Verifying Property of Stochastic Integral

Question

I am trying to verify this simple property for a stochastic integral. Given that f(t,w) is a bounded, nonanticipating function for a given Wiener process $W_t$ show that $E((\int_{0}^{T} f(s,w) dW_s)^4) \le 36T\int_{0}^{T}Ef^4(s,w)ds$ .

I know that $E((\int_{0}^{T} f(s,w) dW_s)^2)=E(\int_{0}^{T} f^2(s,w)ds)$. I tried applying this, but I am at a loss for where the inequality and $36T$ term arrive from.

Answer 1

Applying Itô's formula for the Itô process

$$X_t := \int_0^t f(s) \, dW_s$$

and $f(x) := x^4$ gives

$$X_t^4 = 4 \int_0^t X_s f(s) \, dW_s + \frac{4 \cdot 3}{2} \int_0^t X_s^2 f(s)^2 \, ds.$$

Taking expecation yields

$$\begin{align*} \mathbb{E}(X_t^4) &= 6 \int_0^t \mathbb{E}(X_s^2 f(s)^2) \, ds. \end{align*}$$

Since $X_s^2$ is a submartingale, we have $\mathbb{E}(X_s^2) \leq \mathbb{E}(X_t^2)$ for all $s \leq t$. Combining this with Hölder's inequality, we find

$$\begin{align*} \mathbb{E}(X_t^4) &\leq 6 \int_0^t \sqrt{\mathbb{E}(X_s^4)} \sqrt{\mathbb{E}(|f(s)|^4)} \, ds \leq 6 \sqrt{\mathbb{E}(X_t^4)} \int_0^t \sqrt{\mathbb{E}(f(s)^4)} \, ds.\end{align*}$$

Hence,

$$\begin{align*} \mathbb{E}(|X_t|^4) \leq 36 \left[ \int_0^t \sqrt{\mathbb{E}(f(s)^4)} \, ds \right]^2. \end{align*}$$

Now the claim follows by applying Jensen's inequality.

Remark: The inequality is closely related to the so-called Burkholder-Davis-Gundy inequality.

Answer 2

Here's a different argument. Define

$$X_t = \int_0^t f(s,\omega) dW_s \\ Y_t = X_t^4 \\ Z_t = X_t^2.$$

Apply the Ito formula to $Y_t$ and $Z_t$:

$$Y_t = 4 \int_0^t f(s,\omega) X_s^3 dW_s + 6 \int_0^t f(s,\omega)^2 X_s^2 ds \\ Z_t = 2 \int_0^t f(s,\omega) X_s dW_s + \int_0^t f(s,\omega)^2 ds$$

Now substitute and take expectations, thereby canceling the stochastic integral terms. (There is a Fubini's theorem step in here which I am not writing out.)

$$\mathbb{E}(Y_t) = 6 \mathbb{E} \left ( \int_0^t f(s,\omega)^2 X_s^2 ds \right ) \\ = 6 \mathbb{E} \left ( \int_0^t \int_0^s f(s,\omega)^2 f(u,\omega)^2 du ds \right )$$

The argument of the expectation is an integral of a nonnegative function over a triangle. It only gets larger when extended to the whole square:

$$\mathbb{E}(Y_t) \leq 6 \mathbb{E} \left ( \int_0^t \int_0^t f(s,\omega)^2 f(u,\omega)^2 du ds \right ) = 6 \mathbb{E} \left ( \left ( \int_0^t f(s,\omega)^2 ds \right )^2 \right )$$

Now using Jensen's inequality and then Fubini's theorem:

$$\mathbb{E}(Y_t) \leq 6 t \mathbb{E} \left ( \int_0^t f(s,\omega)^4 ds \right ) = 6t \int_0^t \mathbb{E} \left ( f(s,\omega)^4 \right ) ds.$$

A remark: I am actually fairly sure that this allows us to change the $6$ to a $3$, because I think the integral over the triangle is exactly half the integral over the square, by a symmetry argument. We cannot do any better than $3$, as the simplest possible example $f(t,\omega) \equiv 1$ shows.