"Let $X$ be a discrete random variable on a discrete sample space. Let P be a partition of that sample space.
a) Show that $E[X|P]$ is the unique P-measurable random variable $Y$, ie $Y$ = $E[X|P]$, such that for every $A$ ∈ P, we have $E[1$A$X] = E[1$A$Y]$.
($1$A denotes the indicator function of $A$).
b) Prove that $Y = E[X|P]$ is a minimizer of the function $Y$ → $E[(X-Y)$2] defined on the set of P-measurable random variables."
For a), I'm not sure how to prove uniqueness, though I see how to prove that the equality holds.
For b), we know that this will also be the unique minimizer, I'm not sure if that's of any help here.
Thanks for your help.
Consider the discrete probability space $(\Omega, 2^{\Omega},\operatorname {Pr})$.
In the case of a discrete random variable and that of a discrete partition, the conditional expectation can be directly defined via the corresponding conditional probabilities. Let $P$, the partition be $$P_j, \ \ j=1,2,\dots, \bigcup_{j=1,2,\dots}P_j=\Omega, \ \ P_i\cap P_j=\emptyset\ \text{ if }\ i\not =j.$$
$X$ itself defines a partition: Let $$A_i=\{\omega: X(\omega)=i\}, \ \ i=1,2,\dots \ \ $$
The conditional probabilities mentioned are defined as follows $$\operatorname {Pr}(X=i\mid P_j)=\frac{\operatorname {Pr}(A_i\cap P_j)}{\operatorname {Pr}(P_j)}.\tag 1$$
Then the conditional expectation is defined as follows $$E[X\mid P\ ](\omega)=\sum_{i=1,2,\dots}i\operatorname {Pr}(X=i\ \mid\ P_j)\ \text{ if }\ \omega\in P_j.$$
This is unique and is defined with probability $1$. If for a $j\ $ $\operatorname {Pr}(P_j)=0$ then $(1)$ is not defined. But the probability of that is $0$.