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…

  • $\begingroup$ Ok, now I get it! thanks a lot $\endgroup$
    – Louis
    Apr 20 '17 at 14:20

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.