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Prove that $\mathscr F$ and $\mathscr G$ are independent iff $\forall$ bounded $X_F \in m\mathscr F$ and $\forall$ bounded $X_G \in m\mathscr G$,

$E[X_G X_F] = E[X_G] E[X_F]$

(Hope my iff statement is right)

  1. Why is boundedness needed? Why not just integrability?

  2. What I tried:

'only if'

If $\mathscr F$ and $\mathscr G$ are independent, then $\sigma(X_F)$ and $\sigma(X_G)$ are independent. Then we have $E[X_G X_F] = E[X_G] E[X_F]$. I don't see how integrability doesn't work there. (*)

'if'

Choose $X_F = 1_F, X_G = 1_G \forall F \in \mathscr F, \forall G \in \mathscr G$. Then we have

$$P(F \cap G) = P(F)P(G), \forall F \in \mathscr F, \forall G \in \mathscr G \ QED$$

Afaik, $1_F$ and $1_G$ are integrable $\forall F \in \mathscr F, \forall G \in \mathscr G$

Is that right? If not, how else can I approach this?

(*) Class notes

enter image description here

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BCLC
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1 Answers1

2

Yeah, your proof is correct. Regarding your first question concerning integrability: If you prefer, you can restate the result as follows.

The following statements are equivalent:

  1. $\mathcal{F}$ and $\mathcal{G}$ are independent.
  2. For all bounded $X_F \in m \mathcal{F}$ and bounded $X_G \in m \mathcal{G}$, it holds that $\mathbb{E}(X_G X_F ) = \mathbb{E}(X_G) \mathbb{E}(X_F)$.
  3. For all $X_F \in L^1(\mathcal{F})$ and $X_G \in L^1(\mathcal{G})$ such that $X_F \cdot X_G \in L^1$ it holds that $\mathbb{E}(X_G X_F ) = \mathbb{E}(X_G) \mathbb{E}(X_F)$

The "only if" of your proof shows "(1) $\implies$ (3)" and the "if"-part proves "$(2) \implies (1)$". The remaining implication, i.e. "$(3) \implies (2)$", is obvious (since any bounded random variable is integrable).

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saz
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  • Thanks saz. 1 I think I see it now: integrability of $X_F$ and integrability of $X_G$ are not sufficient because we don't know if $X_F X_G$ is integrable. However if $X_F$ and $X_G$ are independent, they are integrable. Hence, we cannot replace '$X_F, X_G$ bounded' w/ '$X_F, X_G$ integrable'? – BCLC Nov 25 '15 at 16:51
  • 2 $(3) \to (2)$ is obvious? How? Not all integrable functions are bounded right? – BCLC Nov 25 '15 at 16:52
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    @BCLC Yes, to your first comment. The second one: Yes, not all integrale fnctions are bounded, but I didn't claim that this holds. To show "$(3) \implies (2)$, we have to show that $(2)$ holds if $(3)$ holds. So take bounded $X_F$ and bounded $X_G$, then both are integrable and also the product is bounded, hence integrable. Hence, by (3), $$\mathbb{E}(X_G X_F) = \mathbb{E}(X_G) \mathbb{E}(X_F)$, i.e. (2) holds. – saz Nov 25 '15 at 17:06
  • Ah, since it holds for all integrable, then it holds for all bounded. Thanks saz ^-^ Wasn't obvious (which is what you claimed) to me until just now :P – BCLC Nov 25 '15 at 17:20
  • Finishing the $\LaTeX$: $$\mathbb{E}(X_G X_F) = \mathbb{E}(X_G) \mathbb{E}(X_F)$$ – BCLC Nov 25 '15 at 18:15