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let $\mathcal{G}$ be a sigma algebra and $X$,$Y$ be $r.vs$ satisfies $E[X|\mathcal{G}]$，$E[Y|\mathcal{G}]$ exists.Is there hold $E[XY|\mathcal{G}]=E[X|\mathcal{G}]E[Y|\mathcal{G}]$ in general?

the reason why i think above property is right is as follows.

When i prove $E[X_t|\mathcal{F_s}]=X_s$ in the proof that Itô integral of step process $f(t, \omega)$ in $L_{ad}^2([a,b] \times \Omega)$ is a martingale.($X_t=\int_a^tf(t,\omega)dB(t)$)

As $E[X_t|\mathcal{F_s}]=E[\int_a^sf(r,\omega)dB(r)|\mathcal{F_s}] + E[\int_s^tf(r,\omega)dB(r)|\mathcal{F_s}]=X_s+E[\int_s^tf(r,\omega)dB(r)|\mathcal{F_s}]$

we only need to prove $E[\int_s^tf(r,\omega)dB(r)|\mathcal{F_s}] = 0$

without losing generality.Assume $\int_s^tf(r,\omega)dB(r) = \sum_{j=1}^l \eta_{j-1}(B_{t_j} - B_{t_{j-1}})$.

As $\eta_{j-1}\in \mathcal{F_{t_{j-1}}}$ and $B_{t_j} - B_{t_{j-1}}$ independent of $\mathcal{F_{t_{j-1}}}$.

It is obvious that $E[\int_s^tf(r,\omega)dB(r)|\mathcal{F_s}] = 0$ is true if the property claimed in title is right.

As we known.$E[\int_s^tf(r,\omega)dB(r)|\mathcal{F_s}] = 0$.so i guess the property is right boldly.

Can u tell me whether it holds in above condition?

Thanks in advance!

vincen
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  • Why should it ? Do you have any reason to believe it? – Thomas Jun 24 '21 at 14:16
  • @Thomas i don't know whether it is right.i can find a new way to prove Itô Integral is a martingal if it holds. – vincen Jun 24 '21 at 14:24
  • @Thomas Or do you have some intuition about it that you can share with me.Thanks. – vincen Jun 24 '21 at 14:25
  • I will have a better look next days now I am a bit running. Maybe in the meanwhile there are also other contributions... but good update of the post! – Thomas Jun 24 '21 at 15:32
  • @Thomas Okey! have a good rest! – vincen Jun 24 '21 at 15:53
  • Nope unfortunately no. Look at the last answer in https://stats.stackexchange.com/questions/51322/does-independence-imply-conditional-independence – Shashi Jun 24 '21 at 15:57
  • However, you do not need the statement in the title to get what you are actually after. In fact, write $$E[\eta_{j-1}(B_{t_j}-B_{t_{j-1}})\mid \mathcal F_s]=E[E[\eta_{j-1}(B_{t_j}-B_{t_{j-1}})\mid \mathcal F_{t_{j-1}}]\mid \mathcal F_s]$$ using $\mathcal F_s\subset \mathcal F_{t_{j-1}}$. Then pull-out and finish accordingly. – Shashi Jun 24 '21 at 16:06
  • You mention "independence" in the title: did you mean to assume $X$ and $Y$ are independent? – Robert Israel Jun 25 '21 at 04:05

1 Answers1

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No, it's not true. We can modify the example from my answer here.

Consider an experiment with two coin flips, $\Omega = \{HH, HT, TH, TT\}$, where all outcomes have probability $1/4$. Let $A = \{HH, TT\}$ be th event that the coins are the same, and let $\mathcal{G} = \{A,A^c, \Omega, \emptyset\}$. Let $X,Y$ be Rademacher random variables corresponding to the first and second flips respectively, so $X(HH)=X(HT)=1$, $X(TH)=X(TT)=-1$, $Y(HH)=Y(TH)=1$, $Y(HT)=Y(TT)=-1$.

Now you may easily compute that $E[X|A]=E[X|A^c]=E[Y|A]=E[Y|A^c]=0$, so that $E[X|\mathcal{G}] = E[Y |\mathcal{G}] = 0$. On the other hand, $E[XY |A] = 1$ and $E[XY | A^c] = -1$. Thus $$E[XY | \mathcal{G}] = 1_A - 1_{A^c} \ne 0 = E[X| \mathcal{G}] E[Y |\mathcal{G}]$$ disproving the claim.

Nate Eldredge
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