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I'm a little lost on how to show how

$X_{t}=\Phi(\frac{W_{t}}{\sqrt{T-t}})$ $0\leq t\leq T$, where $W_{t}$ is the usual Brownian Motion, is a Uniformly Integrable Martingale?

My goal is to try and show $\mathbb{E}(X_{t}\mid \mathcal{F}_{s})=X_{s}$ for $s\leq t$, but my thought ends fairly quickly trying something like

$\mathbb{E}(\Phi(\frac{W_{t}}{\sqrt{T-t}})\mid\mathcal{F}_{s})=\mathbb{E}(\Phi(\frac{W_{t}-W_{s}+W_{s}}{\sqrt{T-t}})\mid\mathcal{F}_{s})=\mathbb{E}(\Phi(\frac{W_{t}-W_{s}}{\sqrt{T-t}}+\frac{W_{s}}{\sqrt{T-t}})\mid\mathcal{F}_{s})$

Obviously I want to show the last term simplifies to $\Phi(\frac{W_{s}}{\sqrt{T-s}})=X_{s}$, but how should I proceed from the line above?

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  • $\begingroup$ perhaps you could try writing the distribution function as an expectation $\endgroup$
    – seanv507
    May 17, 2023 at 7:22
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    $\begingroup$ also are you sure you dont have a typo and numerator is $W_T−W_t$ as otherwise denominator looks inconsistent $\endgroup$
    – seanv507
    May 17, 2023 at 7:51

1 Answer 1

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Let $f(t,x) := \Phi\left(\frac{x}{\sqrt{T-t}}\right) = \Phi\circ g(t,x)$. By applying the chain rule we get $$\frac{\partial f}{\partial x}(t,x) = \frac{1}{\sqrt{T-t}}\phi\left(\frac{x}{\sqrt{T-t}}\right) $$

$$\frac{\partial^2 f}{\partial x^2}(t,x) = \frac{1}{T-t}\phi_x\left(\frac{x}{\sqrt{T-t}}\right) =\frac{-x}{(T-t)^{3/2}}\phi\left(\frac{x}{\sqrt{T-t}}\right) $$

$$ \frac{\partial f}{\partial t}(t,x) = \frac{x}{2(T-t)^{3/2}}\phi\left(\frac{x}{\sqrt{T-t}}\right)$$

Where $\phi$ denotes the standard normal PDF, whose derivative is known.

Now your process $X_t$ is given by $X_t = f(t,W_t)$, hence by Itô's lemma, we have that

$$\begin{align*} dX_t = df(t,W_t) &= \left(\frac{\partial f}{\partial t} + 0 + \frac 1 2\frac{\partial^2 f}{\partial x^2} \right) dt + \frac{\partial f}{\partial x} dW_t\\ &=0 + \frac{\partial f}{\partial x} dW_t\end{align*} $$

Or in other words, $X_t$ can be written for all $0\le t< T$ as $$ X_t = X_0 + \int_0^t \frac{\partial f}{\partial x}(s,W_s)\ dW_s $$

Now we have the following

Theorem : if

  • $(C(\omega,t))_t$ is adapted to the natural filtration of $(W_t)$ and progressively measurable
  • $\mathbb{E}\left(\int_0^{T^*} C(\omega,s)^2\,ds\right) < \infty$

Then the process $Z_t := \int_0^t C(\omega,s)dW_s$ is a square integrable martingale on $[0,T^*]$

(see this and this blog post by George Lowther and links within for a proof and additional references).

As the assumption of the theorem are satisfied in our case, it follows that $X_t$ is indeed a martingale on $[0,T^*]$ for all $T^*< T$ (note that we can not include $T$ in the interval since the integral explodes as $t$ approaches $t$, making the theorem's assumption fail).

The uniform integrability follows immediately from the square integrability of $(X_t)$, indeed, for all $t\in[0,T^*]$ and any measurable set $A$, we have by Cauchy-Schwarz :

$$\mathbb E[|X_t|\mathbf 1_A]\le \|X_t\|_2 \sqrt{\mathbb P(A)} \le \sup_{0\le t\le T^*}\|X_t\|_2 \sqrt{\mathbb P(A)} $$ Which, since the first factor is bounded, uniformly goes to $0$ as $\mathbb P(A)$ goes to zero.

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    $\begingroup$ Good answer (+1) of application of Ito's lemma. The OP should check the two conditions in your cited theorem are really satisfied by this stochastic process. $\endgroup$
    – Zhanxiong
    May 17, 2023 at 13:56

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