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$\mu_{Y|x} = \mu_{2} + \rho \cdot \frac{\sigma_2}{\sigma_1}\cdot (x-\mu_{1})$

and the variance

$\sigma^{2}_{Y|x} = \sigma^{2}_{2}(1-\rho^2)$

and the conditonal density of $X$ given $Y=y$ is a normal distribution with the mean

$\mu_{X|y} = \mu_{1} + \rho\cdot \frac{\sigma_{1}}{\sigma_{2}}\cdot (y-\mu_{2})$

and the variance

$\sigma^{2}_{X|y} = \sigma^{2}_{1}(1-\rho^2)$

Moreover, one must compare coefficients to each other for the bivariate normal density and the defintion above. Thus

$$\color{red}{(1)} \quad \frac{-1}{2(1-p^2)} =\frac{-1}{102}$$

$$\color{red}{(2)} \quad \left( \frac{x-\mu_1}{\sigma_{1})}\right)^2 =\left( \frac{x+2}{1} \right)^2$$

$$\color{red}{(3)} \quad \left( \frac{y-\mu_1}{\sigma_{1})}\right)^2 =\left( 4(y-1) \right)^2$$

$$\color{red}{(4)} \quad \left( \frac{2\rho \cdot (x-\mu_{1})}{(\sigma_{1})} \frac{ (y-\mu_{2})}{(\sigma_{2})}\right) =\left( 2.8(x+2)(y-1) \right)$$

Although from the information I have provided I cannot derive the solution for $\mu_{1}$ and $\mu_{2}$?

  • $\begingroup$ If $\displaystyle \left( \frac{x-\mu_1}{\sigma_{1}}\right)^2 =\left( \frac{x+2}{1} \right)^2$ is true for all $x$, how do you fail to come to the conclusion that $\mu_1=-2$ and $\sigma_1=1$? You could for example expand and compare coefficients. (Possibly you might find $\sigma_1=\pm1$ but standard deviations are the non-negative square-root of the variance) $\endgroup$ – Henry Nov 21 '16 at 16:52
  • $\begingroup$ I just do not see the correlation after expanding I get $\sigma_{1}^2=-\frac{\mu_1-\mu_1^{2}}{2}$ $\endgroup$ – Ocelot Nov 21 '16 at 20:02
  • $\begingroup$ after expanding I get $\frac{x^2-2x\mu_1-\mu_1^2}{\sigma^2_1} = x^2+4x+4$ $\endgroup$ – Ocelot Nov 21 '16 at 20:08
  • $\begingroup$ So you get $\dfrac{1}{\sigma_1^2}x^2 +\dfrac{-2\mu_1}{\sigma_1^2}x+\dfrac{\mu_1^2}{\sigma_1^2} = x^2+4x+4$. This has to be true for all $x$ so matching coefficients will give you $\dfrac{1}{\sigma_1^2}=1$ and $\dfrac{-2\mu_1}{\sigma_1^2}=4$ and $\dfrac{\mu_1^2}{\sigma_1^2}=4$ which implies $\sigma_1^2=1$ and $\mu_1=-2$. $\endgroup$ – Henry Nov 21 '16 at 22:43
  • $\begingroup$ Sorry friend but according to the book $\sigma_1 =10 $ and $\sigma_2 =5$. I wish I knew why . $\endgroup$ – Ocelot Nov 21 '16 at 23:05

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