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

• Jun 26th 2010, 11:40 AM
oldguynewstudent
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;$\displaystyle \theta$)=$\displaystyle \left(\frac{x}{\theta^{2}}\right)*e^{\frac{-x^{2}}{(2\theta^{2})}}$ for x>0 and 0 otherwise as a model for the X distribution.

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

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

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

Is $\displaystyle e^{-u}\mid_{-\infty}^{\infty}$ = 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 $\displaystyle {\infty}$
• Jun 26th 2010, 12:14 PM
oldguynewstudent
Quote:

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

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

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

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

Is $\displaystyle e^{-u}\mid_{-\infty}^{\infty}$ = 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 $\displaystyle {\infty}$

Just figured it out for myself once I realized the limits error:

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