A 2-dimensional process $(B_t^1,B_t^2),\ t \geq 0 $ is a standard 2-dimensional Brownian Motion if each $B_t$ is a standard Brownian motion and the $B_t^s$ are independent of each other. Let $ X_t = t^2 + \sin B_t^1 + B_t^1B_t^2$. Calculate $dX_t$, and explain why $X_t$ is a semimartingale.
Belwo is m final attempt at the solution. I am unsure of my justification of why it is a semi-martingale.
First $X_t$ is a function of $t$,$B_t^1$,$B_t^2$:
$$X_t=f(t,B_t^1,B_t^2)=t^2 + \sin B_t^1 + B_t^1B_t^2$$
The Ito lemma gives an expression for $df$:
$$df(t,B_t^1,B_t^2)=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial B_t^1}dB_t^1+\frac{\partial f}{\partial B_t^2}dB_t^2+ \frac{1}{2}\frac{\partial^2 f}{\partial^2 B_t^1}d[B_t^1,B_t^1]+\frac{1}{2}\frac{\partial^2 f}{\partial^2 B_t^2}d[B_t^2,B_t^2]+\frac{\partial^2 f}{\partial B_t^1 \partial B_t^2}d[B_t^1,B_t^2]$$
Since:
$ \frac{df}{dt} = 2t, \frac{df}{dB_t^1} = \ \text{cos}B_t^1 + B_t^2, \ \ \frac{df}{dB_t^2} = B_t^1 \frac{d^2f}{dt^2} =2 \ \ \ \ \ \frac{d^2}{d(B_t^1)^2} = -\text{sin}B_t^1 \frac{d^2f}{d(B_t^2)^2} = 0 $
Therefore:
$$df(t,B_t^1,B_t^2)=2tdt+\cos{B_t^1}dB_t^1+B_t^2 dB_t^1+B_t^1 dB_t^2-\frac{1}{2}\sin{B_t^1}d[B_t^1,B_t^1]+d[B_t^1,B_t^2]$$
Gathering like terms we get that:
$ (2t-\frac{1}{2} \ \text{sin} B_t^1)dt + ( \ \text{cos}B_t^1 + B_t^2, B_t^1) \begin{bmatrix} dB_t^1 \\ dB_t^2 \end{bmatrix} $ $= \mu(t,B_t^1,B_t^2)dt + \sigma(t,B_t^1,B_t^2) \begin{bmatrix} dB_t^1 \\ dB_t^2 \end{bmatrix} $
Now
$ X_t = X_0 + \int_0^t(\cos B_s^1 + B_s^2) dB_s^1 + \int_0^t B_s^1dB_s^2- \frac{1}{2} \int_0^t \sin B_s^1 ds + \int_0^t 2sds $ where $ \int_0^t (\cos B_s^1 + B_s^2) dB_s^1 + \int_0^t B_s^1dB_s^2$ is a Martingale. And $ \frac{1}{2} \int_0^t \sin B_s^1 ds + \int_0^t 2sds$ is the Finite Variation process. Hence $X_t$ is a semi-martingale.
First $X_t$ is a function of $t$,$B_t^1$,$B_t^2$:
$$X_t=f(t,B_t^1,B_t^2)=t^2 + \sin B_t^1 + B_t^1B_t^2$$
The Ito lemma gives you a expression to $df$:
$$df(t,B_t^1,B_t^2)=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial B_t^1}dB_t^1+\frac{\partial f}{\partial B_t^2}dB_t^2+\frac{1}{2}\frac{\partial^2 f}{\partial^2 B_t^1}d[B_t^1,B_t^1]+\frac{1}{2}\frac{\partial^2 f}{\partial^2 B_t^2}d[B_t^2,B_t^2]+\frac{\partial^2 f}{\partial B_t^1 \partial B_t^2}d[B_t^1,B_t^2]$$
Therefore:
$$df(t,B_t^1,B_t^2)=2tdt+\cos{B_t^1}dB_t^1+B_t^2 dB_t^1+B_t^1 dB_t^2-\frac{1}{2}\sin{B_t^1}d[B_t^1,B_t^1]+d[B_t^1,B_t^2]$$
How the brownian motion are independent of each other:
$$df(t,B_t^1,B_t^2)=2tdt-\frac{1}{2}\sin{B_t^1}dt+\cos{B_t^1}dB_t^1+B_t^2 dB_t^1+B_t^1 dB_t^2$$
I let to you see why is a semimartingale.