# conditional expectation respect to a discrete random variable

Let $(\Omega, \mathscr F, \mathsf P)$ be a probability space and $\mathscr G$ a sub$\sigma$-algebra of $\mathscr F$. Given a random variable $X$ (that can be assumed to be integrable) one can define $\mathsf E(X|\mathscr G)$. In particular, for any event $A \in \mathscr F$ one puts $\mathsf P(A|\mathscr G)=\mathsf E(1_A|\mathscr G)$.

If $\mathscr G$ is generated by a countable partition $\{\Lambda_i:i \in \mathbb N\}$ of $\Omega$ with $\mathsf P(\Lambda_i)>0$ for every $i$ (so that every member of $\mathscr G$ is an union of some $\Lambda_i$), then one has $$\mathsf E(X|\mathscr G)=\sum_{i=1}^{+\infty}\left(\frac{1}{\mathsf P(\Lambda_i)}\int_{\Lambda_i} X \;\mathrm d \mathsf P\right)\; 1_{\Lambda_i}$$ and $$\mathsf P(A|\mathscr G)=\sum_{i=1}^{+\infty} \frac{\mathsf P(A \cap \Lambda_i)}{\mathsf P(\Lambda_i)}\;1_{\Lambda_i}.$$

If $\mathscr G$ is generated by another random variable $Y$, i.e. $\mathscr G=\sigma(Y)$, one writes $\mathsf E(X|Y)$ instead of $\mathsf E(X|\mathscr G)$. Futhermore from the characterization of $\sigma(Y)$-measurablity, it follows that there exists a borel function $\varphi \colon \mathbb R \to \mathbb R$ such that $\mathsf E(X|Y)= \varphi \circ Y$ and one usually puts $\mathsf E(X|Y=y)=\varphi(y)$. Similarly one defines also $\mathsf P(A|Y=y)=\mathsf E(1_A|Y=y)$.

Now if $Y$ is discrete and takes only the of values $\{y_i: i \in \mathbb N\}$ with positive probability, then $\sigma(Y)$ coincides with the $\sigma$-algebra generated by the partition given by the sets $\{Y=y_i\}$ for $i \in \mathbb N$. So we have that $$\mathsf E(X|Y)=\sum_{i=1}^{+\infty}\left(\frac{1}{\mathsf P(Y=y_i)}\int_{\{Y=y_i\}} X \;\mathrm d \mathsf P\right)\; 1_{\{Y=y_i\}}$$ $$\mathsf P(A|Y)=\sum_{i=1}^{+\infty} \frac{\mathsf P(A \cap \{Y=y_i\})}{\mathsf P(Y=y_i)}\;1_{\{Y=y_i\}}.$$

Question: How can I recognize from the last two formulas the expressions of $\mathsf E(X|Y=y)$ and $\mathsf P(A|Y=y)$? I know that these expressions of the function $\varphi$ turn out to be $$\mathsf E(X|Y=y)=\frac{1}{\mathsf P(Y=y_i)}\int_{\{Y=y_i\}} X \;\mathrm d \mathsf P\:\:\:\:\:\:\:\:\:\mbox{for y=y_i and 0 otherwise}$$ $$\mathsf P(A|Y=y)=\frac{\mathsf P(A \cap \{Y=y_i\})}{\mathsf P(Y=y_i)}\:\:\:\:\:\:\:\:\:\mbox{for y=y_i and 0 otherwise}.$$

This fact would establish the agreement with the naive usage of conditional probability.

I will change your indeces to avoid confusions… You are right with

$$\mathsf P(A|Y)=\sum_{k=1}^{+\infty} \frac{\mathsf P(A \cap \{Y=y_k\})}{\mathsf P(Y=y_k)}\;1_{\{Y=y_k\}}$$

and additionally it holds $P(A|Y=y) = P(A|Y)|_{Y=y}$ So literally $P(A|Y=y)$ is $P(A|Y)$ where we plug in $y$ for $Y$ and we get:

$$P(A|Y=y) = P(A|Y)|_{Y=y} = \sum_{k=1}^{+\infty} \frac{\mathsf P(A \cap \{y=y_k\})}{\mathsf P(y=y_k)}\;1_{\{y=y_k\}}$$

Is $y = y_i$ all summands equal zero insted for $k=i$ and we get:

$$P(A|Y=y) = P(A|Y)|_{Y=y} = \frac{\mathsf P(A \cap \{Y=y_i\})}{\mathsf P(Y=y_i)}$$ if $y = y_i$ and $0$ otherwise

The argumentation for the second term is the same…

• Ok, now I get it! thanks a lot Apr 20 '17 at 14:20