I have $E(X|Y,Z)=0$, $X$ independent of $Y$ and of $Z$ and I want to conclude that $E(X)=0$ ($X,Y,Z$ are real-valued random variables). Okay it seems quite obvious, but if I try to make a strict argument I find a missing step.

I tried to approach this from both sides:

  • From the independence follows, that $X$ is independent of $\sigma(Y) \cup \sigma(Z)$.

  • The first expression is, more exactly $=E(X| \sigma(Y,Z))$. $\sigma(Y,Z)=\sigma(\sigma(Y) \cup \sigma(Z))$.

It is known, that in general $\sigma(\mathcal{A} \cup \mathcal{B}) \neq \mathcal{A} \cup \mathcal{B} (*)$ for $\sigma$-algebras $\mathcal{A}, \mathcal{B}$. Question: How can I conclude $E(X)=0$ anyway?

Maybe for independent random variables the equality $(*)$ does hold? Or there is m maybe an alternative, more convenient expression for conditional expectations on more than one random variables?

  • 1
    $\begingroup$ For every integrable random variable X and every sigma-algebra G, one of the two defining properties which characterize Y=E(X|G) is that E(X;A)=E(Y;A) for every A in G (you know the other one). Use this for A=Omega. $\endgroup$
    – Did
    Aug 4, 2011 at 12:48

2 Answers 2


Elaborating on the comment by Didier Piau. You don't even need an independence, since it's just simply a "tower rule": $$ \mathsf E[\mathsf E[X|\mathcal F]] = \mathsf EX $$ under some conditions on the integrability. In your simple case by the definition of the conditional expectation you have $$ \mathsf E[\eta\cdot \mathbf 1_A] = \mathsf E[X \mathbf 1_A] $$ if $A\in \sigma(Y,Z)$. Here $\eta = \mathsf E [X|Y,Z]$.

You know from the definition of $\sigma$-algebra that $\Omega\in \sigma(Y,Z)$ so $\mathbf 1_\Omega = 1$ and you have $$ 0 = \mathsf E[\eta] = \mathsf E[X] $$

Edited: answering your last comment. First, an example which justifies your intuition. Let $$ X = \begin{cases}1,\quad p=0.5,\\ 0,\quad p = 0.5.\end{cases} $$ and $A\subset\Omega = \{X = 0\}$, $\mathsf P(A) = 0.5$. By the definition of conditional expectation we have $$ \mathsf E[X|A] = \int\limits_\Omega X(\omega)\mathsf P(d\omega|A) = \frac{1}{\mathsf P(A)}\int\limits_A X(\omega)\mathsf P(d\omega) = 0 $$ while $\mathsf E[X] = 0.5$.

On the other hand, let us consider the random variable $\xi = \mathbf 1_A$. Clearly, $\mathcal F_\xi = \{\emptyset,A,A^c,\Omega\}$ and you would like to have the same result as before. But there is a slightly difference. $$ \mathsf E[X\mid\mathcal F_\xi] = 1-\xi $$ and we cannot say that $1-\xi = 0 \quad\mathsf P$-a.s. That's way you don't have that $\mathsf E[X] = \mathsf E[X\mid\mathcal F_\xi] = 0$.

Roughly speaking, the condition $\mathsf E[X\mid \mathcal F_\xi] = 0$ says that it will hold independently of the value of $\xi$ - so in all cases which can be characterized by $\xi$.

  • $\begingroup$ very nice, clear argument $\endgroup$
    – Johannes L
    Aug 5, 2011 at 7:41
  • $\begingroup$ On a second thought, this shows that $E(X | \mathcal{F})=0$ implies $E(X)=0$ and it seems counter-intuitive to me, if I view the first expression as the expectation given the "information" in $\mathcal{F}$ and in the second I don't consider that "information". How would you interpret the result? $\endgroup$
    – Johannes L
    Aug 5, 2011 at 8:38
  • $\begingroup$ @Johannes L: here you are. $\endgroup$
    – SBF
    Aug 5, 2011 at 9:02

See law of total expectation: $$ E(E(X \mid Y)) = E(X). $$ So that does it.


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