Conditional Expectation Orthogonality and Variance Minimization

# Expectations in Infinite Probability Spaces with Sub Sigma-Algebras


# Expectations in Infinite Probability Spaces with Sub Sigma-Algebras












Let $X$ be an (integrable) random variable on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Suppose $\mathcal{G}$ is a sub-$\sigma$-algebra of $\mathcal{F}$ and let $Z=\mathbb{E}(X|\mathcal{G})$.

(a) Show that: $\mathbb{E}(X-Z)=0$

(b) Let $Y$ be an arbitrary $\mathcal{G}$-measurable random variable. Show that: $\mathbb{E}[(X-Z)(Z-Y)|\mathcal{G}]=0$

(c) Show that: $\mathbb{E}[(X-Z)(Z-Y)]=0$

(d) Suppose that $\mathbb{E}(X-Y)=0$. Show that: Var$(X-Z)\le$ Var$(X-Y)$.

Hint: Break up $X-Y = (X-Z)+(Z-Y)$.

The ideas that need to be proven all make sense to me intuitively, but I just don't know how to go about formalizing the actual proof itself in an infinite probability space. I tried approaching it from a finite probability space standpoint, but I don't think it's working.

## Answer by Cettt (score 2)

https://quant.stackexchange.com/a/51582

you are asking four questions at the same time. I will try to answer the first and maybe you can take it from there.

First of all we have to understand the definition of the conditional expectation give a sigma-field $Z=\mathbb{E}(X|\mathcal{G})$. $Z$ is a random variable which satisfies three properties (by definition):

- $Z$ is integrable, i.e. $\mathbb{E}(|Z|) < \infty$.

- $Z$ is $\mathcal{G}$-measurable.

- $\mathbb{E}(Z \cdot 1_G ) = \mathbb{E}(X \cdot 1_G )$ for all $G \in \mathcal G$. This is the most important property.

All of your questions can be solved using this definition.

For your first question we take property 3 and set $G = \Omega$. Then we get that $$ \mathbb{E}(Z) = \mathbb{E}(Z \cdot 1_\Omega ) = \mathbb{E}(X \cdot 1_\Omega ) = \mathbb{E}(X). $$ This is equivalent to $$ 0 = \mathbb{E}(X - Z) = \mathbb{E}(X) - \mathbb{E}(Z). $$

Hope this helps a little.

### Edit (part b)

In order to prove (b) you have to show that the random variable $W = 0$ has all three defining properties of $\mathbb{E}((X-Z)(Z-Y)|\mathcal{G})$. The first two are trivial to show. For the third property you have to show that $$ \mathbb{E}((X-Z)(Y-Z) 1_G) = \mathbb E(W 1_G) = 0, \quad \text{for all} \ G \in \mathcal{G}. $$

This can be done by a classical procedure: first you assume that $Y = 1_A$, then you assume that $Y$ is a simple function, than we assume that $Y$ is a positive random variable (monotone convergence), and finally we can show it for general $Y$.