Expectation of a function of Ito diffusion

Question

Given an Ito Diffusion i.e.:

$$ dX(t) = \mu dt + \sigma dW(t) $$

and a function

$$ k(x) = \lambda x^2 $$

and I want to find the expected value $E[k(X(t)]$ of the function - the only way I know how to do is to average the function evaluated at the solution of the diffusion

$$ E[k(X(t))] = \frac{1}{n} \sum_{1}^{n} k(X_i(t)) $$

where $n$ is the number of the simulated paths.

If the transition density of the diffusion is known i.e.:

$$ p(x,t) = \frac{1}{\sqrt(2 \pi \sigma^2 t)} e^\frac{-(x-x_0-\mu t)^2}{2 \sigma^2 t}$$

Can I say that the expectation is also equal to the following?

$$ E[k(X(t))] = \intop\nolimits_{-\infty}^{\infty} k(x) p(x,t)dx $$

where $k(x) = \lambda x^2$ deterministic

Answer 1

You will need the ito lemma:

For $dX_{t}=\mu _{t}\,dt+\sigma _{t}\,dB_{t}$ and $f=f(t,x)$ , we have $ df=\left({\frac {\partial f}{\partial t}}+\mu _{t}{\frac {\partial f}{\partial x}}+{\frac {\sigma _{t}^{2}}{2}}{\frac {\partial ^{2}f}{\partial x^{2}}}\right)dt+\sigma _{t}{\frac {\partial f}{\partial x}}\,dB_{t} $

and the required parameter you are looking for is $\left({\frac {\partial f}{\partial t}}+\mu _{t}{\frac {\partial f}{\partial x}}+{\frac {\sigma _{t}^{2}}{2}}{\frac {\partial ^{2}f}{\partial x^{2}}}\right)$ .

Answer 2

Yes, you can say the expectation is that integral. This is because if $Y$ is a random variable with PDF $f_{Y}$, then the expected value of a function of $Y$, say $g(Y)$, is given by $\mathbb{E}\left[g(Y)\right] = \int_{-\infty}^{\infty}g(y)f_{Y}(y)\, dy$. See the law of the unconscious statistician.