Differential of a a function of a semimartingale

Question

Let $X_t$ be a continuous semimartingale and $U_t$=$X_t-1/2[X,X]_t$ where [] is the quadratic variation.Is $$dU_t=dX_t-d[X,X]_t/2?$$ If yes how can I prove this from the Ito's Lemma.In particular what is the function f to which I shoild apply the ito's lemma?

Answer 1

You don't need to prove this from Ito's lemma because $d(A_t+B_t)=dA_t+dB_t$ holds for any two stochastic processes $A$ and $B$ by definition. If you still want to use Ito you should apply it to the function $f(x,y)=x-y/2\,:$ \begin{eqnarray*} df(X_t,[X,X]_t)&=&(\partial_xf)\,dX_t+(\partial_yf)\,d[X,X]_t+\underbrace{\frac{1}{2}(\partial_x^2f)\,d[X,X]_t}_{0}\\ &&+\underbrace{\Delta f(X_t,[X,X]_t)-(\partial_xf)(\Delta X_t)-(\partial_yf)(\Delta[X,X]_t)}_{\Delta X_t-\Delta[X,X]_t/2-\Delta X_t+\Delta[X,X]_t/2=0}\\ &=&dX_t-d[X,X]_t/2. \end{eqnarray*} (here $\Delta Y_t$ is the jump of $Y$ in $t$: $\Delta Y_t=Y_t-Y_{t-}$).

In other words: $d$ is linear and the simple formula holds because all terms vanish that usually make the Ito formula for non continuous semimartingales complicated.