0

I have to prove the following inequality: $$ \mathbb{E}\left[ \left( \int_0^t X_s \text{d}B_s \right)^4 \right] \leq 3c^4 t^2 $$ where

  • $(X_t)_{t \geq 0}$ is an elementary process such that $|X_t| \leq c$ for all $t \geq 0$;
  • $(B_t)_{t \geq 0}$ is a standard Brownian motion.

I'm clueless about a way to solve it: I tried expanding the integral with the definition of stochastic integral for elementary process, but I cannot go anywhere.

I thought about using Itô isometry $$ \mathbb{E}\left[ \left( \int_0^t X_s \text{d}B_s \right)^2 \right] = \mathbb{E} \left[ \int_0^t X_s^2 \text{d}s \right] $$ but I didn't know how to get rid of the power $4$ (given the fact that I cannot use Jensen inequality).

Any hint would be appreciated.

user515933
  • 563

1 Answers1

2

Let $Y_t = \int_0^t X_s dB_s$ and $Z_t = Y_t^4$.

We are need to prove that $$\mathbb E[Z_t] \leq 3c^4t^2.$$

Now, we know that $dY_t = X_tdB_t$ and $d\left\langle Y\right\rangle_t = X_t^2dt$, and thus, using Ito's Lemma,

$$dZ_t = 4Y_t^3dY_t + 6Y_t^2d\left\langle Y\right\rangle_t = 4Y_t^3X_tdB_t + 6Y_t^2X_t^2dt.$$

Since $Z_0 = 0$, this means that $$Z_t = 4\int_0^t Y_s^3X_sdB_s + 6\int_0^tY_s^2X_s^2ds.$$

Finally (leaving to you the proof that the expected value of the first term is $0$), this means that $$\mathbb E[Z_t] = 6\int_0^t\mathbb E[Y_s^2X_s^2]ds \leq 6c^2\int_0^t\mathbb E[Y_s^2]ds = 6c^2\int_0^t\mathbb E\Big[\big(\int_0^sX_rdB_r\big)^2\Big]ds = 6c^2\int_0^t\mathbb E\Big[\int_0^sX_r^2dr\Big]ds \leq 6c^2\int_0^t\int_0^sc^2dr ds = 6c^4 \cdot \frac12 t^2 = 3c^4t^2.$$

LucaMac
  • 3,339