Does characteristic function of bounded random variables have any additional properties?

More specifically, let $X$ be a symmetric random variable such that $P[|X|\le a]=1$.

What more can we say about $\phi_X(t)$ ?

  • $\begingroup$ When you say "additional properties", in addition to what do you have in mind? $\endgroup$ – grndl Aug 24 '16 at 15:56
  • $\begingroup$ In addition to what we already to know about characteristic function: continuity, $\phi(0)=1$. For example, can we say something about the tails of $\phi(t)$. $\endgroup$ – Boby Aug 24 '16 at 16:00
  • $\begingroup$ see mathoverflow.net/q/361604/11260 $\endgroup$ – Carlo Beenakker May 28 at 20:43

The boundedness of $X$ implies that moment generating function $\psi(z):=\Bbb E[e^{zX}]$ is an entire function $\psi:\Bbb C\to\Bbb C$. Of course $\phi_X(t) = \psi(it)$ for all real $t$. In particular, $\phi_X$ is a smooth function. The symmetry of $X$ implies (and is indeed equivalent to the fact) that $\phi_X$ is real valued.

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  • $\begingroup$ Would characteristic function have zeros if $X$ is continuous? What if it's discrete? $\endgroup$ – Boby Aug 24 '16 at 23:29
  • $\begingroup$ Examples: $X$ uniformly distributed on $(-1,1)$, hence bounded and symmetric. Then $\phi_X(t) = {\sin t\over t}$ for $t\not=0$ and $=1$ for $t=0$. This characteristic function has many zeros. Similarly, if $X$ takes the two values $\pm1$ with probability $1/2$ each, then $\phi_X(t) =\cos t$. $\endgroup$ – John Dawkins Aug 25 '16 at 16:37
  • $\begingroup$ These examples, are in particularly why I asked the question about zeros. I thought maybe this is a more general behaviour for bound random variables. $\endgroup$ – Boby Aug 25 '16 at 17:08

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