First of all, there are several typos in your calculations (e.g. it should read $\int_0^t W_s^2 \,ds$ instead of $\int_0^t W_t^2 \, ds$). Your calculation goes wrong when you write
$$\mathbb{E} \left( \int_0^t W_s^3 \, dW_s \right) = \frac{\mathbb{E}(W_t^4)}{4} - \frac{3}{2} \int_0^t V(W_s) \, ds = \frac{\color{red}{3t^4}}{4} - \frac{3t^2}{4}.$$
(I don't get what you did in this last step - you want to calculate $\mathbb{E}(W_t^4)$; so why replace it with $3t^4$?)
Note that applying Itô's lemma is overkill: Since $W(_t)_{t \geq 0}$ is a Wiener process, we know that $W_t \sim N(0,t)$ (i.e. $W_t$ is Gaussian with mean $0$ and variance $t$) and the moments of Gaussian random variables can be calculated explicitly. However, if you really want to invoke Itô's formula, then it goes like that: By Itô's formula, we have
$$W_t^4 = 4 \int_0^t W_s^3 \, dW_s + 6 \int_0^t W_s^2 \, ds. \tag{1}$$
Since $(W_s^3)_{s \geq 0}$ is properly integrable, we know that the stochastic integral
$$M_t := \int_0^t W_s^3 \, dW_s$$
is a martingale and therefore $\mathbb{E}M_t = \mathbb{E}M_0=0$. Taking expectation in $(1)$ yields
$$\mathbb{E}(W_t^4) = 6 \int_0^t \mathbb{E}(W_s^2) \, ds$$
by Fubini's theorem. Finally, since $\mathbb{E}(W_s^2)=s$, we get $\mathbb{E}(W_t^4) = 3t^2$.
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) $E[\exp(Y)\mid X] =\exp(TW_0 + (W_T-W_0)/2 + T^3/6-T^2/8)$.
Best Answer
It is well-known and easy to see by Ito's lemma that $$ M_t=e^{W_t-t/2} $$ is a martingale that satsifes the SDE $$ dM_t=M_t\,dW_t\,. $$ Equivalently, $$ M_t=1+\int_0^tM_s\,dW_s\,. $$ The calculation of $\mathbb E[e^{W_t}\int_0^te^{-W_s}\,dW_s]$ then boils down to \begin{align} &e^{t/2}\,\mathbb E\Big[M_t\int_0^te^{-W_s}\,dW_s\Big]= e^{t/2}\,\mathbb E\Big[\underbrace{\Big(\int_0^te^{W_s-s/2}\,dW_s\Big)}_{\textstyle=:X_t}\underbrace{\Big(\int_0^te^{-W_s}\,dW_s\Big)}_{\textstyle=:Y_t}\Big]\\ &=e^{t/2}\mathbb E\big[\big\langle X,Y\big\rangle_t\big]=e^{t/2}\mathbb E\Big[\int_0^te^{W_s-s/2}\,e^{-W_s}\,ds\Big]=e^{t/2}\int_0^te^{-s/2}\,ds\,. \end{align} Can you finish?