Refreshing my stats, I wanted to learn how to derive the bivariate normal distribution, for which I found this source on Wolfram.com.

On that website, the authors derive the joint probability distributions for the jointly Guassian random variables (Eq.'s (10,11)): $$ Y_1 = \mu_1 + \sigma_{11} X_1 + \sigma_{12} X_2 \\ Y_2 = \mu_2 + \sigma_{21} X_1 + \sigma_{22} X_2 $$

Here, $X_1$ and $X_2$ are $\sim \mathcal{N}(0,1)$ for simplicity, and are independent. In Eq. (30), they find an expression for $x_1^2 + x_2^2$, solve it for the $x_i$, and finally use a Jacobian to change variables to the $y_i$, obtaining the final joint PDF Eq. (38)

My problem (and question) is: from Eq. (30), how to I solve for the $x_i$, Eq.'s (32,33)? Particulary, since we only have one equation in two unknowns, we need some more information. I suppose the authors use Eq.'s (10,11) for that?

I am kind of stuck here, and I believe the authors are doing something weird. In particular, in Eq. (31) they introduce the $\rho^\prime$ to write the results more concisely; however, if you look at Eq. (29), you note that $\rho^\prime = 1$, making the whole thing rather redundant!

So how do I obtain do I solve Eq. (30) from the above link for $x_i$? (I typed it below, so that you don't have to refer to the original all the time.)

$$ x_1^2 + x_2^2 = \frac{1}{1-\rho^2} \left[ \frac{(y_1 - \mu_1)^2}{\sigma_1^2} + \frac{(y_2 - \mu_2)^2}{\sigma_2^2} - \frac{2 \rho (y_1 - \mu_1)(y_2 - \mu_2)}{\sigma_1 \sigma_2}\right] $$ where $$ \rho = \frac{\sigma_{11} \sigma_{21} + \sigma_{12}\sigma_{22}}{\sigma_1 \sigma_2} \\ \sigma_1^2 = \sigma_{11}^2 + \sigma_{12}^2 \\ \sigma_2^2 = \sigma_{21}^2 + \sigma_{22}^2 $$


migrated from stats.stackexchange.com Mar 7 '15 at 15:57

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  • $\begingroup$ Please restate your question so that readers do not have to read the original entry. $\endgroup$ – Xi'an Mar 7 '15 at 15:45
  • 1
    $\begingroup$ This is merely equation (21) with $\rho$ used in place of that ratio of covariances. $\endgroup$ – whuber Mar 7 '15 at 15:57
  • $\begingroup$ Yes, you are correct! Thank you for noticing that. This actually answers my question. I should have looked at that part again... :-/ $\endgroup$ – mSSM Mar 7 '15 at 16:47

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