# Jointly normal distribution

I encountered the following sentence while reading a textbook and could not figure out why.

If $x$ and $y$ are jointly normal, then $y$ is a linear (affine) function of $x$: $y = a+bx$.

If someone could explain this to me, that would be great.

If $X$ and $Y$ are jointly normal,

$$\bigl( \begin{array}{cc} x \\ y \\ \end{array} \bigl) \sim N(\bigl( \begin{array}{cc} \mu_{x} \\ \mu_{y} \\ \end{array} \bigl), \bigl( \begin{array}{cc} \sigma_{x}^2&\sigma_{xy} \\ \sigma_{xy}&\sigma_{y}^2 \\ \end{array} \bigl) )$$

then the conditional distribution of $Y$ conditional on $X$ is

$$Y|X \sim N(\mu_{y} + \frac{\sigma_{y}}{\sigma_{x}} \rho(x-\mu_{x}), (1-\rho^2)\sigma_{y})$$

where $\rho=\frac{\sigma_{xy}}{\sigma_{x} \sigma{y}}$. Then the conditional expectation of $Y$ given $X=x$ has the form you described,

$$E(Y|X)=\mu_{y} + \frac{\sigma_{y}}{\sigma_{x}} \rho(x-\mu_{x})=a+bx$$

where $a=\mu_{y}-\frac{\sigma_{y}}{\sigma_{x}} \rho \mu_{x}$ and $b=\frac{\sigma_{y}}{\sigma_{x}} \rho$.

• I thought that the author meant the random variable $y$ not the conditional expectation $E(Y|X)$ in the expression $y=a+bx$. Could you construct an example where $x$ and $y$ are jointly normal, and $y$ is not equal to $a+bx$? Thanks a lot. – Emx Dec 11 '12 at 7:12
• Any nondegenerate binormal distribution would serve that purpose, Dave. The simplest example would be a pair of independent standard normal variates. – whuber Dec 11 '12 at 7:35
• @caburke There are several typos in your answer. The conditional mean is $\mu_y + \cdots$ and not $\mu_x + \cdots$ and your final value for $a$ is incorrect too, since you used $\mu_x$ instead of $\mu_y$ and simplified the RHS. – Dilip Sarwate Dec 11 '12 at 17:37
• @DilipSarwate Thanks! The errors have been corrected. – caburke Dec 11 '12 at 19:35
• @caburke One more typo. In your first displayed equation, the mean vector should be $(\mu_x,\mu_y)$ (transposed), – Dilip Sarwate Dec 12 '12 at 3:09