Statistics: Multivariate = Bivariate normal distribution?

True, only if n = 2.

Formula for the multivariate normal distribution:

If $\ \Sigma$ is non-singular, then the distribution may be described by the following PDF:

$f_X(x_1, \dots, x_N) = \frac {1} {(2\pi)^{N/2}|\Sigma|^{1/2}} \exp \left( -\frac{1}{2} ( x - \mu)^\top \Sigma^{-1} (x - \mu) \right)$

where $\ \Sigma$ represents the variance-covariance matrix.

Formula for the bivariate normal distribution:

In the 2-dimensional nonsingular case, the probability density function (with mean (0,0)) is

$f(x,y) = \frac{1}{2 \pi \sigma_x \sigma_y \sqrt{1-\rho^2}} \exp \left( -\frac{1}{2 (1-\rho^2)} \left( \frac{x^2}{\sigma_x^2} + \frac{y^2}{\sigma_y^2} - \frac{2 \rho x y}{ (\sigma_x \sigma_y)} \right) \right)$

where $ρ$ is the correlation between $X$ and $Y$ . In this case,

$\Sigma = \begin{bmatrix} \sigma_x^2 & \rho \sigma_x \sigma_y \\ \rho \sigma_x \sigma_y & \sigma_y^2 \end{bmatrix}$ .

The above $\ \Sigma$ matrix gives a hint of how to solve the problem. In fact, I had to do this exercise for a class assignment. The proof took 3 pages long, using various statistical definitions and theorems! My proof was just meticulous and long, and I am sure that 1.5 to 2 pages should suffice. It is an amazing feeling to solve such a problem as that.

Have a look at the multivariate normal distribution page on Wikipedia, from which I copied and pasted the above formulae.

. - * I just write * - .

Friday, May 2, 2008

Statistics: Multivariate = Bivariate normal distribution?

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