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I'm currently self-studying to be quant and have been thoroughly enjoying PW's book. I have some questions regarding his derivation of Ito's lemma. Specifically, I can see that the first line in his expansion comes from

$$ \begin{split} (F(X(t + h)) - F(X(t))) + (F(X(t + 2h)) - F(X(t + h))) + . . . + (F(X(t + nh)) -F(X(t + (n - 1)h)))\\ = F(X(t + nh)) -F(X(t + (n - 1)h)) + F(X(t + (n - 1)h)) - ... - F(X(t + 2h)) + F(X(t + 2h))\\ - F(X(t + h)) + F(X(t + h)) - F(X(t))\\ = F(X(t + nh)) - F(X(t))\\ = F(X(t + \delta t)) - F(X(t)) \end{split} $$ By using the definition $$ \begin{split} W(t) &= \int^t_0 f(\tau)dX(\tau)\\ &= \lim_{n\to\infty}\sum_{j=1}^{n} f(t_{j-1})(X(t_j) - X(t_{j-1}))\\ &= \lim_{n\to\infty}\sum_{j=1}^{n} f((j-1)t/n)(X(jt/n) - X((j-1)t/n))\\ \end{split} $$ which I can see that the second line becomes $$ \begin{split} \sum^n_{j=1}(X(t+jh)-X(t+(j-1)h))\frac{dF(X(t+(j-1)h))}{dX} \\ = \sum^n_{j=1}(X(t+j\delta t/n)-X(t+(j-1)\delta t/n))\frac{dF(X(t+(j-1)\delta t/n)))}{dX}\\ = \int^{t+\delta t}_{t} \frac{dF}{dX}dX \end{split} $$ Finally, what I don't quite see if how the final term goes from $$ \frac{1}{2}\frac{d^2F(X(t))}{dX^2}\sum^n_{j=1}(X(t+jh)-X(t+(j-1)h))^2 $$ to $$ \frac{1}{2}\frac{d^2F}{dX^2}(X(t))\delta t $$ to then become $$ \frac{1}{2}\int^{t+\delta t}_{t}\frac{d^2F(X(\tau))}{dX^2}d\tau $$ If in the mean square limit of $n\rightarrow\infty$ for $t_j = jt/n$, $$ \sum^n_{j=1} (X(t_j)-X(t_{j-1}))^2 = \sum^n_{j=1} (X(jt/n)-X((j-1)t/n))^2 = t $$ then I don't quite see how $$ \sum^n_{j=1} (X(t+jh)-X(t+(j-1)h))^2 = \sum^n_{j=1} (X(t+j\delta t/n)-X(t+(j-1)\delta t/n))^2 = \delta t ? $$ don't they differ by a $t + $ term? Perhaps I don't see this due to my maths being rusty. And also finally, I don't see how $\frac{1}{2}\frac{d^2F(X(t))}{dX^2}\delta t$ becomes $\frac{1}{2}\int^{t+\delta t}_{t}\frac{d^2F(X(\tau))}{dX^2}d\tau$ when the integrand is also dependent on $\tau$.

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  • $\begingroup$ Hi: I don't have time to read about the first question. but, for the second, what he's doing is using the fact that, if $d \tau $ is really small, then one can think the integrand as a constant. ( like a tall narrow bar in the reimann integral ) so that integrating is the same as just taking the bar and multiplying it by the length of the horizontal axis. The length of the horizontal axis is the value at the top of the integral - the value at the bottom of the integral.so you end up with the same thing. $\endgroup$
    – mark leeds
    Sep 25 at 16:30
  • $\begingroup$ Does it help to realize that $t_n=t$ ? Hence your equation ending with $=t$. Likewise, for your next equation you are having trouble with: Does it help to realize that $t+nh=t+\delta t$ ? This immediately implies that equation. Both these equations are well known from the quadratic variation of Brownian motion. $\endgroup$
    – Kurt G.
    Sep 27 at 14:53
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Given my earlier comment, the only open question is how $\frac{1}{2}\frac{d^2F(X(t))}{dX^2}\delta t$ becomes $\frac{1}{2}\int^{t+\delta t}_{t}\frac{d^2F(X(\tau))}{dX^2}d\tau\,.$ A more standard proof is this: Writing $$ t_j=t+jh $$ we have

\begin{align} &\sum_{j=1}^n\frac{d^2F}{dX^2}\Big(X(t_{j-1})\Big)\Big(X(t_j)-X(t_{j-1})\Big)^2\\ &=\sum_{j=1}^n\frac{d^2F}{dX^2}\Big(X(t_{j-1})\Big)\Big(X^2(t_j)-X^2(t_{j-1})+2X(t_{j-1})(X(t_{j-1})-X(t_j))\Big)\\ &\to\int_t^{t+\delta t}\frac{d^2F}{dX^2}\Big(X(\tau)\Big)\,dX^2(\tau)-2\int_t^{t+\delta t}\frac{d^2F}{dX^2}\Big(X(\tau)\Big)\,X(\tau)\,dX(\tau)\\ &=\int_t^{t+\delta t}\frac{d^2F}{dX^2}(X(\tau))\,d\tau\,. \end{align} The last line follows from $dX^2=2X\,dX+dt\,.$ Wilmott's approximation $$ \frac{d^2F}{dX^2}(X(t_{j-1}))=\frac{d^2F}{dX^2}(X(t)) $$ is not needed.

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  • $\begingroup$ From line 1 to 2, I don't see how you get $2X(t_{j-1})(X(t_{j-1})-X(t_j))$ instead of $2X(t_j)X(t_{j-1})$? $\endgroup$ Oct 12 at 14:49
  • $\begingroup$ Is it not $(a-b)^2=a^2-2ab+b^2=a^2-2ab-b^2+2b^2=a^2-b^2+2b(b-a)$ ? $\endgroup$
    – Kurt G.
    Oct 12 at 15:17

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