Let $X_1$ and $X_2$ are Jointly Gaussian Random Variables(Which Implies They are Individually Gaussian) i.e.,

$$ X_1 \sim \mathcal{N}(0,\sigma_1^2)$$ and

$$X_2 \sim \mathcal{N}(0,\sigma_2^2)$$, with Correlation Coefficient as $\rho$.

Then we Know that :

$$f_{X_1X_2}(x_1,x_2)=\\ \\ \left(\frac{1}{2 \pi \sigma_1 \sigma_2 \sqrt{1-\rho^2}} \right)\times

\mathbf{e^{\frac{-1}{2(1-\rho^2)}\left(\frac{x_1^2}{\sigma_1^2}-\frac{2 \rho x_1 x_2}{\sigma_1 \sigma_2}+\frac{x_2^2}{\sigma_2^2} \right )}}\\ \\ \;\;\;\; \forall\; x_1,x_2 \in (-\infty\; \infty)\\

\\ $$

Prove that :

$$ E(X_2|X_1)= \frac{\rho x_1 \sigma_2}{\sigma_1} $$

$$ Var(X_2|X_1)=\sigma_2^2\,(1-\rho^2) $$

$$ X_1 \sim \mathcal{N}(0,\sigma_1^2)$$ and

$$X_2 \sim \mathcal{N}(0,\sigma_2^2)$$, with Correlation Coefficient as $\rho$.

Then we Know that :

$$f_{X_1X_2}(x_1,x_2)=\\ \\ \left(\frac{1}{2 \pi \sigma_1 \sigma_2 \sqrt{1-\rho^2}} \right)\times

\mathbf{e^{\frac{-1}{2(1-\rho^2)}\left(\frac{x_1^2}{\sigma_1^2}-\frac{2 \rho x_1 x_2}{\sigma_1 \sigma_2}+\frac{x_2^2}{\sigma_2^2} \right )}}\\ \\ \;\;\;\; \forall\; x_1,x_2 \in (-\infty\; \infty)\\

\\ $$

Prove that :

$$ E(X_2|X_1)= \frac{\rho x_1 \sigma_2}{\sigma_1} $$

$$ Var(X_2|X_1)=\sigma_2^2\,(1-\rho^2) $$

By Simple Algebra, The Random Variable $X_2|X_1$ is also Gaussian Distributed, with Mean and Variance Given above.

ReplyDelete