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

$\phi(t)= \begin{cases} 1,&\text{if $|x|< 1$;}\\ e^{-(|x|-1)^2} ,&\text{if $|x|\geq1$.} \end{cases} $

Can anyone help?

share|cite|improve this question
If this was the cf of a random variable, what would the moments of that rv be? – guy Jan 19 '13 at 21:39

No, any characteristic function that is equal to 1 on an interval around 0 must be equal to 1 everywhere. This can easily be deduced from the the fact that $|\phi(t)|\leq 1$ and the inequality $1-\cos(2t)\leq 4(1-\cos(t))$ which allows you to conclude $1-\Re \phi(2t)\leq 4[1-\Re \phi(t)]$ which is essentially a statement that says the behavior of $\phi(t)$ is regulated by local behavior around 0.

share|cite|improve this answer

Just to iron out the details in my comment, since others have also posted complete answers, $\phi(t)$ is infinitely differentiable at $0$; in fact, $\left.\frac{d^n \phi}{dt^n} \right|_{t = 0} = 0$ and hence if $\phi(t)$ is the cf of some random variable $X$ it must be that $E[X^n] = 0$ for all $n$. In particular, $\mbox{Var}(X) = 0$ and $E[X] = 0$ so that $X = 0$ a.s. But $\phi(t)$ is not the cf of a point mass at $0$.

share|cite|improve this answer

If it was the characteristic function of a random variable, we could find its density as $\phi$ is integrable (it involves inverse formula). But the obtained function is not the density of a random variable.

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.