How to calculate $\mathbb{E}\left(\varPhi^{2}\left(\frac{W_{s}}{\sqrt{1-s}}\right)\right)$?

Question

How can we calculate $$\mathbb{E}\left(\varPhi^{2}\left(\frac{W_{s}}{\sqrt{1-s}}\right)\right)$$ where $W$ denotes a Wiener process, and $\varPhi$ is the CDF of the standard normal random variable... If $\varphi$ denotes the PDF of the standard normal variable, then$$\mathbb{E}\left(\varPhi^{2}\left(\frac{W_{s}}{\sqrt{1-s}}\right)\right)=\int_{\mathbb{R}}\varPhi^{2}\left(\frac{x}{\sqrt{1-s}}\right)\frac{1}{\sqrt{2\pi s}}e^{-\frac{x^{2}}{2s}}dx=\int_{\mathbb{R}}\varPhi^{2}\left(\frac{x}{\sqrt{1-s}}\right)\frac{1}{\sqrt{s}}\varphi\left(\frac{x}{\sqrt{s}}\right)dx=...?$$ I can't continue...

Answer 1

Let $$M_t = \Phi \left( \frac{W_t}{\sqrt{1-t}} \right) = g(t, W_t)$$ where $g(t,x) =\Phi \left( \frac{x}{\sqrt{1-t}} \right)$. We claim that $M_t$ is a martingale. First, observe that $\Phi^\prime (s) =\frac{1}{\sqrt{2 \pi}} \exp \left( \frac{s^2}{2} \right)$ and $\Phi^{\prime \prime}(s) = - \frac{s}{\sqrt{2 \pi}}\exp \left( \frac{s^2}{2} \right)$. Furthermore, $$\begin{align*} g_x(t,x) &= \frac{1}{\sqrt{1 - t}} \Phi^\prime \left( \frac{x}{\sqrt{1 - t}} \right) \\ g_t(t,x) &= \frac{x}{2\sqrt{1 - t}(1-t)} \Phi^\prime \left( \frac{x}{\sqrt{1 - t}} \right) \\ g_{xx}(t,x) &= \frac{1}{1-t} \Phi^{\prime \prime}\left( \frac{x}{\sqrt{1 - t}} \right) \end{align*}$$

From here, you may check that $g_t(t,x) + \frac{1}{2}g_{xx}(t,x) = 0$, so that $M_t$ is driftless and hence a local martingale. It's in fact a true martingale by observing that $$dM_t = \frac{1}{\sqrt{2\pi}\sqrt{1-t}} \exp \left( - \frac{W_t^2}{2(1 - t)} \right) dW_t$$ and noting that the integrand (being bounded) belongs to $L^2 (\Omega \times [0,T])$ for $T < 1$.

The dynamics of $M_t^2$ are given easily by Itô's lemma: $$d(M^2)_t = (dM_t)^2 + 2M_t(dM_t) = \frac{1}{2\pi (1-t)} \exp \left( - \frac{W_t^2}{1-t} \right) dt + \text{martingale}$$

So that $$E(M_t^2) = M_0^2 + \int_0^t \frac{1}{2\pi (1-s)} E\exp\left( -\frac{W_s^2}{1-s}\right) ds$$ The inner expectation evaluates to $$E\exp\left( -\frac{W_s^2}{1-s}\right) = E\exp\left( -\frac{s}{1-s}W_1^2\right) = \left( \frac{1+s}{1-s} \right)^{-1/2} = \sqrt{ \frac{1-s}{1+s} }$$ Where in the second equality we have evaluated the MGF of $W_1^2 \sim \chi^2(1)$ at the point $- \frac{s}{1-s}$.

From this point on, the expectation is a standard calculus exercise which I leave to you.