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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 $t^2<\frac{2!}{m_2}$?

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1 Answer 1

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 $t$.

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