Proving $\mathbb{D}(\xi|\mathcal{D})=\mathbb{E}(\xi^2|\mathcal{D}) -(\mathbb{E}(\xi|\mathcal{D}))^2 $

Question

The conditional dispersion of random variable $\xi$ according $\sigma$-algebra $\mathcal{D}$ is a random variable: $\mathbb{D}(\xi|\mathcal{D})=\mathbb{E}((\xi - \mathbb{E}(\xi|\mathcal{D}))^2|\mathcal{D}).$ I need to show that next to equalities is correct

$\mathbb{D}(\xi|\mathcal{D})=\mathbb{E}(\xi^2|\mathcal{D}) -(\mathbb{E}(\xi|\mathcal{D}))^2 $

and

$\mathbb{D}\xi=\mathbb{E}\mathbb{D}(\xi|\mathcal{D})+\mathbb{D}\mathbb{E}(\xi|\mathcal{D})$.

So how to begin and what properties to use?

Answer 1

First, expand out the square: $$ \mathbb D(\xi | \mathcal D) = \mathbb E \left( \xi^2 - 2 \xi \mathbb E(\xi | \mathcal D) + \mathbb E (\xi | \mathcal D)^2 \mid \mathcal D \right).$$

Then use:

• Linearity:

$$ \mathbb E(X_1 + X_2 | \mathcal D) = \mathbb E(X_1 | \mathcal D) + \mathbb E(X_2 | \mathcal D), \ \ \ \ \mathbb E(aX|\mathcal D)=a\mathbb E(X|\mathcal D).$$

• Pulling out known factors: if $X$ is $\mathcal D$-measurable, then $$ \mathbb E(XY| \mathcal D) = X \ \mathbb E(Y | \mathcal D).$$

(And remember, $\mathbb E(\xi | \mathcal D)$ is $\mathcal D$-measurable, by definition. Hence $\mathbb E(\xi | \mathcal D)^2$ is $\mathcal D$-measurable too.)