Conditional expectation of exponent to the power of BM, representation theorem

Question

Let $T > 0$ and $M(t) = E[e^{W(T)}|F(t)]$ where $\{F(t) : t \geq 0\}$ is a natural filtration generated by W and $t \leq T$ I need to show that M(t) is a martingale and I also need to find a unique square integrable function $\phi$ for the representation $M(t) = M(0) + \int_0^t \phi(s)dW(s)$

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What I did: for the martingale part I checked 3 conditions:

1. It does look adapted, doesn't depend on future values (not sure how to prove this rigorously)

2. Integrability: $E[|E[e^{W(T)}|F(t)]|] \leq (jensen) E[E[e^{|W(T)|}|F(t)]] = E[e^{|W(T)|}] = E[e^{\sqrt{T}*|\epsilon|}] < \infty $ (I can just say that MGF is finite, right? $\epsilon$ is a standard normal random variable)

3. For any $s \leq t: E[M(t)|F(s)] = E[E[e^{W(T)}|F(t)]|F(s)] = E[E[e^{W(T)}|F(s)]|F(t)] = E[e^{W(T)}|F(s)] = M(s) $ Thus $M(t)$ is a martingale

For the representation part:

$E[e^{W(T)}|F(t)] = M(0) + \int_0^t \phi(s)dW(s)$

Now, this is where I'm not sure if I'm doing it right:

$E[e^{W(T)}|F(t)] = E[e^{W(T) - W(t) + W(t)}|F(t)] = $ (take out what is known) $ e^{W(t)}*E[e^{W(T) - W(t)}|F(t)] = $ (independent increments) $ e^{W(t)}*E[e^{W(T)-W(t)}] = e^{W(t)}*e^{\frac{T-t}{2}} = e^{W(t) + 0.5(T - t)} $

$M(0) = e^{0.5T}$, so my representation turns into this:

$e^{W(t) - 0.5t} = \int_0^t \phi(s)dW(s)$

Now, I am going to use Ito on a function $f(x,t) = e^{x - 0.5t}$

$f_x = e^{x - 0.5t}$

$f_{xx} = e^{x - 0.5t}$

$f_t = -0.5e^{x - 0.5t}$

$d(f) = e^{x - 0.5t}dx - 0.5e^{x - 0.5t}dt + 0.5*e^{x - 0.5t}dt = e^{x - 0.5t}dx$

$e^{W(t) - 0.5t} = 1 + \int_0^t e^{W(s)-0.5s}dW(s) = \int_0^t \frac{1}{W(t)} + e^{W(s) - 0.5s}dW(s)$

So my $\phi$ is $\frac{1}{W(t)} + e^{W(s) - 0.5s}$

I need to check that it is square-integrable, but for some reason I'm confused about the $\frac{1}{W(t)}$. Do I treat it as a constant and freely take out of expectation or ... maybe I made a mistake somewhere? I hope someone can give me a helping hand.

Answer 1

You already applied Ito's lemma to find $$ e^{W(T)-T/2}=1+\int_{0}^{T}e^{W(s)-s/2}dW(s). $$ Next, note that \begin{multline*} \mathbb{E}\left[\int_{0}^{T}e^{W(s)-s/2}dW(s)\mid\mathcal{F}(t)\right]=\int_{0}^{t}e^{W(s)-s/2}dW(s)+\mathbb{E}\left[\int_{t}^{T}e^{W(s)-s/2}dW(s)\mid\mathcal{F}(t)\right]\\=\int_{0}^{t}e^{W(s)-s/2}dW(s). \end{multline*} Therefore, applying conditional expectations to the first equation, $$ e^{-T/2}M(t)=1+\int_{0}^{t}e^{W(s)-s/2}dW(s). $$ Multiplying by $e^{T/2}$ yields the desired result.