Let $(M_n)_{n\in\mathbb N_0}$ be a square-integrable martingale on a filtered probability spacce $(\Omega,\mathcal A,(\mathcal F_n)_{n\in\mathbb N_0},\operatorname P)$ with $M_0=0$ almost surely such that $$Y_n:=M_n-M_{n-1}\;\;\;\text{for }n\in\mathbb N$$ is stationary. Below equation (2.36) on p. 53 on this lecture notes, the following claim is made:
I don't understand how the remainder $R_{n,\:k}$ is defined such that it is really of order $o(1/n)$. Moreover, do we need to apply some special version of Taylor's theorem to ensure measurbility?
Best Answer
Let $f : [a, b] \to \mathbf{R}$ be a $d+1$ times continuously differentiable function. Then one can write $$f(b)=f(a)+\frac{(b-a)}{1!}f'(a)+\cdots+\frac{(b-a)^d}{d!}f^{(d)}(a)+\int_a^b \frac{(b-t)^d}{d!}f^{(d+1)}(t)dt.$$ If $f$ is the exponential function this yields : $$e^b = e^a \sum_{j=0}^{d} \frac{(b-a)^j}{j!} + \int_{a}^{b} \frac{(b-t)^d}{d!} e^t dt$$ which gives by a change of variable in the integral remainder : $$e^b = e^a \sum_{j=0}^{d} \frac{(b-a)^j}{j!} + \frac{(b-a)^{d+1}}{d!}\int_{0}^{1} (1-u)^d e^{(b-a)u + a} du$$ and this is valid for any $d\in\mathbf{N}$ as the exponentiel has continuous derivatives of all orders. If $b = x$ and $a = 0$ this gives : $$e^x - 1 = \sum_{j=1}^{d} \frac{x^j}{j!} + \frac{x^{d+1}}{d!}\int_{0}^{1} (1-u)^d e^{xu} du\;\;\;\;\;\; (F)$$ for any $d\in\mathbf{N}$.
Now without even thinking which optimal order $d$ you should take (you'll do it yourself properly later) simply plug $x = \frac{ip}{\sqrt{n}} Y_k + \frac{\sigma^2 p^2}{2n}$ inside the previous equation (F). You have : $$e^{\frac{ip}{\sqrt{n}} Y_k + \frac{\sigma^2 p^2}{2n}} - 1 = \frac{ip}{\sqrt{n}} Y_k + \frac{\sigma^2 p^2}{2n} + \frac{1}{2} \left(\frac{ip}{\sqrt{n}} Y_k + \frac{\sigma^2 p^2}{2n}\right)^2 + \sum_{j=3}^d \frac{\left(\frac{ip}{\sqrt{n}} Y_k + \frac{\sigma^2 p^2}{2n}\right)^j}{j!} + \frac{\left(\frac{ip}{\sqrt{n}} Y_k + \frac{\sigma^2 p^2}{2n}\right)^{d+1}}{d!} \int_0^1 (1-u)^d e^{u\left(\frac{ip}{\sqrt{n}} Y_k + \frac{\sigma^2 p^2}{2n}\right)} du$$ and developping the square and factorizing gives : $$e^{\frac{ip}{\sqrt{n}} Y_k + \frac{\sigma^2 p^2}{2n}} - 1 = \frac{ip}{\sqrt{n}} Y_k - \frac{p^2}{2n} \left(Y_k^2 - \sigma^2\right) + \frac{1}{2}\left(\frac{\sigma^4 p^4}{4n^2} + 2i \frac{p Y_k}{\sqrt{n}} \frac{\sigma^2 p^2}{2n}\right) + \sum_{j=3}^d \frac{\left(\frac{ip}{\sqrt{n}} Y_k + \frac{\sigma^2 p^2}{2n}\right)^j}{j!} + \frac{\left(\frac{ip}{\sqrt{n}} Y_k + \frac{\sigma^2 p^2}{2n}\right)^{d+1}}{d!} \int_0^1 (1-u)^d e^{u\left(\frac{ip}{\sqrt{n}} Y_k + \frac{\sigma^2 p^2}{2n}\right)} du$$
The quantity $\frac{ip}{\sqrt{n}} Y_k - \frac{p^2}{2n} \left(Y_k^2 - \sigma^2\right)$ is the one appearing in the lectures notes you quote.
The quantity $$\frac{1}{2}\left(\frac{\sigma^4 p^4}{4n^2} + 2i \frac{p Y_k}{\sqrt{n}} \frac{\sigma^2 p^2}{2n}\right) + \sum_{j=3}^d \frac{\left(\frac{ip}{\sqrt{n}} Y_k + \frac{\sigma^2 p^2}{2n}\right)^j}{j!}$$ is clearly a $o\left(\frac{1}{n}\right)$ as $n \to +\infty$. Same for the integral part -- I leave you work this out alone.
All of this is for any $d \geq 3$ so you should choose $d = 3$.