Prove that $\mathbb{E}\left[ \exp\left(\int_0^t X_s dW_s - \int_0^t X_s^2 ds \right)\right] \leq 1$ $\forall t > 0$

Question

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.

Answer 1

You showed $dY_t = e^{Z_t}X_t dW_t - \frac 12 e^{Z_t}X_t^2 dt$, so $$Y_t = Y_0 + \int_0^t e^{Z_s}X_s dW_s - \frac 12 \int_0^t e^{Z_s}X_s^2ds.$$

If we take for granted that $\mathbb{E}[\int_0^t e^{Z_s}X_s dW_s] = 0$ (we don't actually need this assumption, but it simplifies the computation a little bit), then \begin{align*} \mathbb{E}[Y_t] = Y_0 - \frac 12 \int_0^t \mathbb{E}[e^{Z_s}X_s^2]ds. \end{align*} You are correct that evaluating $\mathbb{E}[e^{Z_s}X_s^2]$ is hard, but the only important thing is that it is non-negative, so $\mathbb{E}[Y_t] \le Y_0 = 1$.

If you didn't want to assume $\mathbb{E}[\int_0^t e^{Z_s}X_s dW_s] = 0$, note that $dY_t = e^{Z_t}X_t dW_t - \frac 12 e^{Z_t}X_t^2 dt$ implies $Y$ is a local supermartingale. Then, because $Y$ is non-negative, one can use Fatou's lemma to show $\mathbb{E}[Y_t] \le Y_0$.