A systematic way to do this:
- Compute the joint distribution of the two Gaussian variables $Y:=\int_0^T W_t dt$ and $X:=W_T$ and then
- Evaluate the conditional distribution, which we know is obtained from the following least squares regression:
$$ Y = \alpha + \beta X + Z, \label{LSQ}\tag{1}$$
where $Z$ is zero mean Gaussian variable independent of $X$ with variance
$$\sigma^2_Z = \mathrm{Var}(Y-\beta X)=\mathrm{Var}(Y)-\beta^2\mathrm{Var}(X),\label{sigZ}\tag{2}$$
$\beta$ is the least squares slope coefficient,
$$\beta = \frac{\mathrm{Cov}(X,Y)}{\mathrm{Var}(X)},\label{beta}\tag{3}$$
and $\alpha$ is the so-called intercept, chosen to make the mean of $Z$ zero,
$$ \alpha = E[Y]-\beta E[X].\label{alpha}\tag{4}$$ In summary this gives the conditional distribution formula
$$ Y\mid X \sim N(\alpha +\beta X,\sigma^2_Z).\label{Y|X}\tag{5}$$
- Finally evaluate the conditional expectation using the moment generating function formula for a Gaussian random variable
$$E[\exp(Y)\mid X] =\exp(\alpha +\beta X + \sigma^2_Z/2).\label{MGF}\tag{6}$$
It remains to compute the individual ingredients.
1a) $Y$ is rewritten using (stochastic) integration by parts,
$ Y = TW_0+\int_0^T (T-t)dW_t.$ This gives $E[Y]=TW_0$ and $\mathrm{Var}(Y)=\int_0^T (T-t)^2dt=T^3/3$.
1b) $E[X] = W_0$ and $\mathrm{Var}(X)=T$ by standard properties of BM.
1c) $\mathrm{Cov}(X,Y) = \mathrm{Cov}(\int_0^T dW_t,\int_0^T (T-t)dW_t)
= \int_0^T (T-t)dt =T^2/2$.
Calculate all the parameters of the regression (\ref{LSQ}) using formulae (\ref{sigZ}-\ref{alpha}), starting with (\ref{beta}):
2a) $\beta = T^2/(2T^2)=1/2$, $\alpha = TW_0 - W_0/2$.
2b) $\sigma^2_Z = T^3/3-T^2/4$.
2c) From (\ref{Y|X}) $Y\mid X \sim N(TW_0 + (W_T-W_0)/2, T^3/3-T^2/4)$.
Finally put everything together in (\ref{MGF}):
3a) $E[\exp(Y)\mid X] =\exp(TW_0 + (W_T-W_0)/2 + T^3/6-T^2/8)$.
I have not been able to find bibliography on this, but I think the result is correct. I also think your proof is fine, you use properties of the Brownian motion very nicely.
Here is just an alternative proof which I personally find clearer. Using Fubini (and assuming $f$ is nice enough):
\begin{align}
g_T - g_0 = & \int_0^T \big(f(s, T) - f(s, s) \big) dW_s + \int_0^T f(t, t) dW_t, \\
= & \int_0^T\int_s^T \frac{\partial f}{\partial t}(s, t) dt dW_s + \int_0^T f(t, t) dW_t, \\
= & \int_0^T \int_0^t \frac{\partial f}{\partial t}(s, t) dW_s dt + \int_0^T f(t, t) dW_t,
\end{align}
which is just the integral form of your SDE.
Best Answer
Use the other author's hint, $f(w,t)=w^3$. Then, compute the partial derivatives needed for Ito's lemma: $$\partial_{w}f=3w^2$$ $$\partial_{w^2}f=6w$$ $$\partial_{t}f=0$$ Plug in Ito's formula, $$d(W_t^3)=3W_t^2dW_t+3W_tdt$$ Then integrate both sides from $0$ to $t$ (introduce dummy variable $s$), $$W_t^3=3\int_{0}^{t}W_s^2dW_s+3\int_{0}^{t}W_sds$$ Reaarange, done. Of course, there are other $f(w,t)$'s that you can plug in.