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

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.

• What have you tried so far? Do you have any thoughts on how to show this? Mar 24 at 14:59
• I thought I was completely wrong, so initially I didn't add my attempts. I added them now. Mar 24 at 15:23

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 $$\mathbb{E}[Y_t] \le Y_0$$.