I am trying to understand the proof of Ito's Formula for polynomial functions: $$ f(X_t) = f(X_0) + \int_0^t f'(X)\,dX + \frac{1}{2}\int_0^t f''(X)\, d\langle X \rangle $$ where $X$ is a continous semimartingale and f a twice continuously differentiable function on $\mathbb{R}$ (The first Integral is a stochastic (ito) integral for semimartingales as integrators and the second integral is just a Lebesgue-Stieltjes integral). Right now i only consider functions $f(x)=x^n$ for $n\in\mathbb{N}$ and try to prove by induction. By using the Integration by Parts Theorem for Semimartingales we get: $$ X_t^n = X_0^n + \int_0^t X^{n-1}\, dX + \int_0^t X \,dX^{n-1} + \langle X, X^{n-1} \rangle_t \tag{1} $$ and by assumption, if we say the formula holds for $n-1$, we get $$ X_t^{n-1} = X_0^{n-1} + \int_0^t (n-1)X^{n-2}\, dX + \frac{1}{2} \int_0^t (n-1)(n-2)X^{n-3} \,d\langle X\rangle. \tag{2} $$ From (1) we get $\langle X, X^{n-1} \rangle_t = \int_0^t (n-1) X^{n-2}\, d \langle X \rangle$. The next step would be to insert (2) into (1) and show that the formula holds, however I don't understand how. If I insert (2) into (1) I get double integrals and I can't transform the term in a way, such that the formula will be true. Any help is much appreciated
2026-04-05 20:50:42.1775422242
Ito's Formula for semimartingales proof
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It might be simpler to write (1) and (2) in differential form: \begin{align} dX^n_t&=X^{n-1}_t\,dX_t+X_t\,\color{red}{dX^{n-1}_t}+d\langle X,X^{n-1}\rangle_t\,,\tag{1d}\\[2mm] \color{red}{dX^{n-1}_t}&=(n-1)X^{n-2}_t\,dX_t+\frac{1}{2}(n-1)(n-2)X_t^{n-3}\,d\langle X\rangle_t\,.\tag{2d} \end{align} Inserting $dX^{n-1}_t$ into (1d) is easy and you already found out that $$\tag{3} d\langle X,X^{n-1}\rangle_t=(n-1)X^{n-2}_t\,d\langle X\rangle_t\,. $$ This gives \begin{align}\require{cancel} dX^n_t&=\cancel{X^{n-1}_t\,dX_t}+(n-\cancel{1})X_t^{n-1}\,dX_t\\[2mm] &~+(n-1)X_t^{n-2}\,d\langle X\rangle_t+\frac{1}{2}(n-1)(n-2)X_t^{n-2}\,d\langle X\rangle_t\tag{4}\\[2mm] &=nX^{n-1}_t\,dX_t+\frac{1}{2}n(n-1)X_t^{n-2}\,d\langle X\rangle_t\tag{5} \end{align} as it should.