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March 11th, 2015, 11:01 PM  #1 
Newbie Joined: Mar 2015 From: USA Posts: 1 Thanks: 0  Standard Deviation of Squared Gaussian Distribution
Hello Everyone, I have a quick question. Assuming we have a random variable described by a Gaussian probability density function: f(x) = 1/((2*pi)^0.5*s)*exp(x^2/(2s^2)) If a new random variable is defined as: y = x*x. What would be a probability density function describing it and what would be its standard deviation? Thank you, AeroX 
March 12th, 2015, 02:00 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,727 Thanks: 688 
Let $\displaystyle Y=X^2$ where X is N(0,s), call it f(x). The distribution function for Y (P(Y<y)) is given by: $\displaystyle g(y)=P(Y<y)=P(X^2<y)=P(\sqrt y<X<\sqrt y)=2\int_{0}^{\sqrt y}f(u)du$ The density function is g'(y). You can compute the mean and standard deviation from this. 
March 12th, 2015, 02:59 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,727 Thanks: 688 
If you don't need the density function, the mean and standard deviation can be computed directly: $\displaystyle E(Y)=E(X^2)=s^2,\ E(Y^2)=E(X^4)=3s^4$ Therefore the standard deviation of Y is $\displaystyle \sqrt 2 s^2$ 

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deviation, distribution, gaussian, squared, standard 
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