how can i calculate z^2 distibution?

Hi, I am currently trying to calculate z^2 distribution when z is folloing (0,a) gaussian distribution

how can i calculate z^2 distribution?

[Hernia Baby]

Do you want to calculate Chi-square distribution?

https://en.wikipedia.org/wiki/Chi-square_distribution

The distribution depends on the degree of freedom.

Answer 1

[David Goodmanson]

Hello JS

By gaussian I assume you mean the normal distribution, mean 0, std deviation 'a'

f(z) =  N*exp(-(z/a)^2/2)
with N = 1/(sqrt(2)*sqrt(pi)*a)

y = z^2 has one sign while z can have either sign. Since the normal distribution is symmetric, to make life easier assume the initial distribution is one-sided, with z>=0 only, and double the height of the distribution to keep the normalization correct:

f(z) = 2*N*exp(-(z/a)^2/2)      z>=0

Now that z >= 0 we can use z = sqrt(y). The the idea with pdfs is to find a function g(y) such that

f(z) dz = g(y) dy
with     z = y^(1/2)
which means that  dz = (1/2) y^(-1/2) dy     

plug that in

2*N*exp(-(y/(2*a^2))) (1/2) y(-1/2) dy = g(y) dy
g(y) = N*exp(-(y/(2*a^2)))/sqrt(y)

which is a chi-squared probability distribution function (with 1 degree of freeedom) as noted by Jeff. This function has the disadvantage of being unbounded as y --> 0, although its area is still 1 as required.

If you meant y = (z/a)^2 instead of y = z^2, then the expression would be the standard chi-squared

g(y) = N*exp(-(y/2))/sqrt(y)     N =  1/(sqrt(2)*sqrt(pi))
 

[jinseok shin]

It helped a lot Thank you

Answer 2

[Jeff Miller]

The distribution of z^2/a^2 is a chi-square with 1 df.

[Jeff Miller]

To normalize. If y is normal(0,1), then the distribution of y^ is chi-square with 1 df.

For your z ~ N(0,a), y = z/a is N(0,1), so y^2 = z^2/a^2 is chi-square(1).

[jinseok shin]

Then how can i make pdf? Can I divide the Gaussian distribution by the variance?