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Suppose we have a non-negative random variable $X$ with density $p(x)$,and its characteristic function, evaluated at a complex number $z$, being $\phi(z)=E[e^{z X}]=\int_{0}^{\infty}e^{zx}p(x)dx$.

It seems that if $z$ has strictly negative real part, then $|\phi(z)|<1$. My question is, given any such $z$ with a strictly negative real parts, can we estimate how small the value $|\phi(z)|<1$ is as compared with one? Also how does the magnitude $|\phi(z)|<1$ affected by the properties of the density $p(x)$?

I am no expert in Fourier or Complex analysis, and would be very grateful if someone can point to the relevant part of the theory. Many many thanks.

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searching for "bounds of characteristic functions" gives quite a number of hits that seem relevant, like… – Carlo Beenakker Aug 29 '12 at 21:22

The inequality is best possible: take $p=\delta$ (delta-function). Then $\phi\equiv 1$. If $a>0$ is the minimum of the support of $p$, we have $|\phi(z)|\leq \exp(a \Re z)$, and this is also best possible.

That $\delta$ is not a density in the usual sense does not matter, it can be approximated with smooth densities as close as you wish.

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$|\phi(z)| \le E\left[e^{-\text{Re}(z) X}\right]$. If $\alpha > 0$ and $P(X > \alpha) = \beta > 0$, we then have $|\phi(z)| \le 1 - \beta + \beta e^{-\text{Re}(z) \alpha}$.

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