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March 11th, 2015, 11:01 PM   #1
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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
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March 12th, 2015, 02:00 PM   #2
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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.
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March 12th, 2015, 02:59 PM   #3
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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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