I have a stochastic differential equation for which I have solved the process X$_t$. The SDE is as follows:
$$ dX_t = \left( r\mu X_t + \frac{r(r-1)} 2 \sigma^2 X_t \right) \, dt + r\sigma X_t\,dB_t, $$
in which $X_0 = x$ and $x > 0$.
Here is what I did to solve for the process $X_t$:
$$ dX_t=\left(r\mu+\frac{1}{2}r(r-1)\sigma^2\right)X_t\,dt+r\sigma X_t \,dB_t $$ Using Ito's lemma $$d(\ln X_t)=\frac{1}{X_t}dX_t+\frac{1}{2}\left(\frac{-1}{X_t^2}\right) \, d[X_t,X_t]$$ and then I solved as follows $$d(\ln X_t)=\left(r\mu+\frac{1}{2}r(r-1)\sigma^2\right)\,dt+r\sigma \, dB_t-\frac{1}{2}r^2\sigma^2 \, dt$$
$$d(\ln X_t)=\left(r\mu-\frac{1}{2}r\sigma^2\right)\,dt+r\sigma \, dB_t$$
$$\ln X_t-\ln x_0=\left(r\mu-\frac{1}{2}r\sigma^2\right)t+r\sigma \, B_t$$
$$X_t=x_0\exp\left(\left[r\mu-\frac{1}{2}r\sigma^2\right]t+r\sigma \, B_t\right)$$
If I have $f(x) = (x - K)^2$ for some $K > 0$, how do I compute the expectation $\operatorname{E}[f (X_t)]$? Can someone please show me how to do this? Thank you.
$$\mathbb{E}[f(X_t)]=\mathbb{E}[X_t^2-2KX_t+K^2]=\mathbb{E}[X_t^2]-2K\,\mathbb{E}[X]+K^2\tag 1$$ set $Y=\ln X_t$, therefore $$Y=\ln X_t\sim N\left( \ln x_0+\left( \mu -\frac{1}{2} r\sigma^2 \right)t\,\,,\,\,r^2 \sigma^2t \right)$$ We have $$\mathbb{E}[X_t^2]=\mathbb{E}\left[e^{2\ln X_t}\right]=\mathbb{E}\left[e^{2Y}\right]=M_{Y}(2)=\exp\left(2\mu_Y+\frac 12\sigma_Y^2(2)^2\right)$$ Indeed $$\mathbb{E}[X_t^2]=\exp\left(2\ln x_0+\left( 2\mu -r\sigma^2 \right) t+2r^2\sigma^2t\right) \tag 2$$ similarly $$\mathbb{E}[X_t]=\mathbb{E}\left[e^{\ln X_t}\right] = \mathbb{E} \left[e^Y\right]=M_Y(1)=\exp\left(\mu_Y+\frac 12\sigma_Y^2(1)^2\right)$$ in other words $$\mathbb{E}[X_t]=\exp\left(\ln x_0+\left( \mu -\frac{1}{2}r \sigma^2 \right) t+\frac{1}{2}r^2\sigma^2t\right) \tag 3$$ $(1)$,$(2)$ and $(3)$ $$\mathbb{E}[f(X_t)]=x_0^2\exp\left(\left( 2\mu -r\sigma^2 \right) t +2r^2\sigma^2t\right) -2x_0 K\,\exp\left(\left( \mu -\frac{1}{2}r \sigma^2 \right)t+\frac{1}{2}r^2\sigma^2t\right)+K^2$$