Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have a one dimensional standard brownian motion $W$ defined under a stochastic basis with probability $\mathbf{Q}$ and filtration $\left(\mathscr{F}\right)_{t\in{\mathbf{R}}_{+}}$, and I want to calculate $\mathbf{E}\left[ W_{({W_t})^2} \right]$ where $\mathbf{E}$ is the expectation operator under the probability $\mathbf{Q}$ and where $W_{({W_t})^2}$ is the random variable defined by $W_{({W_t (\omega)})^2}(\omega)$ on each "path" $\omega$.

So far, I have tried two things. First was to defined for each $n\in\mathbf{N}$ a random variable $\tau_n = (W_t)^2 \wedge n$ that is obviously a bounded stopping time for the filtration $\left(\mathscr{F}\right)_{t\in{\mathbf{R}}_{+}}$ because $W$ is adapted to this filtration, and then to look at $W_{\tau_n}$. As $W$ is a martingale and as $\tau_n$ is finite, optional stopping theorem yields that $\mathbf{E}\left[ W_{\tau_n} \right] = \mathbf{E}\left[ W_{0} \right] = 0$, and as $W_{\tau_n}$ tends to $W_{({W_t})^2}$ almost surely when $n\rightarrow +\infty$, I would like to show somehow that $\mathbf{E}\left[ W_{\tau_n} \right]$ tends to $\mathbf{E}\left[ W_{({W_t})^2} \right]$ as $n\rightarrow +\infty$, and show thereby that $\mathbf{E}\left[ W_{({W_t})^2} \right] = 0$. Unfortunately, no standard convergence theorems seem to apply (as far as I know) and if we apply Fatou's lemma with $\underline{\lim}$ and $\overline{\lim}$ to $W_{\tau_n}$, we finally find that $\mathbf{E}\left[ W_{({W_t})^2} \right]$ is $\geq 0$ and $\leq 0$ which then shows that $\mathbf{E}\left[ W_{({W_t})^2} \right] = 0$ but... the $W_{\tau_n}$ are not a.s. positive, so that one can't apply Fatou...

The other thing I tried is to start from the caracterization of the law of $W_t$ saying that for every "correct" function $\varphi : \mathbf{R}\rightarrow \mathbf{R}$ one has :

$\mathbf{E}\left[ \varphi (W_t) \right] = \int_{\Omega} \varphi ( W_t (\omega) ) d\mathbf{Q} (\omega) = \int_{\mathbf{R}} \varphi (x) e^{-\frac{x^2}{2t}} \frac{dx}{\sqrt{2\pi t}} = \int_{\mathbf{R}} \varphi (u\sqrt{t}) e^{-\frac{u^2}{2}} \frac{dx}{\sqrt{2\pi }}$

and then to plug $W_t (\omega')$ in $t$, and then to integrate over $\Omega$ against $\mathbf{Q}$ one more time. At the end I found that

$\mathbf{E}^{\mathbf{Q}\otimes\mathbf{Q}}\left[ \varphi( X ) \right] = \int_{\mathbf{R}^2} \varphi(u|v|\sqrt{t}) e^{-\frac{u^2 + v^2}{2}} \frac{du dv}{2\pi}$

where $X$ is the random variable on $\Omega\times\Omega$ defined by $X(\omega,\omega') = W_{({W_t (\omega)})^2}(\omega')$, which show by non parity that the expectation of $X$ is $0$, but I cannot relate $\mathbf{E}^{\mathbf{Q}\otimes\mathbf{Q}}\left[ X \right]$ in a useful manner to $\mathbf{E}\left[ W_{({W_t})^2} \right]$.

If anyone has ideas on how to calculate $\mathbf{E}\left[ W_{({W_t})^2} \right]$, I would be glad to know.

Kind Regards,

MEF

share|improve this question
1  
Hum, what I've tried first is false : $\tau_n$'s are not stopping times as they are not $\mathscr{F}$-mesurable... –  MisesEnForce Oct 11 '13 at 17:31
    
Is $W_{(W_t)^2}$ a random variable? –  Did Oct 11 '13 at 19:42
    
@Did: Certainly, why shouldn't it be? –  Nate Eldredge Oct 11 '13 at 19:56
1  
@NateEldredge "why not?" is rarely considered as a proof. –  Did Oct 11 '13 at 20:15
2  
@Did: Specifically: We can view $W$ as a jointly measurable map from $\Omega \times [0,\infty)$ to $\mathbb{R}$. Then $W_{W_t}^2(\omega) = W(\omega, W(\omega, t)^2)$ which is clearly a composition of measurable functions. I also apologize if I came across as rude. –  Nate Eldredge Oct 11 '13 at 20:25
show 2 more comments

1 Answer 1

By the symmetry of Brownian motion, replacing $W$ by $-W$ will not change the expectation. So we have $$\mathbf{E}\left[W_{(W_t)^2}\right] = \mathbf{E}\left[-W_{(-W_t)^2}\right] = - \mathbf{E}\left[W_{(W_t)^2}\right]$$ and hence $\mathbf{E}\left[W_{(W_t)^2}\right] = 0$.

share|improve this answer
    
I know that processes $W$ and $-W$ have the same law, but why does this imply that the random variables $W_{(W_t)^2}$ and $-W_{(W_t)^2}$ have the same law ? (This is what you are using.) Actually I thought of this "proof" but conclued that is was somehow as difficult as to calculate the law of $W_{(W_t)^2}$, which would straightfowardly give the wanted expectation to me... –  MisesEnForce Oct 11 '13 at 17:50
    
@MisesEnForce: If you think of $F$ as the function on paths defined by $F(\omega) = \omega(\omega(t)^2)$, then what you are looking for is $\mathbb{E}[F(W)]$, i.e. the expectation of a measurable function of $W$. So it only depends on the law of $W$, and will be the same for any other process with the same law. –  Nate Eldredge Oct 11 '13 at 17:58
    
Put another way, this doesn't tell you what the law of $W_{W(t)^2}$ is, but it does tell you that it's symmetric about 0, so its expectation must vanish. –  Nate Eldredge Oct 11 '13 at 18:08
    
Your first comment was enough, thx, but I don't get it perfectly ;-) The map $F$ takes a process X and maps it to the random variable defined by $\omega\mapsto X_{(X_t (\omega)^2} (\omega)$ right ? And you're saying that this map F is measurable, right ? If so, what $\sigma$-algebra do you put on the set of $\mathbf{R}$-valued random variables to make the map $F$ measurable ? (I had the same idea than you, and I feel that $F$ must be "measurable" somehow, but as I'm not a probabilist and as I stopped doing measure stuff a long time ago, I'm not that much sure of my intuition anymore...) –  MisesEnForce Oct 11 '13 at 18:58
    
@MisesEnForce: We can view $W$ as a measurable map from your sample space $(\Omega, \mathcal{F}, \mathbf{Q})$ to the space $C([0,\infty))$ of all continuous paths $\omega : [0,\infty) \to \mathbb{R}$. $C([0,\infty))$ has the topology of uniform convergence on compact sets, and the Borel $\sigma$-algebra generated by that topology. Then $F : C([0,\infty)) \to \mathbb{R}$ is measurable (in fact, continuous). Out of curiosity, how did you come across this question? –  Nate Eldredge Oct 11 '13 at 19:31
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.