I need some help with this exercise.
Given the following stochastical differential equation:
$dX(t)=\frac{-1}{4}(X(t))^3\;dt+\frac{1}{2}(X(t))^2\;dW(t)$
$X(0)=\frac{1}{2}$
I have to obtain $E[X(t)]$ and $E[(X(t))^3]$ using Itô's lemma.
I applied Itô's lemma for $f(t,x)=x$ and $g(t,x)=x^3$, and taking expectations of the resulting equations, I obtained the following pair of differential equations, but they don't seem to get me to the result:
$\bullet\;E'[X(t)]=\frac{-1}{4}E[(X(t))^3]\;,\;\;\;\;E[X(0)]=\frac{1}{2}$
and
$\bullet\;E'[(X(t))^3]=\frac{-3}{4}E[(X(t))^5]+\frac{3}{2}E[(X(t))^3]\;,\;\;\;\;E[(X(0))^3]=\frac{1}{8}$
Any ideas?
Thanks a lot for any help!
There is a mistake in the second equation: Instead of
$$\mathbb{E}'(X_t^3) = - \frac{3}{4} \mathbb{E}(X_t^5) + \frac{3}{2} \mathbb{E}(X_t^3)$$
it should read
$$\mathbb{E}'(X_t^3)=0$$
Indeed: Applying Itô's formula for $g(x) := x^3$ yields
$$X_t^3-X_0^3 = 3\int_0^t X_s^2 \, dX_s + \frac{6}{2} \int_0^t X_s \, d \langle X \rangle_s. \tag{1}$$
By definition, we have
$$dX_s = - \frac{1}{4} X_s^3 \, ds + \frac{1}{2} X_s^2 \, dW_s \qquad d \langle X \rangle_s = \frac{1}{4} X_s^4 \, ds. \tag{2}$$
Combining $(1)$ and $(2)$ we find
$$X_t^3 - X_0^3 = \underbrace{\frac{3}{2} \int_0^t X_s^4 \, dW_s}_{=:M_t} + \underbrace{\int_0^t \left( - \frac{3}{4} X_s^5 + \frac{6}{8} X_s^5 \right) \, ds}_{=0}.$$
As $(M_t)_{t \geq 0}$ is a martingale, hence satisfying $\mathbb{E}M_t=0$, we conclude
$$\mathbb{E}(X_t^3) = \mathbb{E}(X_0^3) = \frac{1}{8}.$$