Usually when we try to show a function is not a characteristic function, we would prove it is not uniformly continuous. I am wondering if there is any other way to show $\cos(t^2)$ is not a characteristic function. $%fooling edit check$
Showing $\cos(t^2)$ is not a Characteristic Function
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probability
probability-theory
probability-distributions
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0For a characteristic function $\varphi$, isn't the set of points $t$ with $\varphi(t)=1$ supposed to be contained in a discrete subgroup of $\mathbb R$ ? But in this case, the group generated is dense. – 2011-09-17
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0If we can show that $E(X^2)=0$, how does that imply $\phi$ is not a char fun? thanks but $$-\varphi^{\prime\prime}(t)|_{t=0}=0.$$ is true since $\phi''=2\sin(t^2)+4t^2 \cos(t^2)$ – 2011-09-17
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0show that it is not positive definite – 2011-09-17