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Let $X$ be a random variable with $P(X=1) = P(X=-1) = \frac{1}{2}$. Using the law of total expectation, we have: $$ 0 = E(X) = E(X | X = 1) P(X=1) + E(X | X = -1)P(X=-1) = \frac{1}{2} E(X | X=1) + \frac{1}{2}E(X | X=-1) $$

The first term equals $\frac{1}{2}E(1 | X=1) = \frac{1}{2}$. Thus: $$ 0 = \frac{1}{2} + \frac{1}{2} E(X | X=-1). $$ Taking expectations on both sides, and using the Law of total expectation again gives: $$ 0 = E(0) = E\left(\frac{1}{2} + \frac{1}{2}E(X | X=-1)\right) = \frac{1}{2} + \frac{1}{2}E\left(E(X|X=-1)\right) = \frac{1}{2} + \frac{1}{2} E(X) = \frac{1}{2}. $$ Where is the mistake?

druduche
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  • Isn't $E(X | X = -1) = -1$? – Gregory Oct 20 '20 at 17:21
  • It is indeed. Still, why can't I apply the Law of total expectation on it? – druduche Oct 20 '20 at 17:31
  • You can, but it is a constant, $E(X | X = \pm 1)$ is constant, it doesn't depend on $X$. – Gregory Oct 20 '20 at 18:36
  • I agree. Yet, if I set $Y=1_{X=-1}$, the indicator of the event ${X=-1}$, I have $E(E(X|X=-1)) = E(E(X|Y)) = E(X) = 0$ applying the law of total expectation and the argument of https://math.stackexchange.com/questions/1427760/conditional-expectation-given-an-event-is-equivalent-to-conditional-expectation. I don't understand what goes wrong in this application, yet the result is obviously wrong. It should be -1, as you say. – druduche Oct 20 '20 at 19:07

1 Answers1

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You have to distinguish between the conditional expectation with respect to a $\sigma$-algebra and the conditional expectation with respect to an event.

Let $X$ be a random variable on some probability space $(\Omega,\mathcal{A}, \mathbb{P})$ (the $\sigma$-algebra $\mathcal{A}$ being the set of all events) such that $\mathbb{E}[|X|]<\infty$. Then, if $A\in \mathcal{A}$ is an event with $\mathbb{P}(A)>0$, the conditional expectation of $X$ given the event $A$ is defined by

$\mathbb{E}[X|A]=\frac{\mathbb{E}[X 1_A]}{\mathbb{P}(A)}$.

In your case, $A=\{X=1\}$ and $\mathbb{E}[X|X=1]=1$ $( A=\{X=-1\}$ and $\mathbb{E}[X|X=-1]=-1$, respectively). Note that $\mathbb{E}[X|A]$ is a scalar quantity.

The law of total expectation that you are wrongly using refers to conditional expectations with respect to a sub-$\sigma$-algebra, and not to some specific event.

If $\mathcal{F}\subseteq \mathcal{A}$ is a sub-$\sigma$-algebra, then the conditional expectation given $\mathcal{F}$, denoted by $\mathbb{E}[X|\mathcal{F}]$, is again a random variable. In this case, the law of total expectation reads

$\mathbb{E}[\mathbb{E}(X|\mathcal{F})]=E[X] $.

If we set $\mathcal{F}:=\sigma(X)$ (the $\sigma$-algebra generated by $X$) in your example, then one can show that

$\mathbb{E}(X|\mathcal{F})(\omega)=\mathbb{E}[X|X=1]\cdot 1_{\{X=1\}}(\omega)+\mathbb{E}[X|X=-1]\cdot 1_{\{X=-1\}}(\omega)$ for all $\omega \in \Omega$.

Applying the law of total expectation leads then to the formula you were first deriving:

$E[X]=\mathbb{E}[\mathbb{E}(X|\mathcal{F})]=\mathbb{E}[X|X=1]\cdot \mathbb{P}(\{X=1\})+\mathbb{E}[X|X=-1]\cdot \mathbb{P}(\{X=-1\})=1\cdot \frac{1}{2}+(-1)\cdot \frac{1}{2}=0$.

ym94
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  • Thanks a lot, that's exactly what I wanted to know. I thought it worked also for events, since in https://math.stackexchange.com/questions/1427760/conditional-expectation-given-an-event-is-equivalent-to-conditional-expectation they prove that $E[X|H] = E[X|\sigma(H)]$ where $H$ is an event. The catch is, they don't actually prove that in general, they only prove that $E[X|H]= EX|\sigma(H)$ for all (or almost all) $\omega \in H$. – druduche Oct 21 '20 at 12:35
  • Actually , they prove $EX|\sigma(H)=E[X|H]1_H(X(\omega))+E[X|H^c]1_{H^c}(X(\omega))$, but I see what your saying. Glad, that I could help :) – ym94 Oct 21 '20 at 13:22