    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by bluebiro View Post
    Good point. I'm just worried I'll have to differentiate something.
    One bird to another....
    You can't be afraid to make a mistake, i.e, fly on your own.

    Show your work here and people will check it for you.
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    Quote Originally Posted by matheagle View Post
    One bird to another....
    You can't be afraid to make a mistake, i.e, fly on your own.

    Show your work here and people will check it for you.
    OK. But just to check, instead of $\displaystyle f(x) = pe^{-px} $ will f(x) be equal to $\displaystyle xp^{-p{x-1}}.e^{-p^{x}}$?
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  12. #12
    MHF Contributor matheagle's Avatar
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    $\displaystyle F(x)=1-e^{-x^p}$, so $\displaystyle f(x)=px^{p-1}e^{-x^p}$

    (where x>0, so the cow doesn't have a cow)

    It's $\displaystyle x^p$ not $\displaystyle p^x$

    And you don't need f(x), just plug F(x) into the equation.
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  13. #13
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    Quote Originally Posted by matheagle View Post
    $\displaystyle F(x)=1-e^{-x^p}$, so $\displaystyle f(x)=px^{p-1}e^{-x^p}$

    (where x>0, so the cow doesn't have a cow)

    It's $\displaystyle x^p$ not $\displaystyle p^x$

    And you don't need f(x), just plug F(x) into the equation.
    I get:

    $\displaystyle 1+e^{-x^{p+t}}$ divided by $\displaystyle 1+e^{-x^{p}}$

    Which then leads to
    $\displaystyle e^0 +e^{-x^{p+t}}$ divided by $\displaystyle e^0 +e^{-x^p}$

    Which equals
    $\displaystyle e^{-x^{p+t-p}}$

    =

    $\displaystyle e^{-x^{p}}$

    Almost certain I'm wrong,
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