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Thread: Verify that f(x;theta) is a legitimate pdf

  1. #1
    Senior Member oldguynewstudent's Avatar
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    Verify that f(x;theta) is a legitimate pdf

    Can anyone help with the verification that the final integral over infinity = 1?

    Let X denote the vibratory stress (psi) on a wind turbine blade at a particular wind speed in a wind tunnel. The article “Blade Fatigue Life Assessment with Application to VAWTS” (J. Solar Energy Engr., 1982: 107-111) proposes the Rayleigh distribution, with pdf

    f(x; \theta)= \left(\frac{x}{\theta^{2}}\right)*e^{\frac{-x^{2}}{(2\theta^{2})}} for x>0 and 0 otherwise as a model for the X distribution.

    a) Verify that f(x; \theta) is a legitimate pdf.

    Rule 1: f(x) \geq0\:\forall x. All powers of e are positive everywhere so it is left to show \left(\frac{x}{\theta^{2}}\right)is positive. \theta^{2} must be positive and x is always greater than or equal to zero. Therefore the function is non-negative.

    Rule 2: \intop_{-\infty}^{\infty}\left(\frac{x}{\theta^{2}}\right)*  e^{\frac{-x^{2}}{(2\theta^{2})}}dx let u= \frac{-x^{2}}{2\theta^{2}} then du= \frac{-2x}{2\theta^{2}}dx so \intop_{-\infty}^{\infty}-e^{-u}du = e^{-u}\mid_{-\infty}^{\infty}

    Is e^{-u}\mid_{-\infty}^{\infty} = 1?? Or did I screw up royally? Or both?

    The problem states that x is non-negative so the integral should go from 0 to {\infty}
    Last edited by mash; Mar 11th 2012 at 11:22 PM.
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  2. #2
    Senior Member oldguynewstudent's Avatar
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    Quote Originally Posted by oldguynewstudent View Post
    Can anyone help with the verification that the final integral over infinity = 1?

    Let X denote the vibratory stress (psi) on a wind turbine blade at a particular wind speed in a wind tunnel. The article “Blade Fatigue Life Assessment with Application to VAWTS” (J. Solar Energy Engr., 1982: 107-111) proposes the Rayleigh distribution, with pdf

    f(x; \theta)= \left(\frac{x}{\theta^{2}}\right)*e^{\frac{-x^{2}}{(2\theta^{2})}} for x>0 and 0 otherwise as a model for the X distribution.

    a) Verify that f(x; \theta) is a legitimate pdf.

    Rule 1: f(x) \geq0\:\forall x. All powers of e are positive everywhere so it is left to show \left(\frac{x}{\theta^{2}}\right)is positive. \theta^{2} must be positive and x is always greater than or equal to zero. Therefore the function is non-negative.

    Rule 2: \intop_{-\infty}^{\infty}\left(\frac{x}{\theta^{2}}\right)*  e^{\frac{-x^{2}}{(2\theta^{2})}}dx let u= \frac{-x^{2}}{2\theta^{2}} then du= \frac{-2x}{2\theta^{2}}dx so \intop_{-\infty}^{\infty}-e^{-u}du = e^{-u}\mid_{-\infty}^{\infty}

    Is e^{-u}\mid_{-\infty}^{\infty} = 1?? Or did I screw up royally? Or both?

    The problem states that x is non-negative so the integral should go from 0 to {\infty}
    Just figured it out for myself once I realized the limits error:

    let u= \frac{x^{2}}{2\theta^{2}} then du= \frac{2x}{2\theta^{2}}dx so \intop_{0}^{\infty}e^{-u}du = -e^{-u}\mid_{0}^{\infty} = 0 + 1 = 1
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