Prove $A_t := W_t^3-3t W_t$ a martingale

Question

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?

Unable to identify $E[tW_t|\mathscr{F}_s]$? Well, we have a problem of learning curve here... — Did, Oct 07 '14 at 18:24
@saz I assumed you were leading me to realize that I could take out t? — BCLC, Oct 09 '14 at 17:51
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. — saz, Oct 09 '14 at 17:56

score 11 · Accepted Answer · edited Dec 12 '18 at 13:34

11

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).

edited Dec 12 '18 at 13:34

Nico Albers

195

answered Oct 07 '14 at 19:17

saz

120,083

Satisfying the martingale property makes it martingale? Aren't there other conditions or something? – BCLC Oct 07 '14 at 22:29
Like integrable or something? Oh and thanks! – BCLC Oct 07 '14 at 22:30
2

@BCLC Yes, it has to be integrable. Note that $\mathbb{E}(W_t^k)<\infty$ for any $k$ as $W_t \sim N(0,t)$. – saz Oct 08 '14 at 05:50
Why is $E[(W_t-W_s)^3|\mathcal{F}_s] = E[(W_t-W_s)^3]=0$? – user428487 Apr 03 '18 at 08:09
2

@user428487 Because $W_t-W_s$ is independent from $\mathcal{F}_s$. – saz Apr 03 '18 at 09:08

Surb · Answer 2 · 2019-05-03T08:33:54.170

3

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.

edited May 03 '19 at 08:33

answered Apr 26 '19 at 16:10

Surb

55,662

A quick question, why can we interchange E and the integral? Do we need to invoke DCT or something similar? – jmac May 31 '23 at 00:38
@jmac: As written, it's Fubini's theorem... – Surb May 31 '23 at 04:52
I apologize. Just to be clear: We can use Fubini when the double integral converges and thus switch the order of the integrals. But why does the double integral converge? I can see the inner integral (wrt time) clearly is finite and well-defined (positive continuous, integrand over a finite interval) but why should the double integral itself be finite? – jmac May 31 '23 at 16:28

score 2 · Answer 3 · answered Jan 14 '16 at 15:23

2

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}

answered Jan 14 '16 at 15:23

Math-fun

9,507

Thanks Math-fun! ^-^ strange notation though? Is a random variable conditioned on a sigma-algebra also another random variable? :| – BCLC Jan 15 '16 at 00:12
1

Interesting, I thought till now that we define conditional probabilities with respect to sigma fields. or maybe I do not understand :-) – Math-fun Jan 15 '16 at 13:38
Well I'll assume it's just shorthand. :P – BCLC Jan 15 '16 at 14:02

Prove $A_t := W_t^3-3t W_t$ a martingale

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