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Given prob space $(\Omega, \mathscr{F}, P)$ and a Wiener process $(W_t)_{t \geq 0}$, define filtration $\mathscr{F}_t = \sigma(W_u : u \leq t)$

Let $(A_t)_{t \geq 0}$ where $A_t = W_t^3 - 3tW_t$. Show that $E[A_t|\mathscr{F}_s] = A_s$ whenever $s < t$.

I think this all comes down to manipulation since there are martingales somewhere

My attempt:

Splitting up into $E[W_t^3|\mathscr{F}_s] - 3E[tW_t|\mathscr{F}_s]$ doesn't do anything since those guys aren't martingales? So, I tried splitting it up into:

$E[W_t(W_t^2 - 3t)|\mathscr{F}_s]$

$= E[W_t(W_t^2 - t -2 t)|\mathscr{F}_s]$

$= E[W_t(W_t^2 - t) -2 tW_t)|\mathscr{F}_s]$

$= E[W_t(W_t^2 - t)|\mathscr{F}_s] -2E[ tW_t|\mathscr{F}_s]$

$W_t$ is not $\mathscr{F}_s$-measurable, so we can't take that out...

$tW_{1/t}$ is Brownian and thus a martingale, but I don't know about $tW_t$...

$cW_{t/c^2}$ is Brownian and thus a martingale, but I don't think we can set c = t...

Help please?

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    $\begingroup$ Unable to identify $E[tW_t|\mathscr{F}_s]$? Well, we have a problem of learning curve here... $\endgroup$ – Did Oct 7 '14 at 18:24
  • $\begingroup$ @Did Is it Wiener? $\endgroup$ – BCLC Oct 7 '14 at 18:25
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    $\begingroup$ @BCLC What's the variance of $t W_t$? $\endgroup$ – saz Oct 7 '14 at 19:10
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    $\begingroup$ @BCLC ?!?!?!?!?! $\endgroup$ – saz Oct 9 '14 at 17:51
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    $\begingroup$ That's what Did was after. My thought was different: Since $\text{var} \, (t W_t) = t^3$, $(tW_t)_{t \geq 0}$ cannot be a Brownian motion. $\endgroup$ – saz Oct 9 '14 at 17:56
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Writing

$$W_t^3 = ((W_t-W_s)+W_s)^3 = (W_t-W_s)^3+ 3 W_s (W_t-W_s)^2 + 3 W_s^2 (W_t-W_s) + W_s^3$$

we find using the independence of the increments

$$\begin{align*} \mathbb{E}(W_t^3 \mid \mathcal{F}_s) &= \underbrace{\mathbb{E}((W_t-W_s)^3)}_{\mathbb{E}(W_{t-s}^3)=0} + 3W_s \underbrace{\mathbb{E}(W_t-W_s)^2}_{\mathbb{E}(W_{t-s}^2)=t-s} + 3W_s^2 \underbrace{\mathbb{E}(W_t-W_s)}_{0}+W_s^3 \\ &= 3(t-s) W_s + W_s^3. \end{align*}$$

Consequently,

$$\mathbb{E}(A_t \mid \mathcal{F}_s) = \mathbb{E}(W_t^3 - 3t W_t \mid \mathcal{F}_s) = 3(t-s)W_s+ W_s^3 - 3t \underbrace{\mathbb{E}(W_t \mid \mathcal{F}_s)}_{W_s} = A_s.$$

A general remark: If you want to prove a process of this form a martingale, it is always a good idea to write $$W_t = (W_t-W_s)+W_s$$ and separate both terms, since this allows us to use the independence of the increments (which in turn makes calculation of conditional expectations much more easier).

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  • $\begingroup$ Satisfying the martingale property makes it martingale? Aren't there other conditions or something? $\endgroup$ – BCLC Oct 7 '14 at 22:29
  • $\begingroup$ Like integrable or something? Oh and thanks! $\endgroup$ – BCLC Oct 7 '14 at 22:30
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    $\begingroup$ @BCLC Yes, it has to be integrable. Note that $\mathbb{E}(W_t^k)<\infty$ for any $k$ as $W_t \sim N(0,t)$. $\endgroup$ – saz Oct 8 '14 at 5:50
  • $\begingroup$ Why is $E[(W_t-W_s)^3|\mathcal{F}_s] = E[(W_t-W_s)^3]=0$? $\endgroup$ – user428487 Apr 3 '18 at 8:09
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    $\begingroup$ @user428487 Because $W_t-W_s$ is independent from $\mathcal{F}_s$. $\endgroup$ – saz Apr 3 '18 at 9:08
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This is a settled question but I like to add another solution :-)

We have $W_t \mid \mathcal{F}_s \sim \mathcal{N}(W_s,t-s)$ hence using this we have $$\mathbb{E}(W_t^3 \mid \mathcal{F}_s)=W_s^3+3W_s(t-s).$$

Therefore \begin{align} \mathbb{E}(W_t^3 - 3t W_t \mid \mathcal{F}_s)&=\mathbb{E}(W_t^3 \mid \mathcal{F}_s)-3t\mathbb{E}( W_t \mid \mathcal{F}_s)\\ &=W_s^3+3W_s(t-s)-3tW_s\\ &=W_s^3-3sW_s. \end{align}

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  • $\begingroup$ Thanks Math-fun! ^-^ strange notation though? Is a random variable conditioned on a sigma-algebra also another random variable? :| $\endgroup$ – BCLC Jan 15 '16 at 0:12
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    $\begingroup$ Interesting, I thought till now that we define conditional probabilities with respect to sigma fields. or maybe I do not understand :-) $\endgroup$ – Math-fun Jan 15 '16 at 13:38
  • $\begingroup$ Well I'll assume it's just shorthand. :P $\endgroup$ – BCLC Jan 15 '16 at 14:02
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An easy way would be to use Itô Formula : $$\mathrm d (W_t^3-3tW_t)=3W_t^2\,\mathrm d W_t+3W_t\,\mathrm d t-3t\,\mathrm dW_t-3W_t\mathrm dt,$$ i.e. $$W_t^3-3tW_t=\int_0^t 3(W_t^2-t)\,\mathrm d W_t.$$

By Fubini, $$\mathbb E\int_0^T(W_t^2-t)^2\,\mathrm d t=\int_0^T\mathbb E[(W_t^2-t)^2]\,\mathrm d t<\infty ,$$ and thus, it's indeed a martingale.

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