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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$\begingroup$
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
2 Answers
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Hint: Put $\varphi(t)=\cos(t^2)$. If $X$ is a random variable with this "characteristic function", we'd have $$\mathbb{E}(X^2)=-\varphi^{\prime\prime}(t)|_{t=0}=0.$$
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1I know this is a very old post, but I was looking for answers to this same question and I have a doubt with this argument. Couldn't it be that $X$ doesn't have a second order moment? [Related](http://math.stackexchange.com/questions/793788/continuous-probability-distribution-with-no-first-moment-but-the-characteristic) – 2015-04-06
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0@Duronman All moments exist since $cos(t^2)$ is infinitely differentiable at 0. – 2015-05-19
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0How is E(X^2)=0 imply that X doesn't have a second moment? Isin't zero a finte value? – 2016-09-02
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Introduce the notation $c(t)=\cos(t^2)$. By Bochner's theorem, every determinant $D(t)$ should be nonnegative, where $$ D(t)=\text{det}\begin{pmatrix}c(0) & c(t) & c(2t)\\ c(-t) & c(0) & c(t)\\ c(-2t) & c(-t) & c(0)\end{pmatrix}. $$ Since $c(t)=c(-t)$ for every $t$, $D(t)=(1-c(2t))E(t)$ with $E(t)=1-2\cos(t^2)^2+\cos(4t^2)$. For $t=1$, $E(1)=1-2\cos(1)^2+\cos(4)=-0.24...$ hence $D(1)<0$.
Alternatively, $E(t)=-6t^4+O(t^8)$ when $t\to0$. Or, $E(t)=-2\sin^2(t^2)(1+2\cos(2t^2))$.