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

1. ## 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 ${\infty}$

2. Originally Posted by oldguynewstudent
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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