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If $X$ and $Y$ are two real random variables such that: $\exists \delta>0;\forall x \in [-\delta,\delta],\varphi_{X}(x)=e^{-\frac{x^2}{2}}$ and $\varphi_Y(x)=1$. Prove that $X$ is normally distributed and that $Y=0$ a.s.

For $Y$ I used the inequality : $1-\mathrm{Re}(\varphi_Y(x)) \leq 4(1-\mathrm{Re}(\varphi(\frac{x}{2^k})))$ for $|x|>\delta$ and $k=\lfloor\frac{\ln(\frac{|x|}{\delta})}{\ln(2)}\rfloor+1$ and we get $ \forall x \in \mathbb{R},\varphi_Y(x)=1$ and $Y=0$ a.s.

For $X$, I tried to find an inequality, similar to above, to prove that $\forall x \in \mathbb{R},\varphi_X(x)=e^{-\frac{x^2}{2}}$

So, do you know any inequalities for $\varphi_X$?

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One way: You know $\varphi_X$ is real-analytic near $0$, hence has a positive radius of convergence. So $X$ is completely determined by its moments (see, e.g., Billingsley Probability and Measure Chapter 30), and so just compare.

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  • $\begingroup$ Is there another way? $\endgroup$
    – Ken
    Jun 1, 2019 at 15:44

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