# Distribution Function of Standard Normal is a U.I. Martingale?

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?

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

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.

• 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. May 17, 2023 at 13:56