Product of gaussian random variable with itself

AI Thread Summary
The discussion centers on the product of a Gaussian random variable with itself, specifically examining what happens when X, a Gaussian variable, is squared (X^2). It is clarified that while the product of two independent Gaussian variables remains Gaussian, squaring a Gaussian variable results in a non-negative variable that cannot be Gaussian. Instead, the outcome is related to the chi-square distribution, particularly when considering the properties of the squared variable. Participants suggest investigating the origins of chi-square distributions for deeper understanding. The conversation emphasizes the importance of probability density functions in this context.
architect
Messages
30
Reaction score
0
Hi,

I am interested in the product of a Gaussian random variable with itself. If X is Gaussian then what is X^2? We know that the resultant variable of the product of two independent Guassian variables is still Gaussian but I am afraid that this is not true when you multiply it with itself. Is it a chi-square? Any clarifications will be appreciated.

BW,

Alex
 
Last edited:
Physics news on Phys.org
architect said:
Hi,

I am interested in the product of a Gaussian random variable with itself. If X is Gaussian then what is X^2? We know that the resultant variable of the product of two independent Guassian variables is still Gaussian but I am afraid that this is not true when you multiply it with itself. Is it a chi-square? Any clarifications will be appreciated.

BW,

Alex

Partial answer to get you thinking more: If, as a specific case, X is standard Gaussian, notice that Y = X^2 will take on only non-negative values so certainly could not be standard Guassian. If I were you I'd investigate the origins of chi-square distributions (both central and non-central chi-square).
 
If p(x) is the probability density function of x, then conservation of probability tells us that

p(x) dx = q(y) dy

where y is any known function of x. Solve the equation for q(y) to get the probability density function for y.
 
thanks for your replies. I will give it a try!
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
Back
Top