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

Let $X$ be a random variable with continuous density $\rho(x)$. Assume that $X$ is symmetric and $\vert X\vert<L$. Since it has a bounded support, all moments of $X$ are well-defined. Let $m_i$ denote the moment $i$ of $X$, i.e. $$ m_i = \int_{-L}^{L} x^i \rho(x) dx. $$ Let $\phi(t)=\int e^{jtx}\rho(x) dx$ be the characteristic function of $X$. Consider the Taylor expansion of $\phi(t)$, $$ \phi(t) = 1 - \frac{t^2}{2!}m_2 + \frac{t^4}{4!}m_4 - \frac{t^6}{6!}m_6+\cdots $$ I would like to lower bound $\phi(t)$ by $1 - \frac{t^2}{2!}m_2$. Particularly, can we say $\phi(t)\geq 1-\frac{t^2}{2!}m_2 $ for $t^2<\frac{2!}{m_2}$?

share|cite|improve this question

Assume only that $X$ is square-integrable with $\mathrm E(X)=0$ and $\mathrm E(X^2)=m_2$. Since $|\phi''(t)|\leqslant m_2$ for every $t$ and $\phi'(0)=0$, the mean value theorem for vector-valued functions shows that $|\phi'(t)|\leqslant m_2|t|$ for every $t$. Since $\phi(0)=1$, a second application of the mean value theorem yields $|\phi(t)-1|\leqslant\frac12m_2t^2$, hence $\phi(t)\geqslant1-\frac12m_2t^2$ for every $t$.

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.