Let $(X_t, t \geq 0)$ be a predictable square integrable process. Let \begin{align*} Y_t := \exp\left(\int_0^t X_s dW_s - \int_0^t X_s^2 ds \right) \end{align*} Prove that $\mathbb{E}[Y_t] \leq 1$, $\forall t > 0$.
I was trying to get the Ito process form for $Y_t$, because, if I'm correct, if $dY_t = g_tdt + f_tdW_t$, then $\mathbb{E}[Y_t]$ is dependent only on the first part (because $\int_0^t f_tdW_t = 0$ is a martingale almost always, so its expectation should equal to zero). Let $Z_t := \int_0^t X_s dW_s - \int_0^t X_s^2 ds $. Then \begin{align*} dY_t &= e^{Z_t} dZ_t + \frac{1}{2}e^{Z_t}d[Z_t] = \\ &= e^{Z_t}X_t dW_t - e^{Z_t}X^2_t dt + \frac{1}{2}e^{Z_t}X^2_t dt = \\ &= e^{Z_t}X_t dW_t - \frac{1}{2}e^{Z_t}X^2_t dt \end{align*} I have no idea, what should I do next, because how to evaluate $\mathbb{E}[\frac{1}{2}e^{Z_t}X^2_t]$?
Any help would be appreciated.