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Math Help - Probability Density Function

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    Probability Density Function

    I'm working on a practice midterm and this question has me a bit stumped. Can anyone please help in working it out? Thank you!!

    f(x) = ae^(-ax) with 0<equalto x < infinity

    where a is constant, show that the mean and standard deviation are both 1/a

    Thank you.
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    Quote Originally Posted by bloomsgal8 View Post
    I'm working on a practice midterm and this question has me a bit stumped. Can anyone please help in working it out? Thank you!!

    f(x) = ae^(-ax) with 0<equalto x < infinity

    where a is constant, show that the mean and standard deviation are both 1/a

    Thank you.
    Mean: \mu = a \int_0^{+\infty} x \, e^{-ax} \, dx.

    Do this integral using integration by parts.


    Variance: E(X^2) = a \int_0^{+\infty} x^2 \, e^{-ax} \, dx.

    Do this integration by using integration by parts - twice.

    Then \sigma^2 = E(X^2) - \mu^2.
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  3. #3
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    EDIT: Please ignore this post.

    Quote Originally Posted by bloomsgal8 View Post
    I'm working on a practice midterm and this question has me a bit stumped. Can anyone please help in working it out? Thank you!!

    f(x) = ae^(-ax) with 0<equalto x < infinity

    where a is constant, show that the mean and standard deviation are both 1/a

    Thank you.
    The mean value of x would be defined thusly:

    \int_{0}^{\bar{x}}ae^{-ax}dx=\int_{\bar{x}}^{\infty}ae^{-ax}dx

    Solving:

    [-e^{-ax}]_{0}^{\bar{x}}=[-e^{-ax}]_{\bar{x}}^{\infty}

    -e^{-a\bar{x}}-(-e^{-a(0)})=[\lim_{x\to\infty}-e^{-ax}]-(-e^{-a\bar{x}})

    -e^{-a\bar{x}}+1=0+e^{-a\bar{x}}

    1=2e^{-a\bar{x}}

    \frac{1}{2}=e^{-a\bar{x}}

    \ln{\frac{1}{2}}=\ln{e^{-a\bar{x}}}

    -\ln{2}=-a\bar{x}\ln{e}

    -\ln{2}=-a\bar{x}

    \frac{\ln{2}}{a}=\bar{x}

    But you require the mean value of y. Can you figure it out from the above?
    Last edited by hatsoff; May 29th 2008 at 10:29 PM.
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    Quote Originally Posted by hatsoff View Post
    The mean value of x would be defined thusly:

    \int_{0}^{\bar{x}}ae^{-ax}dx=\int_{\bar{x}}^{\infty}ae^{-ax}dx

    Solving:

    [-e^{-ax}]_{0}^{\bar{x}}=[-e^{-ax}]_{\bar{x}}^{\infty}

    -e^{-a\bar{x}}-(-e^{-a(0)})=[\lim_{x\to\infty}-e^{-ax}]-(-e^{-a\bar{x}})

    -e^{-a\bar{x}}+1=0+e^{-a\bar{x}}

    1=2e^{-a\bar{x}}

    \frac{1}{2}=e^{-a\bar{x}}

    \ln{\frac{1}{2}}=\ln{e^{-a\bar{x}}}

    -\ln{2}=-a\bar{x}\ln{e}

    -\ln{2}=-a\bar{x}

    \frac{\ln{2}}{a}=\bar{x}

    But you require the mean value of y. Mr F asks: Where has y come from?
    [snip]
    You've calculated the median value of X.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    You've calculated the median value of X.
    So I have, so I have.

    Well, that's why I'm here--to get my fill of idiotic mistakes like that.

    Mr F asks: Where has y come from?
    The x-y plane.
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