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

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