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Consider for the trvial $\sigma$ - field $\mathcal{F}_0 = \{\emptyset , \Omega\}$, What is Conditional expectation of the following in the following cases when $A = \emptyset$ and $A = \Omega$ ???

? Can someone please help me fill in the ?? below, as this would help improve my understanding a lot ?

$$\int_? E[X | \mathcal{F}_0]1_A dP = ? \;\; \forall A \in \mathcal{F}_0$$

Question 2: And What if I just condition on the $\sigma$-field,

$$ \int_? E[X | \mathcal{F}] dP = ? $$

For the second question, I guess it is = X right? Since X is already $\mathcal{F}$- measurable by definition of random variable, if given the $\sigma$ - feld $\mathcal{F}$, everything is known, there is no randomness in X.

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I take it that you mean $E[X\mid\mathcal{F}]$ instead of $E[X\mid\Omega]$ and also $E[X\mid \{\emptyset,\Omega\}]$ instead of $E[X\mid\emptyset]$, where $\mathcal{F}$ is the sigma field on your probability space? –  Stefan Hansen Dec 1 '12 at 12:34
    
Yes, I forgot to put brackets {} sorry –  user1769197 Dec 1 '12 at 12:49
    
I m so sorry guys. I have made a notation problem in my question. As I only a student starting to pick up probability, please bear with me. –  user1769197 Dec 1 '12 at 12:58
    
@user1769197 To your edited question: As you can see from the answers below, $E[X|\mathcal{F}_0]=E[X]$, hence a constant. Therefore $\int E[X]\mathbf1_AdP=E[X]$ if $A=\Omega$ or $0$ if $A=\emptyset$. In the second case you know $E[X|\mathcal{F}]=X$ hence $\int E[X|\mathcal{F}]1_AdP=\int_A XdP$ for all $A\in \mathcal{F}$ –  math Dec 1 '12 at 13:10

3 Answers 3

up vote 3 down vote accepted

Let $(\Omega,\mathcal{F},P)$ be a probability space. Suppose $X$ is an integrable random variable and let $\mathcal{G}$ be a sub-sigma-field of $\mathcal{F}$. The conditional expection $E[X\mid\mathcal{G}]$ is the unique random variable that satisfies:

1) $E[X\mid\mathcal{G}]$ is $\mathcal{G}$-measurable.

2) $\int_A E[X\mid\mathcal{G}]\,\mathrm dP = \int_A X\,\mathrm dP$ for all $A\in\mathcal{G}$.

I'm assuming you want to find expressions for $E[X\mid\mathcal{F}]$ and $E[X\mid\{\emptyset,\Omega\}]$. For the first conditional expectation, try showing that $X$ satisfies 1) and 2), and for the last conditional expectation, try with $E[X]$.


One some times sees $E[X\mid\mathcal{G}]$ as our best guess of $X$ given the information contained in $\mathcal{G}$. Try holding this up with the expectations when $\mathcal{G}=\mathcal{F}$ and $\mathcal{G}=\{\emptyset,\Omega\}$.

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The conditional expectation over the trivial sigma field would be the Expectation of the random variable. To see this, let $\mathcal{F}$ be the trivial sigma field. Then by def of conditional expectation, $$E[E[X|\mathcal{F}]1_A] = E[X1_A] \quad \forall A \in \mathcal{F}$$ But A can take on only two values i.e. $\{\emptyset,\Omega\}$. Hence substitute $A=\Omega$ to get $E[X|\mathcal{F}] = E[X]$ as a candidate solution which works.

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What if you substitute A = emptyset, does E[X | sigma-feld] = E[X] or 0 ?? –  user1769197 Dec 1 '12 at 12:56
    
In that case you get 0=0. This doesn't tell you anything. However when you put omega, E[X] is the only thing a.e that will work. –  Gautam Shenoy Dec 1 '12 at 12:57
    
I know this is a silly question . But why does the LHS becomes 0 when A = emptyset ? Is it because there can be nothing in the empty set, so 1_A = 0 ??? –  user1769197 Dec 1 '12 at 13:09
    
$1_{\phi} = 0$ always because by definition, it is 1 if $\omega \in \phi$. But $\phi$ is empty so $\omega \notin \phi$. Hence 0. –  Gautam Shenoy Dec 1 '12 at 13:12
    
ok Thanks. I get it now. Thanks a lot –  user1769197 Dec 1 '12 at 13:27

As Stefan Hansen commented, I think you mean $E[X|\mathcal{F}]$ and $E[X|\{\emptyset,\Omega\}]$. Note that by definition, we have: For a r.v. $X$ (integrable) on a probability space $(\Omega,\mathcal{F},P)$ and a sub $\sigma-$algebra $\mathcal{G}$ of $\mathcal{F}$, the conditional expectation is the unique $\mathcal{G}$ measurable and integrabel r.v. $Z$ such that

$$E[X\mathbf1_G]=E[Z\mathbf1_G]$$

for all $G\in \mathcal{G}$. Usually one writes $Z:=E[X|\mathcal{G}]$. It follows immediately, in your case, that $E[X|\mathcal{F}]=X$ and $E[X|\{\emptyset,\Omega\}]=E[X]$. For the first, just check the definition.

For the second you may use that if you have $G_i\in \mathcal{F}$ for $1\le i\le N\le \infty$ pairwise disjoint with $P(G_i)>0$ and $\bigcup G_i=\Omega$ and let $\mathcal{G}=\sigma(G_i,1\le i\le N)$ then $E[X|\mathcal{G}]=\sum_{i=1}^NE[X|G_i]\mathbf1_{G_i}$. Apply it for $N=1$, i.e. $G_1=\Omega$.

EDIT: By the special structure of $\mathcal{G}$ you know $E[X|F_i]=E[X\mathbf1_{G_i}]/P(G_i)$

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Note that there are "easier" ways to prove $E[X|\mathcal{F}_0]=E[X]$. But I wanted to mention the example of having a partition of $\Omega$. This is one of the most important application in discrete things (e.g. Binomial models for stocks etc.) –  math Dec 1 '12 at 13:28
    
so what are some easier ways to prove $E[X | \mathcal{F}_0] = E[X]$??? –  user1769197 Dec 2 '12 at 20:39
    
@user1769197 just prove it directly, as it was done by Guatam Shenoy –  math Dec 3 '12 at 9:29

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