5
$\begingroup$

I am confused between these two equations which use the law of total expectation: $$E(X|Y)=E(X|Y,Z)P(Z)+E(X|Y,Z')P(Z')$$ $$E(X|Y)=E(X|Y,Z)P(Z|Y)+E(X|Y,Z')P(Z'|Y)$$ where Z' is the complement of Z. The first one makes sense to me because if one defines the random variable $A=X|Y$, then it is simply using the law of total expectation. The other also makes sense as it is as if we are applying the law of total probability on $X$ but then reducing the universe to the "given Y" subspace. Which one is right?

In general, is thinking of X given Y and Z the same as X given Y, given Z? $$P(X|Y,Z)=P((X|Y)|Z)$$

$\endgroup$
1
  • 4
    $\begingroup$ What does “not $Z$” mean? If $Z$ is a real valued random variable then the logical negation of $Z$ is nonsense. If $Z$ is an event, then $Z’$ is the complement of this event with respect to the sample space. (Further, many authors discourage the notation $A=X|Y$ to denote a “conditional RV”. Pedantically there is no such thing as a “conditional RV” rather we talk about RVs having a conditional distribution conditional on some event, some RV, or in general, some $\sigma$-algebra.) $\endgroup$ Nov 15, 2019 at 20:47

1 Answer 1

7
$\begingroup$

Your second equation is correct, and the first one is wrong. Here is an example.

In my example, $X,Y,Z$ are events, i.e. subsets of the sample space $\Omega$. If you wish to have them be random variables, then consider their indicator variables. As is well known, $P(A) = E[1_A]$

  • Pick a number $\in \Omega = \{1,2,3\}$, uniformly randomly, i.e each with equal prob ${1 \over 3}$.

  • $X = \{1\}, Y = \{1,2\}, Z = \{2,3\}$

  • $Y\cap Z = \{2\}$ and $Y \cap Z' = \{1\}$

  • $P(X\mid Y) = 1/2$

  • $P(X\mid Y,Z) = 0$ and $P(X \mid Y,Z') = 1$

  • 1st RHS $= P(X \mid Y,Z) P(Z) + P(X\mid Y,Z') P(Z') = 0\times \frac23 + 1 \times \frac13 = \frac13 \neq P(X \mid Y)$

  • 2nd RHS $= P(X \mid Y,Z) P(Z \mid Y) + P(X\mid Y,Z') P(Z' \mid Y) = 0 \times \frac12 + 1 \times \frac12 = \frac12 = P(X \mid Y)$


More generally, lets expand:

$$ \begin{array}{} P(X \mid Y) &= {P(X, Y) \over P(Y)}\\ &={P(X,Y,Z) + P(X,Y,Z') \over P(Y)}\\ &={P(X \mid Y,Z) P(Y,Z) + P(X \mid Y,Z') P(Y, Z') \over P(Y)}\\ &=P(X\mid Y,Z) {P(Y,Z) \over P(Y)}+ P(X \mid Y,Z') {P(Y,Z') \over P(Y)}\\ &=P(X \mid Y,Z) P(Z \mid Y) + P(X \mid Y,Z') P(Z' \mid Y) \end{array} $$

The way I remember is that conditioning on $Y$ creates its own probability law, so the correct equations look like Law of Total Probability/Expectation but within the event $Y$, i.e. every term is conditioned on $Y$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .