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Suppose I have two continuous semimartingales $X$ and $Y$. By Ito's formula for semimartingales, we have

$$ X_tY_t=X_0Y_0+\int_0^t X_sdY_s+\int_0^tY_sdX_s+\langle X,Y\rangle_t $$ where $\langle X,Y\rangle_t$ is the predictable cross-variation between $X$ and $Y$.

The definition of $\langle X,Y\rangle_t$ that I am working with is the adapted process such that $X_tY_t-\langle X,Y\rangle_t$ is a martingale.

Suppose now $f,g\in C^2(\mathbb R)$ and $X_t:=f(W_t)$ and $Y_t:=g(W_t)$.

I want to know the explicit form of $\langle X,Y\rangle_t$.

From lecture notes, the professor wrote that, by Ito lemma,

$d(X_tY_t) =X_tdY_t+Y_tdX_t+d\langle X,Y\rangle_t= f(g'dW_t+\frac{1}{2}g''dt)+g(f'dW_t+\frac{1}{2}f''dt)+f'g'dt$.

My question is, why does the second equality hold?

I know, by Ito lemma, $X_tdY_t=f(g'dW_t+\frac{1}{2}g''dt)$, and $Y_tdX_t=g(f'dW_t+\frac{1}{2}f''dt)$. If the second equality above is correct, then it somehow implies that $d\langle X,Y\rangle_t=f'g'dt$. How do we know this is true? We are still on the way deriving the expression of $\langle X,Y\rangle_t$. Quite confusing. Thanks for help.

Sam Wong
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1 Answers1

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By Ito's lemma \begin{align} dX_t&=f'(W_t)\,dW_t+\tfrac12 f''(W_t)\,dt\,,&dY_t&=g'(W_t)\,dW_t+\tfrac12 g''(W_t)\,dt\,. \end{align} To $\langle X,Y\rangle_t$ only the two $dW_t$-terms contribute. Since $d\langle W,W\rangle_t=dt$ we get $$ d\langle X,Y\rangle_t=f'(W_t)g'(W_t)\,dt\,. $$

Kurt G.
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  • Did you implicitly use the an equality like $d\langle X, Y\rangle _t=\langle dX_t, dY_t \rangle$? (If so, can you give me a reference for the proof of such an equality? Thanks.) Alternatively, how shall we substitute $dX_t$ and $dY_t$ into $d\langle X, Y\rangle _t$ to get $f'(W_t)g'(W_t)dt$? Thanks! – Sam Wong Nov 26 '23 at 16:44
  • @SamWong I do not use such equalities of stochastic differentials. All my equations are short forms of integral equations. In pretty much any standard book about stochastic integration you should find the formula $$\left\langle\int_0^.A_s,dX_s,\int_0^. B_s,dY_s\right\rangle_t=\int_0^tA_sB_s,d\langle X,Y\rangle_s,.$$ Karatzas & Shreve would be my first reference. – Kurt G. Nov 26 '23 at 16:49
  • Thanks Kurt. Can you tell me which page of their book $\textit{Brownian Motion and Stochastic Calculus}$ has the formula? I tried to find it but failed because there is too much content in the book. Thanks. – Sam Wong Nov 27 '23 at 09:31
  • @SamWong I don't have any such books here today. As a healthy exercise I recommend to take the definition of quadratic covariation and apply it naively to two $dW_s$-integrals. What do you get? If you see the point this is 95% of a rigorous proof. – Kurt G. Nov 27 '23 at 10:39