# Degenerate random variable and characteristic functions

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?

• You presumably want to require $x_k\ne0,\forall k$ ? Commented Aug 21, 2020 at 21:52
• Consider the additive subgroup of $\mathbb R$ generated by the set of $x_k$ values. Commented Aug 21, 2020 at 22:13
• I have posted an answer after a long time. I will be happy to provide more details if needed. Commented Aug 22, 2020 at 23:32
• 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 Commented Aug 23, 2020 at 23:10
• We can also use the inversion formula for lattice distribution Commented Aug 23, 2020 at 23:12

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.
• You used independence to remove $(y_n)_n$ (defined as above), right? Commented Aug 24, 2020 at 11:38
• 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 Commented Aug 24, 2020 at 11:41