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Let $X$ be a random variable and $(x_n)_n$ a sequence of $\mathbb{R}^*$ such that $\lim_nx_n=0,\forall k \in \mathbb{N},|\varphi_{X}(x_k)|=1.$

Prove that $X$ is degenerate.

To show that $X$ is degenerate, it's sufficient to prove that $\forall x \in \mathbb{R},|\varphi_X(x)|=1,$ or for two values $p,q$ such that $p/q \in \mathbb{R}-\mathbb{Q},|\varphi_X(p)|=|\varphi_X(q)|=1.$

Do you have any suggestions?

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  • $\begingroup$ You presumably want to require $x_k\ne0,\forall k$ ? $\endgroup$ Commented Aug 21, 2020 at 21:52
  • $\begingroup$ Consider the additive subgroup of $\mathbb R$ generated by the set of $x_k$ values. $\endgroup$ Commented Aug 21, 2020 at 22:13
  • $\begingroup$ I have posted an answer after a long time. I will be happy to provide more details if needed. $\endgroup$ Commented Aug 22, 2020 at 23:32
  • $\begingroup$ Thank u for your reply. There is another way, without using independence, $\forall n \in \mathbb{N}, |\varphi_{X}(x_n)|=1$ which implies the existence of a sequence $(y_n)_n$ (chose $-\pi/|x_n| \leq y_n \leq \pi/|x_n|$ ), this implies $\forall \in \mathbb{N},P_X(y_n+\frac{2\pi}{|x_n|}\mathbb{Z})=1,$ so, for $\epsilon>0,$ $P(|X-y_n|>\epsilon) \leq P(X \neq y_n) \leq P(|Xx_n|>\pi/2)+P_X(\mathbb{R}-(y_n+\frac{2\pi}{|x_n|}\mathbb{Z}))=P(|Xx_n|>\pi/2)$ which converges to $0,$ this implies that $\forall x \in \mathbb{R},|\varphi_X(x)|\lim_n|\varphi_{X-y_n}(x)|=1$ which means that $X$ is degenerate $\endgroup$
    – Kurt.W.X
    Commented Aug 23, 2020 at 23:10
  • $\begingroup$ We can also use the inversion formula for lattice distribution $\endgroup$
    – Kurt.W.X
    Commented Aug 23, 2020 at 23:12

1 Answer 1

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Let $Y$ be independent of $X$ with the same distribution as $X$ and $Z=X-Y$. The characteristic function $g$ of $Z$ is given by $g(t)=|\phi (t)|^{2}$. We have $g(x_n)=1$ for all $n$. This gives $\int [1-\cos (x_nZ)]dP =0$ so $P(x_n Z \in 2\pi \mathbb Z)=1$. This implies that $P((Z=0)\cup |Z| \geq \frac {2 \pi} {|x_n|})=1$. If $N$ is any positive integer then $\frac {2 \pi} {|x_n|} >N$ for some $n$. Hence $P((Z=0) \cup (|Z| >N)=1$. But the intersection of the events $(Z=0) \cup (|Z| >N)$ over all $N$ is $Z=0$ so we get $P(Z=0)=1$. This implies that $X=Y$ a.s. In particular $X$ is independent of itself, so it is almost surely constant.

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  • $\begingroup$ You used independence to remove $(y_n)_n$ (defined as above), right? $\endgroup$
    – Kurt.W.X
    Commented Aug 24, 2020 at 11:38
  • $\begingroup$ Yes. The process of symmetrization (i.e. forming $Z=X-Y$ with $X,Y$ i.i.d. ) simplifies many such questions about characteristic functions. @Kurt.W.X $\endgroup$ Commented Aug 24, 2020 at 11:41

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