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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,


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Hum, what I've tried first is false : $\tau_n$'s are not stopping times as they are not $\mathscr{F}$-mesurable... –  Robert Green 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
@NateEldredge "why not?" is rarely considered as a proof. –  Did Oct 11 '13 at 20:15
@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

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 $\mathbf{E}\left[W_{(W_t)^2}\right] = 0$.

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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... –  Robert Green 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...) –  Robert Green 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

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