My Math Forum Conditional distribution of X given Y when both X and Y are gaussian

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 June 14th, 2017, 03:16 AM #1 Newbie   Joined: Jun 2017 From: Singapore Posts: 1 Thanks: 0 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?
 June 14th, 2017, 07:48 AM #2 Senior Member   Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics 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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