# Bivariate normal distribution question

If I have $(X,Y)$ with joint density $f(x,y)$ and $A$ is an invertible $2\times 2$ matrix, then for the random vector $(W,V)$ defined by:

$$\begin{pmatrix} W\\ V \\ \end{pmatrix} = A\begin{pmatrix} X \\ Y \\ \end{pmatrix}$$

the joint density $g(w,v)$ of $(W,V)$ is given by:

$$g(w,v)=f\left(A^{-1}\begin{pmatrix} w \\ v \end{pmatrix}\right) \frac1{|\det A|}$$

If $X$ and $Y$ are independent standard normal variables and we define $Z_1=X$ and $Z_2=\rho X+\sqrt{1-\rho^2}Y$ for $-1<\rho<1$, show that $(Z_1,Z_2)$ has a standard bivariate normal distribution with parameter $\rho$.

And therefore show that if $(Z_1,Z_2)$ is a std bivariate normal random vector, the correlation of $(Z_1,Z_2)$ is $\rho$.

I'm really unsure of how I would approach this question, as I've never encountered bivariate normal distribution with matrices. I think I understand bivariate normal distributions, but I'm not sure how to apply it here. Any help would be great because I'm very, very stuck!

• I was going to say in my answer that your assertion about the joint density is incorrect. But now Dilip Sarwate has corrected it. You should notice what he did. – Michael Hardy May 29 '14 at 12:41

You have $$\begin{bmatrix} Z_1 \\ Z_2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ \rho & \sqrt{1-\rho^2} \end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix}.$$ The determinant of this matrix is $\sqrt{1-\rho^2}$.

You have the density $$f_{X,Y}(x,y) = \frac{1}{2\pi} \exp\left( \frac{-1}{2}(x^2+y^2) \right)$$ and $$\begin{bmatrix} 1 & 0 \\ \rho & \sqrt{1-\rho^2} \end{bmatrix}^{-1} = \begin{bmatrix} 1 & 0 \\ \frac{-\rho}{\sqrt{1-\rho^2}} & \frac{1}{\sqrt{1-\rho^2}} \end{bmatrix}$$ and the determinant of this matrix is $\sqrt{1-\rho^2}$.

That and your assertion about the density will give you the joint density of $W$ and $V$.

If you're looking for the correlation, you can read the covariance and the two variances out of the density function, but that should not be necessary. If you have two random variables $X,Y$ whose covariance matrix is $M$, and you've got $$\begin{bmatrix} W \\ V \end{bmatrix} = A \begin{bmatrix} X \\ Y \end{bmatrix},$$ then the covariance matrix of $\begin{bmatrix} W \\ V \end{bmatrix}$ is $$AMA^T.$$ In this case that is $$\begin{bmatrix} 1 & 0 \\ \rho & \sqrt{1-\rho^2} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & \rho \\ 0 & \sqrt{1-\rho^2} \end{bmatrix} = \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}.$$ That gives you $\operatorname{cov}(W,V)$ and the two variances, and since both variances are $1$, the correlation is the covariance.

• Thanks! That was helpful! Just as a side note, how would I prove that the conditional distribution of $Z_2|Z_1=x$ is normally distributed? I know that it is I'm just not sure how I'd go about showing it. – killianj12 May 29 '14 at 14:20
• One way to show that would be by looking at the form of the density function restricted to $z_1=x$. It's the exponential of a quadradic polynomial in $z_2$. – Michael Hardy May 29 '14 at 17:43
• In this instance, it is easier to find that $$\operatorname{cov}(Z_1,Z_2)=\operatorname{cov}\left(X,\rho X+\sqrt{1-\rho^2}Y\right)=\operatorname{cov}(X,\rho X)=\rho$$ and $$\operatorname{var}(Z_2)=\rho^3\operatorname{var}(X) + (1-\rho^2)\operatorname{var}(Y)=1$$ leading to the correlation coefficient being $\rho$ – Dilip Sarwate May 29 '14 at 18:58

Hints:

You know $f_{X,Y}(x,y)$ since $X$ and $Y$ are independent standard normal random variables. You are given the matrix $A = \begin{pmatrix}1&0\\\rho&\sqrt{1-\rho^2}\end{pmatrix}$. So, find $A^{-1}$ and use the formula given to you. Since you claim that you know about bivariate normal distributions, you should have no difficulty in recognizing the result.

Oh, by the way, if what you get from $A^{-1}\begin{pmatrix}w\\v\end{pmatrix}$ is $\begin{pmatrix}\alpha\\\beta\end{pmatrix}$, then substitute $\alpha$ for $x$ and $\beta$ for $y$ in $f_{X,Y}(x,y)$. It would have been a lot simpler visually if your book had used row vectors instead column vectors so that you would have ended up with $f_{X,Y}\left(\begin{pmatrix}\alpha&\beta\end{pmatrix}\right)$ which you would have interpreted as $f_{X.Y}(\alpha,\beta)$. I knew there was a reason I hate column vectors....