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Thread: What does this distribution mean?

  1. #1
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    What does this distribution mean?




    When it says:

    the marginal distribution of Y is given by

    1/Y ~ Exponential(lamda)

    does that mean the distribution is written has:

    1/(lamda)e^(-lamda)(y)

    ??

    Edit: Nevermind i think my first thought process is wrong, i believe its:

    (lamda)e^(-lamda)(1/y)

    if im right.
    Last edited by c2q; 11-26-2010 at 07:06 PM.

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    Re: What does this distribution mean?

    if the probability density function of X = \frac {1} {Y} is

    f_X(x;\lambda) = \lambda e^{-\lambda x} , x > 0, \lambda > 0

    Then it is easy to show that the probability density function of Y is

    f_Y(y;\lambda) = \lambda \frac {1} {y^2} e^{-\lambda \frac {1} {y}},y > 0, \lambda > 0

    Note: The expected value of Y is may not exist / finite.

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    Re: What does this distribution mean?

    Quote Originally Posted by BGM View Post
    if the probability density function of X = \frac {1} {Y} is

    f_X(x;\lambda) = \lambda e^{-\lambda x} , x > 0, \lambda > 0

    Then it is easy to show that the probability density function of Y is

    f_Y(y;\lambda) = \lambda \frac {1} {y^2} e^{-\lambda \frac {1} {y}},
y > 0, \lambda > 0

    Note: The expected value of Y is may not exist / finite.
    just wondering, how did you get the 1/y^2 in the second equation of the PDF of Y? I thought its basically like that without the 1/y^2 if you can explain that.

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    Re: What does this distribution mean?

    Basically you can try the CDF approach or the Jacobian Transformation on pdf
    with 1-to-1 transformation.

    1. Consider

    F_Y(y) = \Pr\{Y \leq y\} 
= \Pr\left\{\frac {1} {Y} \geq \frac {1} {y} \right\}
= 1 - \Pr\left\{X \leq \frac {1} {y} \right\}
= 1 - F_X\left(\frac {1} {y} \right)

    Since we can get the pdf by differentiating the CDF, we have

    f_Y(y) = \frac {dF_Y(y)} {dy} 
= \frac {d} {dy} [ 1 - F_X(y^{-1}) ]
= - \frac {d} {dy^{-1}} F_X(y^{-1}) \frac {dy^{-1}} {dy}
= \frac {1} {y^2} f_X(y^{-1})
= \frac {1} {y^2} \lambda e^{-\lambda \frac {1} {y}}

    2. Since y = \frac {1} {x} \iff x = \frac {1} {y}

    we have the absolute value of the Jacobian determinant

    |J| = \left|\frac {dx} {dy}\right| 
= \left|-\frac {1} {y^2} \right| = \frac {1} {y^2}

    and thus

    f_Y(y) = |J|f_X(x) = \frac {1} {y^2} f_X(y^{-1})
= \frac {1} {y^2} \lambda e^{-\lambda \frac {1} {y}}

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    Re: What does this distribution mean?


    Thanks that makes sense.

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