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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

Ethan Davitt
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1 Answers1

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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

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

Thomas Kojar
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  • Can you elaborate more on how do you know that $\int_{0}^{t}\phi_{x}(X_{s})dX_{s}$ is a martingale?. I know what conditions guarantee that the integral is a martingale, I just don't see how it is applied here – Ethan Davitt Jan 01 '24 at 20:56
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    @EthanDavitt SDE solutions are adapted i.e. $X_{t}\in \mathcal{F}{t}$ and the discrete approximation is $\sum \phi(X{t_{i}}) (B_{t_{i+1}}-B_{t_{i}})$ and so $$E[\sum \phi(X_{t_{i}}) (B_{t_{i+1}}-B_{t_{i}})]=\sum E\phi(X_{t_{i}}) 0=0$$. – Thomas Kojar Jan 01 '24 at 21:10