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This question already has an answer here:

I'm trying to figure out the solution to the following:

Given:

a) $Z_1$ and $Z_2$ are independent standard normal random variables

b) $Y_1=Z_1$ and $Y_2 = Z_2\sqrt{1-\rho^2} + \rho Z_1$

Find: joint probability density function ($ f_{Y_1 Y_2}(y_1,y_2)$) and marginal probability density functions of $Y_1$ and $Y_2$

For the joint I found the following using the jacobian determinant: $$(1/(2\pi((1-p^2)^{1/2})))(e^{-((y-px)/((1-p^2)^{1/2}))^2}/2)) e^{-y^2/2}$$

I'm not sure if this is the right answer as it looks too complicated. Thus, it has been hard for me to find the probability density function of Y2, as integrating the joint density function is hard.

I would greatly appreciate any help

Ps. I will be working on putting the density function I got in mathjax language. Bu you could insert the one I put there into wolfram alpha and get a better visual.

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marked as duplicate by BruceET, John B, choco_addicted, Stefan Mesken, Silvia Ghinassi Mar 13 '16 at 2:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Please check here: http://math.stackexchange.com/questions/1687795/. Also, link there. $\endgroup$ – BruceET Mar 12 '16 at 21:01
  • $\begingroup$ @BruceET thank you for the reference. However, in that problem the transformation is the inverse (between Z and Y2). For this reason my problem is not fitting well there. $\endgroup$ – user320559 Mar 12 '16 at 21:12
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I would first try to find the expected values, the variances, and the covariances, and then use those to find the joint density. Since $Y_1,Y_2$ are both linear combinations of the same independent standard-normally distributed random variables, the pair $(Y_1,Y_2)$ is jointly normally distributed, so just applying what is known about the bivariate normal p.d.f. will do it.

The expected value and variance of $Y_1$ should be clear. $$ \operatorname{E}(Y_2) = \sqrt{1-\rho^2}\operatorname{E}(Z_2) + \rho \operatorname{E}(Z_1) = 0 + 0. $$ $$ \operatorname{var}(Y_2) = \left(\sqrt{1-\rho^2}\,\right)^2 \operatorname{var}(Z_2) + \rho^2 \operatorname{var}(Z_1) = (1-\rho^2) + \rho^2 = 1. $$ \begin{align} \operatorname{cov}(Y_1,Y_2) & = \operatorname{cov}\left(Z_1, \sqrt{1-\rho^2} Z_2 + \rho Z_1 \right) \\[10pt] & = \sqrt{1-\rho^2} \operatorname{cov}\left(Z_1, Z_2 \right) + \rho \operatorname{cov}(Z_1,Z_1) \\[10pt] & = 0 + \rho\cdot1 = \rho. \end{align}

Thus the matrix of covariances is $S=\begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}$.

The density is then $$ \frac 1 {2\pi\det S} \exp\left( \frac{-1} 2 [y_1, \, y_2] S^{-1} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} \right). $$ If you find the inverse of $S$ and then do the matrix multiplication, then you'll have the density expressed without the use of matrices.

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