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Given that $X$ is a random variable such that $E(|X|^{\alpha})<\infty$, where $\alpha \in (2,3)$, show that $|E(e^{itX})-(1+iE(X)t-E(X^{2})\frac{t^2}{2})|<K|t|^{\alpha}$ for some $K<\infty$ and any real number, $t$.

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3 Answers 3

The result holds for every $\alpha$ in $[2,3]$, with $\color{red}{K=\mathrm E(|X|^{\alpha})}$.

Let $u(x)=\mathrm e^{\mathrm ix}-1-\mathrm ix+\frac12x^2$. Then $u(0)=u'(0)=u''(0)=0$ and $|u^{(3)}(x)|=1$ for every real number $x$. By Taylor's theorem, $|u(x)|\leqslant\frac16|x|^3$ for every $x$ and in particular for every $|x|\leqslant1$.

Let $v(x)=\mathrm e^{\mathrm ix}-1-\mathrm ix$. Then $v(0)=v'(0)=0$ and $|v''(x)|=1$ for every real number $x$. By Taylor's theorem, $|v(x)|\leqslant\frac12|x|^2$ for every $x$ hence $|u(x)|\leqslant|v(x)|+\frac12|x|^2\leqslant|x|^2$ for every $x$ and in particular for every $|x|\geqslant1$.

Let $\alpha$ in $[2,3]$. For every $|x|\geqslant1$, $|x|^2\leqslant|x|^{\alpha}$. For every $|x|\leqslant1$, $\frac16|x|^3\leqslant|x|^{\alpha}$. Hence the two upper bounds on $|u|$ can be combined into the fact that $|u(x)|\leqslant|x|^\alpha$ for every $x$.

Applying this upper bound to $|u(tX)|$ and integrating the resulting inequality yields $$ \left|\mathrm E(\mathrm e^{\mathrm itX})-1-\mathrm it\mathrm E(X)+\tfrac12t^2\mathrm E(X^2)\right|\leqslant\mathrm E(|u(tX)|)\leqslant\mathrm E(|X|^{\alpha})\,|t|^\alpha. $$

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+1. Very nice and unexpected proof. –  Ashok Nov 19 '11 at 10:54

If you meant $\alpha\in \{2,3\}$, here is a proof.

Using the fact that $\frac{d^r}{dx^r}[e^{ix}]=\left[\frac{d^r}{dx^r} \cos x\right]+i \left[\frac{d^r}{dx^r} \sin x\right]=i^r e^{ix}$, and using the Taylor's expansion of $\sin x$ and $\cos x$ about $0$ we get $$\left|e^{ix}-\sum_{k=0}^{r-1}\frac{(ix)^k}{k!}\right|\le \frac{|x|^r}{r!}$$

Hence, also $$\begin{eqnarray*}\left|e^{ix}-\sum_{k=0}^{r-1}\frac{(ix)^k}{k!}\right|&\le& \left|e^{ix}-\sum_{k=0}^{r-2}\frac{(ix)^k}{k!}\right|+\frac{|x|^{r-1}}{(r-1)!}\\ &\le& \frac{2|x|^{r-1}}{(r-1)!}\end{eqnarray*}$$

Hence $$\begin{eqnarray*}\left|E(e^{itX})-\sum_{k=0}^{r-1}\frac{E((itX)^k)}{k!}\right|&\le&E\left|e^{iX}-\sum_{k=0}^{r-1}\frac{(itX)^k}{k!}\right| \\&\le&\min\left\{\frac{E|X|^r}{r!}|t|^r, \frac{2E|X|^{r-1}}{(r-1)!}|t|^{r-1}\right\} \end{eqnarray*}$$ $r=3$ gives you the answer.

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Consider $f(x) = \mathrm{e}^{i x} - \left(1 + i x - \frac{1}{2} x^2 \right)$.

Notice that, $\forall x\in \mathbb{R}$: $$ \vert f(x) \vert \le \vert \mathrm{e}^{i x} \vert + 1 + \vert x \vert + \frac{\vert x \vert^2}{2} = 2 + \vert x \vert + \frac{\vert x \vert^2}{2} = \frac{1}{2} \left( \vert x \vert +1 \right)^2 + \frac{3}{2} $$ Therefore, for $2 < \alpha <3$, $\lim_{\vert x \vert \to \infty} \frac{\vert f(x) \vert}{\vert x \vert^\alpha} = 0$. Using Taylor series for $f(x)$, $\lim_{\vert x \vert \downarrow 0} \frac{\vert f(x) \vert}{\vert x\vert^\alpha}$ = 0

Additionally $\frac{\vert f(x) \vert}{\vert x\vert^\alpha} \ge 0$ for all $x \not 0$. It follows therefore, that there exists a bounded function $g: \mathbb{R} \rightarrow \mathbb{R}$ such that $\vert f(x) \vert = \vert x\vert^\alpha g(x ) $. Let $K_1$ be such that $\forall x \in \mathbb{R}$, $g(x) \le K_1$.

Then $$ \vert \mathbb{E}(f(t X)) \vert \le \mathrm{E}( \vert f(X t) \vert ) = \mathrm{E}( \vert t X \vert^\alpha g( t X ) \le \vert t \vert^\alpha K_1 \mathrm{E}( \vert X \vert^\alpha ) = K \vert t \vert^\alpha $$

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