As well known, Ito formula is an equality that $$f(W_t)=f(W_0)+\int_o^t f'(W_s)dW_s +\frac{1}{2}\int_o^t f''(W_s)ds,$$ where $f''$ is countunous and $(W_t)_{t>0}$ is a standard brownian motion.
We know any Ito integral of a general stochastic process is defined by the $L^2$ limit of Ito integrals of simple processes. So it convinces me that this Ito formula holds in the $L^2$ sense. However, my professor didn't mention what kind of convergence in the Ito formula is.
The general stagetegy of proving this formula is first expanding $f(W_t)$ by Taylor extension and then do some approximations to get terms on the RHS. My professor used convergence in probability for some term in the Taylor expansion to get the term $\int_o^t f''(W_s)ds$ on the RHS, for which I am also quite confused. We don't have bounded condition and so convergence in probability can not imply convergence in $L^2$. Hence in this sense, it seems like the Ito formula he proved doesn't hold in the $L^2$ sense.
So here is the point I am painful about. We first expand $f(W_t)$, and then he has shown some term converges to $\int_o^t f''(W_s)ds$ in probability and some term converges to $\int_o^t f'(W_s)dW_s$ in $L^2$. Then what does the equal sign in the Ito formula mean? Thanks for any explanation.
Edit: Added more details of his confusing proof
My professor didn't prove the general case and instead he tried to prove the case where $f(W_t)=W_t^3$.
We let $(\Delta W)_i:=W_{t_{i+1}}-W_{t_i}.$
So, $$f(W_t)=W_t^3=\sum_{i=0}^{n-1}(f(W_{t_{i+1}})-f(W_{t_i}))\\=\sum_{i=0}^{n-1} \{ f'(W_{t_i})(\Delta W)_i +\frac{1}{2} f''(W_{t_i})(\Delta W)_i^2 +f'''(\gamma_i)(\Delta W)_i^3 \}\\= 3\sum_{i=0}^{n-1} W_{t_i}^2(\Delta W)_i +3\sum_{i=0}^{n-1} W_{t_i}(\Delta W)_i^2+6\sum_{i=0}^{n-1}(\Delta W)_i^3,$$ where $\gamma_i$ comes from Taylor expansion.
It is easy to see the last term converges to zero in $L^2$.
Then he used the fact that $\sum_{i=0}^{n-1}W_{t_i}(\Delta W)_i^2$ converges to $\sum_{i=0}^{n-1}W_{t_i}(t_{i+1}-t_i)$ in probability to conclude that this term converges to $\int_0^t W_sds$ (without mentioning what kind of convergence it is). Actually this is the step which confuses me. In this step he used convergence in probability and pointwise convergence(for deriving the Riemann integral).
The convergence of Ito integral in $L^2$ is by definition and trivial.
Best Answer
The key theorem is, for a sequence of random variables $(X_n)$,
is equivalent to
For your problem, you can rewrite the expansion as $3 \sum W_{t_i}(\Delta W)_i^2 = W_t^3 - 3 \sum W_{t_i}^2 (\Delta W)_i - 6 \sum (\Delta W)_i^3$. As all the terms on the right hand side converge in $L^2$, the term on the right converges in $L^2$, which is equivalent to $\{(W_t^3 - 3 \sum W_{t_i}^2 (\Delta W)_i - 6 \sum (\Delta W)_i^3)^2\}$ is uniformly integrable. Hence $\{(3 \sum W_{t_i}(\Delta W)_i^2)^2\}$ is also uniformly integrable. Since $3 \sum W_{t_i}(\Delta W)_i^2$ converges to $3\int_0^t W_s ds$ in probability, this uniform integrability implies that the convergence also holds in $L^2$.
The equality in $$f(W_t) = f(W_0)+\int_0^t f'(W_s)dW_s + \frac 12 \int_0^t f''(W_s)ds =: X$$ is in the almost sure sense. You show the limits in $L^2$, i.e. you show $(X_n) \rightarrow X$ in $L^2$ where the $(X_n)$s are the Taylor approximations. But each $X_n$ is also equal to $f(W_t)$, so you trivially also have $(X_n) \rightarrow f(W_t)$ in $L^2$. By uniqueness of $L^2$ limits, $f(W_t) = X$ a.s.