# A Deriving the standard normal distribution

1. Apr 9, 2016

### rabbed

I've calculated the joint distribution, XY_PDF(x,y) of random
variables X and Y (both coming from a distribution N(n) = C*e^(-K*n^2)).

I use XY_PDF(x,y) to calculate the joint distribution AR_PDF(a,r)
of the random variables A (angle) and R (radius), with the PDF
method and the Jacobian.

Since AR_PDF(a,r) = A_PDF(a)*R_PDF(r) and I want A_PDF(a) = 1/(2*pi),
I can find that C = 1/(2*pi)^(1/2) in N(n), since the other factors in
AR_PDF(a,r) calculated from XY_PDF(x,y) are related to R.

If I do the same in 3D (using the longitude/latitude/radius distributions
for producing a uniform surface distribution), I get C = 1/(2*pi)^(1/3)
after discarding the factors related to the longitude distribution and the

The correct answer (for a multivariate gaussian) should be 1/(2*pi)^(1/2)
here also, right?

Is my reasoning to find this constant C wrong? Is there a better way?

Last edited: Apr 9, 2016
2. Apr 9, 2016

### andrewkirk

You don't need to perform the Cartesian to Polar transformation to calculate C. Just observe that the pdf N is that of a normal random variable with zero mean and variance $\sigma^2=\frac{1}{2K}$. Then use the formula for a normal distribution to conclude that
$$C=\frac{1}{\sigma\sqrt{2\pi}}=\sqrt\frac{2K}{2\pi}$$
The formula you calculated above is missing the factor $\sqrt{2K}$.

3. Apr 10, 2016

### rabbed

Is there no way to derive the standard normal distribution (not the normal distribution with concepts like mean and variance) from just basic knowledge of:
- The Jacobian
- How to calculate one joint distribution from another joint distribution using the PDF method
- Random vectors in 2D/3D (deriving the distributions of independent variables (polar/spherical coordinates) to get a uniform surface distribution by calculating the joint distribution of their dependent cartesian coordinates)
- Isotropy, wanting another kind of random vector by letting the cartesian coordinates be independent by coming from the same unknown distribution, N, as the radius

4. Apr 12, 2016

### zinq

For a standard normal distribution in n dimensional Euclidean space Rn, I like to think of it just as the joint distribution of n independent 1-dimensional standard normal random variables. As such, its n-dimensional probability density function is just the product

(1/√(2π)) e-x12/2 ⋅ ... ⋅ (1/√(2π)) e-xn2/2

which is

(2π)-n/2 e-(x12 + ... + xn2)/2

or in other words

(2π)-n/2 e-r2/2,

where r is the distance from the origin in Euclidean n-space Rn.

I know this doesn't answer all parts of your question, but it's one easy way to remember the constant factor occurring in the n-dimensional standard normal density.

5. Apr 12, 2016

Thanks, zinq