# $\lim\limits_nx_n=0,\forall n \in \mathbb{N},\phi_X(x_n)=e^{-\sigma^2x^2_n} \implies \forall x \in \mathbb{R}, \phi_X(x)=e^{-x^2\sigma^2}$

$$X$$ is a random variable such that there exist a sequence $$(x_n)_n$$ in $$\mathbb{R}^*,\lim_nx_n=0,$$ and for all $$n \in \mathbb{N},\phi_X(x_n)=e^{-\sigma^2x^2_n}, \sigma \in \mathbb{R}.$$

Prove that $$X$$ is normally distributed (possible degenerate).

The case where $$\sigma=0$$ (degenerate distribution) was asked here.

Any suggestions on how to deal with $$\sigma \neq 0$$?

• Hint: define $X_n=X+x_n$. – Martín Vacas Vignolo Mar 4 at 7:02
• $X_n$ converges in distribution to $X,$ and $\phi_{X_n}(x)=e^{ixx_n} \phi_X(x),\phi_{X_n}(x_n)=e^{x^2_n(i-\sigma^2)},$ why introducing $X_n$? – Kurt.W.X Mar 4 at 17:11
• Any regularity assumption on $X$ (moments, analyticity of $\phi_X$, ...) ? – Gabriel Romon Mar 8 at 20:56
• No, nothing is assumed on the moments, analyticity of $\phi_X.$ (Unless we succeed to prove them) – Kurt.W.X Mar 8 at 21:26

Let $$Y$$ be a random variable with characteristic function $$\phi_Y:t\mapsto e^{-\sigma^2t^2}$$, i.e. $$Y$$ is Gaussian (possibly degenerate).

The hypothesis in the problem states that $$\phi_X$$ and $$\phi_Y$$ coincide along a sequence $$(x_n)$$ of non-zero reals such that $$x_n\to 0$$. If $$\phi_X$$ were analytic in a neighborhood of $$0$$, so would be $$\phi_X - \phi_Y$$ and since the zeroes of an analytic function are isolated, this would imply $$\phi_X - \phi_Y = 0$$, as wanted.

Instead of showing directly that $$\phi_X$$ is analytic, we show that $$X$$ has moments of all orders. This will imply that $$\phi_X$$ is infinitely differentiable. We show next that $$\forall n \geq 0$$, $$\phi_X^{(n)}(0)=\phi_Y^{(n)}(0)$$. Since $$Y$$ is uniquely determined by its moments (see here or here), this will yield $$\phi_X=\phi_Y$$.

For $$n\geq 1$$, let $$\displaystyle f_n:t\mapsto \frac{2-e^{ix_n t}-e^{-ix_n t}}{x_n^2}$$ and note that it is real-valued, non-negative and by a Taylor expansion $$\forall t\in \mathbb R, f_n(t) \to t^2.$$ By Fatou's lemma, \begin{align} E(X^2) &\leq \liminf_n E(f_n(X)) \\ &=\liminf_n \frac{2-\phi_X(x_n)-\phi_X(-x_n)}{x_n^2} \\ &=\liminf_n \frac{2-\phi_X(x_n)-\overline{\phi_X(x_n)}}{x_n^2} \\ &=\liminf_n \frac{2-\phi_Y(x_n)-\phi_Y(-x_n)}{x_n^2} \\ &= -\phi_Y''(0) \end{align} where the last equality follows from a Taylor expansion of $$\phi_Y$$. Hence $$X$$ has a second moment, and by a classical result $$\phi_X$$ is twice-differentiable over $$\mathbb R$$.

Since $$\phi_Y$$ is even we may assume WLOG that all the $$x_n$$ are positive, and also that $$(x_n)$$ is strictly decreasing (by taking a subsequence if needed). By Rolle's theorem applied to $$\phi_X-\phi_Y$$, there is some sequence $$(y_n)$$, strictly decreasing, with $$y_n\to 0$$ and $$\forall n\geq 1, \phi_X'(y_n) = \phi_Y'(y_n)$$. Another application of Rolle's provides some sequence $$(z_n)$$, strictly decreasing, with $$z_n\to 0$$ and $$\forall n\geq 1, \phi_X''(z_n) = \phi_Y''(z_n)$$.

Defining a new $$f_n$$ and performing a similar reasoning, we prove successively that $$E(X^4)<\infty$$, $$\phi_X^{(3)}$$ and $$\phi_Y^{(3)}$$ coincide along a sequence that goes to $$0$$ and similarly for $$\phi_X^{(4)}$$ and $$\phi_Y^{(4)}$$. This continues indefinitely. $$\phi_X$$ is therefore $$C^\infty$$, each $$\phi_X^{(m)}$$ is continuous and taking limits along the aforementioned sequences we find that $$\forall m\geq 1$$, $$\phi_X^{(m)}(0)=\phi_Y^{(m)}(0)$$.

Note that the proof follows through if $$Y$$ is replaced by any random variable that is uniquely determined by its moments. For example, it is sufficient that $$\limsup_n \frac{E(|Y|^n)^{1/n}}{n} <\infty$$.