Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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, $t$.

share|cite|improve this question

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 $$ \left|\mathrm E(\mathrm e^{\mathrm itX})-1-\mathrm it\mathrm E(X)+\tfrac12t^2\mathrm E(X^2)\right|\leqslant\mathrm E(|u(tX)|)\leqslant\mathrm E(|X|^{\alpha})\,|t|^\alpha. $$

share|cite|improve this answer
+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.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.