I am trying to compute the expectation and variance of the following stochastic process: $$ Z_t = \exp \left( \frac{1}{2} \int_0^t W_s \, dW_s \right) $$ where $W_t$ is a standard Brownian motion. I have tried the following to compute the expectation:
Let $Y_t = \int_0^t W_s \, dW_s$, then $dY_t = W_t \ dW_t$, and applying the Ito formula to $Z_t = \exp\left( \frac{1}{2} Y_t \right) $ I get $$ dZ_t = \frac{1}{2} Z_t W_t \, dW_t + \frac{1}{8} Z_t W_t^2 \, dt $$ Writing this in integral form and taking expectations, the $dW_t$ integral vanishes and I find that $$ \mathbb{E}(Z_t) = 1 + \frac{1}{8} \int_0^t \mathbb{E}(Z_s W_s^2) \, ds $$
So it seems like in order to compute $ \mathbb{E}(Z_t) $, I need to compute $ \mathbb{E}(Z_t W_t^2)$. That involves expressing $ Z_t W_t^2$ as an Ito process, which in turn involves expressing $Z_t W_t^n$ as an Ito process for some higher power of $n$, and this process does not seem to terminate. Is there some trick I can use to simplify the computation of this expectation?
I am running into the same issues when computing $\mathbb{E}(Z_t^2) $ to find the variance as well, so suggestions for that computation would also be very welcome.