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{\bf Question}

Suppose that $X$ and $Y$ is
$N_2(\mu_x,\mu_y,\sigma_x,\sigma_y,\rho)$ for which

$E(X|Y=y)=3.7-0.15y,E(Y|X=x)=0.4-0.6x, {\rm var}(Y|X=x)=3.64.$

Find all the five parameters.



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{\bf Answer}

We know $$E(X|Y=y)=\mu_x + \rho
\frac{\sigma_x}{\sigma_y}(y-\mu_y)$$ $$E(Y|X=x)=\mu_y + \rho
\frac{\sigma_y}{\sigma_x}(x-\mu_x)$$ $${\rm
var}(Y|X=x)=\sigma^2_y(1-\rho^2)$$

Here: $$E(X|Y=y)=3.7-0.15y$$ $$E(Y|X=x)=0.4-0.6x$$ $${\rm
var}(Y|X=x)=3.64$$

Note that the coefficient of $y$ in $E(X|Y=y)$ is
$\rho\frac{\sigma_x}{\sigma_y}$

Note that the coefficient of $x$ in $E(Y|X=x)$ is
$\rho\frac{\sigma_y}{\sigma_x}$

Multiplying the two we get $\rho^2$.

Therefore $\rho^2=(-0.15)(-0.6)=0.09$

Therefore $\rho=-\sqrt{0.09}=-0.3$

Negative sign because coefficient of $y$ in $E(X|Y=y)$ is
$=-0.15=\rho\frac{\sigma_y}{\sigma_x}$ and $\sigma_x$ and
$\sigma_y$ are positive.

\begin{eqnarray*} {\rm var}(Y|X=x) & = & \sigma^2_y(1-\rho^2)=3.64\\ &
\Rightarrow & \sigma^2_y(1-0.09)=3.64\\ & \Rightarrow &
\sigma^2_y=4 \end{eqnarray*}

Now \begin{eqnarray*}\frac{\rho \sigma_x}{\sigma_y}=-0.15 &
\Rightarrow & \frac{(-0.3)\sigma_x}{2}=-0.15\\ & \Rightarrow &
\sigma_x=1 \end{eqnarray*}

$\left.\begin{array}{rrcc} {\rm Now} & \mu_x - 0.15(-\mu_y) & = &
3.7\\{\rm and} & \mu_y - 0.6(-\mu_x) & = & 0.4
\end{array}\right\}$ solve for $\mu_x, \mu_y$.

Final answer: $\mu_x=4,\ \mu_y=2,\ \sigma_x=1,\ \sigma_y=2,\
\rho=-0.3$


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