Martingale representation theorem application

Question

Let $X = \exp(W_{T/2}+W_T)$. I try to figure the adapted process $g(s)$ such that according to the MRT we have $$X = \mathbb{E}[X]+\int^T_0 g_s dW_s.$$ I can figure out $X = \exp(2W_{T/2}+W_{T-T/2})$ and therefore by knowing that $$\exp(2W_{T/2})=\mathbb{E}[\exp(2W_{T/2})]+\int^{T/2}_0 2 \exp(2W_s-2s+T)dW_s,$$ and $$exp(W_{T/2})=\mathbb{E}[\exp(W_{T/2})]+\int^{T/2}_0 \exp(W_s-\frac{1}{2}s+T/4)dW_s.$$ However how to combine them to obtain $g(s)$?

Answer 1

Remark: What follows is a different argument from my initial solution attempt. Here, each of the steps in the derivation hold almost surely.

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We proceed by a sequence of steps.

Step 1: Define the process $f_t:=f(W_{t/2}, W_t)=W_{t/2}+W_t$ and consider that, by an application of Itô's lemma, we deduce the quadratic variation for the process as $$\langle f(W, W) \rangle_{t/2,t} = \langle W \rangle_{t/2}+\langle W \rangle_t+2\langle W, W \rangle_{t/2,t}=\frac{t}{2}+t+t=\frac{5}{2}t\tag{1}$$

Step 2: Consider the martingale $X_te^{-\frac{1}{2}\langle f\rangle_{t/2,t}}=e^{f_t-\frac{1}{2}\langle f\rangle_{t/2,t}}$ to which we, again, apply Itô's lemma, yielding

$$ \text d\Big(X_te^{-\frac{1}{2}\langle f\rangle_{t/2,t}}\Big)=X_te^{-\frac{1}{2}\langle f\rangle_{t/2,t}}\text df_t\,.\tag{2} $$

Consequently, (2) immediately implies

$$X_t = e^{\frac{1}{2}\langle f\rangle_{t/2,t}}+e^{\frac{1}{2}\langle f\rangle_{t/2,t}}\int_0^t X_se^{-\frac{1}{2}\langle f\rangle_{s/2,s}}\text df_s\,\tag{3}$$

Step 3: We are almost done. Note that, for some independent Brownian-motions $W^1_t$ and $W^2_t$, we have $W_t=W^1_t$ and $\frac{1}{\sqrt{2}}\Big(\sqrt{2}W_{t/2}\Big)=\frac{1}{2}\Big(W^1_t+W^2_t\Big)$. You can check that the quadratic covariations are consistent. Therefore, $$ \text df_t=\text d(W_{t/2}+W_t)=\text d\Big(\frac{3}{2}W^1_t+\frac{1}{2}W^2_t\Big)\tag{4} $$ Using (4) in (3) gives a pair of previsible processes as defining "$g_s$":

$$ g_{s,u} = (\alpha^1_s, \alpha^2_u)= e^{\frac{1}{2}\langle f\rangle_{t/2,t}}\Big( \frac{3}{2}X_se^{-\frac{1}{2}\langle f\rangle_{s/2,s}},~\frac{1}{2}X_ue^{-\frac{1}{2}\langle f\rangle_{u/2,u}} \Big)$$

Answer 2

Obviously,

$$\ln X = \int_0^T \sigma(s) \, dW_s \qquad \text{for} \quad \sigma(s) := 1_{[0,T/2]}(s) + 1_{[0,T]}(s). \tag{1}$$

Define an Itô process $(Y_t)_{t \geq 0}$ by

$$Y_t := \int_0^t \sigma(s) \, dW_s - \frac{1}{2} \int_0^t \sigma(s)^2 \, ds. \tag{2}$$

Then, by Itô's formula,

$$\begin{align*} \exp(Y_t)-1 &= \int_0^t \exp(Y_s) \, dY_s + \frac{1}{2} \int_0^t \exp(Y_s) \, d\langle Y \rangle_s \\ &= \int_0^t e^{Y_s} \sigma(s) \, dW_s; \tag{3} \end{align*}$$

here we have used that the quadratic variation $(\langle Y \rangle_s)_{s \geq 0}$ equals by $(2)$ $$d\langle Y \rangle_s = \sigma^2(s) \, ds.$$

Consequently,

$$\begin{align*} X &= \exp(Y_T) \exp \left( \frac{1}{2} \int_0^T \sigma(s)^2 \, ds \right) \\ &\stackrel{(3)}{=} \left( 1+ \int_0^T e^{Y_s} \, \sigma(s) \, dW_s \right) \cdot \exp \left( \frac{1}{2} \left( \frac{4T}{2}+\frac{T}{2} \right) \right) \\ &= \exp \left( \frac{5}{4} T \right) + \int_0^T \underbrace{\exp \left( \frac{5}{4} T + Y_s \right) \sigma(s)}_{=:g(s)} \, dW_s \end{align*}$$

(Note that, using $(1)$ and $(2)$, we can calculate $g$ explicitly. I leave this to the reader.)