# How to derive the multivariate normal distribution

1. Jul 18, 2009

### jone

If the covariance matrix $\mathbf{\Sigma}$ of the multivariate normal distribution is invertible one can derive the density function:

$f(x_1,...,x_n) = f(\mathbf{x}) = \frac{1}{(\sqrt(2\pi))^n\sqrt(\det(\mathbf{\Sigma)}}\exp(-\frac{1}{2}(\mathbf{x}-\mathbf{\mu})^T\mathbf{\Sigma}^{-1}(\mathbf{x}-\mathbf{\mu}))$

So, how do I derive the above?

2. Jul 18, 2009

### John Creighto

Start with a normal distribution where all the variables are independent and then do a change of variables.

3. Jul 18, 2009

### jone

I was on that track before, make use of the CDF and then differentiate back to get the PDF. This is how far I get: Let Y be a standard i.i.d. Gaussian vector. Then use the transformation

$\mathbf{X} = \mathbf{A}\mathbf{Y} + \mathbf{\mu}$

$P(\mathbf{X} < \mathbf{x}) = P(\mathbf{A}\mathbf{Y} + \mathbf{\mu} < \mathbf{x}) = P(\mathbf{Y} < \mathbf{A}^{-1}(\mathbf{x}-\mathbf{\mu}))$
Now I differentiate this to get the PDF

$f_{\mathbf{X}}(\mathbf{x}) = f_{\mathbf{Y}}(\mathbf{A}^{-1}\mathbf{x-\mu})\det(\mathbf{A}^{-1}) = f_{\mathbf{Y}}(\mathbf{A}^{-1}\mathbf{x-\mu})\frac{1}{\det(\mathbf{A})} = \frac{1}{(2\pi)^{n/2}\det(A)}\exp\left(\frac{1}{2}(\mathbf{x-\mu})^{T}(\mathbf{AA^T})^{-1}(\mathbf{x-\mu})\right)$

So $\det(\mathbf{A})}$ pops out in the denominator, instead of $\det(\mathbf{AA^T})}$ it as it should be. Something is wrong in my differentiation here but I can't figure it out.

4. Jul 19, 2009

### John Creighto

Why do you think the denominator should be $\det(\mathbf{AA^T})}$.

That would give you something analogies to the variance while the denominator of the Gaussian function is the standard deviation.

You want:

$\sqrt{|\mathbf{AA^T}|}=\sqrt{|\mathbf{A}|}\sqrt{|\mathbf{A^T}|}=|\mathbf{A}|$

5. Jul 19, 2009

### jone

Ok, so now it works out. $\mathbf{\Sigma} = \mathbf{A}\mathbf{A}^T$ is the covariance matrix. Thank you for your help!

6. Jul 19, 2009