Let $X_1,X_2,\ldots$ be a martingale and $Y_n$ be the martingale differences, which satisfy \begin{align} \mathbb{E}[Y_n^2\mid\mathcal{F}_{n-1}]=1,n=1,2,\ldots,\sup_{n\geq 1}\mathbb{E}[|Y_n|^3\mid\mathcal{F}_{n-1}]=K \end{align}

Let $a_{n,N}=\mathbb{E}(\exp(i \sigma X_n/\sqrt{N}))$. Then I want to show that \begin{align} \left|a_{n,N}-[1-\sigma^2/2N]a_{n-1,N}\right|&\leq K\sigma^3 /6N^{3/2} \end{align}

I'm looking at equation (2.23) of prop 2.2 on p. 4 of http://www.math.lsa.umich.edu/~conlon/math625/chapter2.pdf. I can't figure out why it's true.


First observe that $\exp\left(i\sigma\frac{X_{n-1}}{\sqrt N}\right)$ is $\mathcal F_{n-1}$-measurable hence $$ a_{n,N}=\mathbb E\left[\mathbb E\left[\exp\left(i\sigma\frac{X_n}{\sqrt N}\right)\mid \mathcal F_{n-1}\right]\right]=\mathbb E\left[\exp\left(i\sigma\frac{X_{n-1}}{\sqrt N}\right)\mathbb E\left[\exp\left(i\sigma\frac{Y_n}{\sqrt N}\right)\mid \mathcal F_{n-1}\right]\right]. $$ It follows that $$ a_{n,N}-[1-\sigma^2/2N]a_{n-1,N}=\mathbb E\left[\exp\left(i\sigma\frac{X_{n-1}}{\sqrt N}\right)\left(\mathbb E\left[\exp\left(i\sigma\frac{Y_n}{\sqrt N}\right)\mid \mathcal F_{n-1}\right]-[1-\sigma^2/2N] \right)\right] $$ and since $\exp\left(i\sigma\frac{X_{n-1}}{\sqrt N}\right)$ is bounded by $1$, we get $$ \left|a_{n,N}-[1-\sigma^2/2N]a_{n-1,N}\right|\leqslant\left\lvert \mathbb E\left[\exp\left(i\sigma\frac{Y_n}{\sqrt N}\right)\mid \mathcal F_{n-1}\right]-[1-\sigma^2/2N]\right\rvert. $$ Then use Taylor's formula and the assumption on the conditional moments to get the wanted inequality.

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  • $\begingroup$ Thanks. How do we eliminate the expectations of the higher order powers in the Taylor expansion of the exponential? $\endgroup$ – mlstudent Jun 19 '18 at 22:03
  • $\begingroup$ You need a Taylor formula for $e^{ix}$ where the remainder is estimated in absolute value by a constant times $x^3$. $\endgroup$ – Davide Giraudo Jun 20 '18 at 9:25

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