# Is this function - characteristic function of a random variable?

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

Can anyone help?

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If this was the cf of a random variable, what would the moments of that rv be? – guy Jan 19 '13 at 21:39
– Byron Schmuland Jan 20 '13 at 0:55

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.
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 $0$.
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.