Expected Value of Ito process

Question

I am trying to find the 3rd moment of the a process $X_{t}$ that satisfies the following differential equation:

$$dX_{t} = (AX_{t} + a)dt + (BX_{t} + b)dWt$$

where A,a,B,b are constants. I came across the following formula for the moments of a process that satisfy a SDE: denote $f = AX_{t} + a$ and $L = (BX_{t} + b)$, the moments of the differential equation are given by:

$$\frac{dE(X_{t}^{n})}{dt} = nE(X_{t}^{n-1}f(x)) + \frac{n(n-1)}{2}E(X_{t}^{n-2}L^{2}(x))$$

The proof of the equation relies on applying Ito's lemma to $X^{n}$ and the fact that the expected value of an integral with respect to the brownian motion is $0$.

There are a couple of things I don't understand about this:

• Why can we move the 'd' outside the expectation?
• Why is the expected value of the integral with respect to the Brownian motion always zero in this case?

This equation can be found on page 73 of https://users.aalto.fi/~ssarkka/pub/sde_book.pdf

Answer 1

For the zero expectation of the Itô-term see Itō Integral has expectation zero or more directly here

Theorem 5 An adapted integrable process $X$ is

• a martingale if and only if $\mathbb E\left[\int_0^\infty \xi\,dX\right]=0$ for all bounded elementary processes $\xi$.

For the derivative question, first we note that the "derivative" in Itô formula is only formal because we actually integrate

$$\phi(X_{t})=\phi(X_{0})+\int_{0}^{t} \phi_{x}(X_{s})dX_{s}+\int_{0}^{t} \frac{1}{2}\phi_{xx}(X_{s})d[X]_{s}$$

and so by taking expectation we are left with

$$E\left[\phi(X_{t})\right]=\phi(X_{0})+E\left[\int_{0}^{t} \phi_{x}(X_{s})b(X_{s})ds\right]+E\left[\int_{0}^{t} \frac{1}{2}\phi_{xx}(X_{s})\sigma^{2}(X_{s})ds\right].$$

Here we apply Fubini

$$E\left[\phi(X_{t})\right]=\phi(X_{0})+\int_{0}^{t} E\left[\phi_{x}(X_{s})b(X_{s})+\frac{1}{2}\phi_{xx}(X_{s})\sigma^{2}(X_{s})\right]ds$$

and we see that the RHS is now differentiable in $t$ and so we have

$$\frac{d}{dt}E\left[\phi(X_{t})\right]=E\left[\phi_{x}(X_{t})b(X_{t})+\frac{1}{2}\phi_{xx}(X_{t})\sigma^{2}(X_{t})\right].$$