I'm trying to compute a conditional expectation. If $(\Omega, \mathcal{F}, P)$ denotes a probability space, and let $A, B\in\mathcal{F}$ with $0<P(B)<1$ and let $\mathcal{G}=\{ B, \Omega\backslash B, \Omega, \emptyset\}$. I am then asked to compute $E(1_A\mid \mathcal{G})$.

I figure this is doable, simply computing the $1_B$- and $1_{\Omega\backslash B}$-parts separately. This would yield

$$E(1_A\mid \mathcal{G})=\frac{P(A\cap B)}{P(B)}1_B+\frac{P(A\cap (\Omega\backslash B))}{P(\Omega\backslash B)}1_{\Omega\backslash B}.$$

Now, this leaves me with two questions: First of all, is this even correct reasoning? And more importantly, if this is correct, how would I go about explaining my reasoning to the professor? Isn't there a more explicit way of computing this expectation?


If you could 'guess' that the conditional expectation $E[1_A\mid \mathcal{G}]$ should be of the form (here $B^c=\Omega\setminus B$) $$ E[1_A\mid \mathcal{G}]=\frac{P(A\cap B)}{P(B)}1_B+\frac{P(A\cap B^c)}{P(B^c)}1_{B^c}, $$ then you could simply check that the right-hand side fulfills the requirements for being the conditional expectation of $1_A$ given $\mathcal{G}$. That is:

  • Is the right-hand side $\mathcal{G}$-measurable,
  • Is it $P$-integrable,
  • Is $E[\text{RHS}\cdot 1_C]=E[1_A\cdot 1_C]$ for all $C\in\mathcal{G}$.

This is pretty straight-forward to check.

If you can't 'guess' that the conditional expectation should be of that form, then you have to start by looking at $\mathcal{G}$. A variable $X$ is measurable with respect to $\mathcal{G}$ if and only if it is constant on $B$ and $B^c$. That is $$ E[1_A\mid\mathcal{G}]=a\cdot 1_B+b\cdot 1_{B^c} $$ for some constants $a,b\in\mathbb{R}$. This is also clearly $P$-integrable, so we have to use bullet number 3 above to determine $a$ and $b$. Now, bullet number 3 states that for all $C\in\mathcal{G}$ we have $$ P(A\cap C)=E[1_A\cdot 1_C]=E[(a\cdot 1_B+b\cdot 1_{B^c})\cdot 1_C]=aP(B\cap C)+bP(B^c\cap C). $$ Now let $C$ 'walk through' $\mathcal{G}$ and determine $a$ and $b$.


this is correct and your reasoning is right. Conditional expectionas are just the classical conditional probabilities when the sample space can be partitioned into 'building blocks'. This is the most explicit thing you could have written down. What else could you have wanted?

If you want, you can verify your solution. You know: the definition of conditional expectation says that

when you take expection of your original random variable and its conditional expectation on $\mathcal{G}$ measurable set, they must agree

the coditional expectation is $\mathcal{G}$-measurable.

Well it is pretty clear to me that is the case with your solution. since conditional expectations are unique, you are done!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.