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Conditional Normal Distribution (view source)
Revision as of 18:55, 7 April 2010
, 18:55, 7 April 2010New page: If <math>X \sim N(\mu_x,\sigma_x^2)\,</math> and <math>Y \sim N(\mu_y,\sigma_y^2)\,</math>, then we can use Bayes' Rule: :<math>f(X|Y) = \frac{f(XY)}{f(Y)} = \frac{f(Y|X)f(X)}{f(Y)}\,</...
If <math>X \sim N(\mu_x,\sigma_x^2)\,</math> and <math>Y \sim N(\mu_y,\sigma_y^2)\,</math>, then we can use
Bayes' Rule:
:<math>f(X|Y) = \frac{f(XY)}{f(Y)} = \frac{f(Y|X)f(X)}{f(Y)}\,</math>
where <math>f(XY)\,</math> will be bivariate normal:
:<math>
f(x,y)
=
\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}{\sigma_x})^2 +
(\frac{y\mu_y}{\sigma_y})^2 -
2 \rho (\frac{x-\mu_x}{\sigma_x})(\frac{y\mu_y}{\sigma_y})
\right]
\right)
</math>
where <math>\rho = \frac{\mathbb{E}XY}{\sigma_x \sigma_y}\,</math>, and
:<math>
\Sigma =
\begin{bmatrix}
\sigma_x^2 & \rho \sigma_x \sigma_y \\
\rho \sigma_x \sigma_y & \sigma_y^2
\end{bmatrix}.
</math>
To give that:
:<math>f(X|Y) \sim N (\mu_x + \rho \sigma_x \frac{y-\mu_y}{\sigma_y},\sigma_x \sqrt{1-\rho^2})\,</math>
Bayes' Rule:
:<math>f(X|Y) = \frac{f(XY)}{f(Y)} = \frac{f(Y|X)f(X)}{f(Y)}\,</math>
where <math>f(XY)\,</math> will be bivariate normal:
:<math>
f(x,y)
=
\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}{\sigma_x})^2 +
(\frac{y\mu_y}{\sigma_y})^2 -
2 \rho (\frac{x-\mu_x}{\sigma_x})(\frac{y\mu_y}{\sigma_y})
\right]
\right)
</math>
where <math>\rho = \frac{\mathbb{E}XY}{\sigma_x \sigma_y}\,</math>, and
:<math>
\Sigma =
\begin{bmatrix}
\sigma_x^2 & \rho \sigma_x \sigma_y \\
\rho \sigma_x \sigma_y & \sigma_y^2
\end{bmatrix}.
</math>
To give that:
:<math>f(X|Y) \sim N (\mu_x + \rho \sigma_x \frac{y-\mu_y}{\sigma_y},\sigma_x \sqrt{1-\rho^2})\,</math>
