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June 14th, 2017, 03:16 AM   #1
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Conditional distribution of X given Y when both X and Y are gaussian

Two Random variables X and Y are gaussian.Is the conditional distribution of X given Y that is X|Y also a gaussian when Y lies in certain range from a<=y<=b?I know its true when y has a specific value but what if it lies in a certain range as mentioned above?
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June 14th, 2017, 07:48 AM   #2
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I'm not quite sure what you mean Y being in a certain range... If Y is Gaussian, its support should be $\displaystyle \mathbb{R}$, so if you know Y lies in a proper subset of $\displaystyle \mathbb{R}$, it probably can't be Gaussian any more...

Without the restriction, the conditional distribution X | Y would be

$\displaystyle
\frac{
\frac{1}{2 \pi \sigma_X \sigma_Y \sqrt{1-\rho^2}}
\exp\left(
-\frac{1}{2(1-\rho^2)}\left[
\frac{(x-\mu_X)^2}{\sigma_X^2} +
\frac{(y-\mu_Y)^2}{\sigma_Y^2} -
\frac{2\rho(x-\mu_X)(y-\mu_Y)}{\sigma_X \sigma_Y}
\right]
\right)
}{\frac{1}{\sqrt{2\pi}\sigma_Y}\exp\left(
-\frac{(y-\mu_Y)^2}{2\sigma_Y^2}
\right)}
$

which simplifies to

$\displaystyle \frac{1}{\sqrt{2 \pi} \sigma_X \sqrt{1-\rho^2}}
\exp\left(
-\frac{1}{2(1-\rho^2)}\left[
\frac{(x-\mu_X)^2}{\sigma_X^2} +
\frac{(y-\mu_Y)^2}{\sigma_Y^2} -
\frac{2\rho(x-\mu_X)(y-\mu_Y)}{\sigma_X \sigma_Y} \right]+\frac{(y-\mu_y)^2}{2\sigma_Y^2}
\right)$

and then
$\displaystyle \frac{1}{\sqrt{2 \pi} \sigma_X \sqrt{1-\rho^2}}
\exp\left(
-\frac{1}{2(1-\rho^2)\sigma_X^2}\left[
(x-\mu_X)^2 +
\frac{\rho^2\sigma_X^2(y-\mu_Y)^2}{\sigma_Y^2} -
\frac{2\sigma_X\rho(x-\mu_X)(y-\mu_Y)}{ \sigma_Y} \right]
\right)$

which finally leaves us with

$\displaystyle \frac{1}{\sqrt{2 \pi} \sigma_X \sqrt{1-\rho^2}}
\exp\left(
-\frac{1}{2(1-\rho^2)\sigma_X^2}\left[
x-\mu_X - \frac{\rho\sigma_X(y-\mu_Y)}{\sigma_Y} \right]^2
\right)$

which means $\displaystyle X | Y \sim N(\mu_X + \frac{\rho\sigma_X(y-\mu_Y)}{\sigma_Y}, (1-\rho^2)\sigma_X^2)$.

Phew! That took me a long time to type.

Last edited by 123qwerty; June 14th, 2017 at 08:04 AM.
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