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I know that if $$\mathbb E[e^{\xi X}]=e^{\frac{\xi^2}{2}}$$ for all $\xi\in\mathbb R$, then $X\sim \mathcal N(0,1)$. Now, if $$\mathbb E[e^{\xi X}]=e^{\frac{\xi^2}{2}}$$ for all $\xi >0$ does $X\sim \mathcal N(0,1)$ as well ?

I really can't find a counter example, and I know that to have $X\sim Y$ we need that $\mathbb E[e^{tX}]=\mathbb E[e^{tY}]$ for all $|t|\leq t_0$ for some $t_0>0$. So, I'm not so sure if $\mathbb E[e^{\xi X}]=e^{\frac{\xi^2}{2}}$ for all $\xi>0$ is enough to have $X\sim N(0,1)$.

Any idea ?

John
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1 Answers1

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Yes, it does imply $X \sim N(0,1)$.

Since $\mathbb{E}\exp(\xi X) = \exp(\xi^2/2)$ for all $\xi \geq 0$, it follows by differentiation (w.r.t. $\xi$) that $X$ has the same moments as the standard Gaussian distribution, i.e.

$$\mathbb{E}(X^n) =\mathbb{E}(U^n)\tag{1}$$

for all $n \in \mathbb{N}$, where $U \sim N(0,1)$ is standard Gaussian. In particular, $$\mathbb{E}(|X|^{2n}) =\mathbb{E}(|U|^{2n}), \qquad n \in \mathbb{N}. \tag{2}$$ Noting that

$$\mathbb{E}(|X|^{2n+1}) \leq 1+ \mathbb{E}(|X|^{2n+2}),\tag{3}$$

we get

\begin{align*} \mathbb{E}\exp(|\xi| \cdot |X|) &= \mathbb{E} \left( \sum_{k=0}^{\infty} \frac{|\xi|^k \, |X|^k}{k!} \right) \\ &= \sum_{n=0}^{\infty} \frac{|\xi|^{2n+1} \mathbb{E}(|X|^{2n+1})}{(2n+1)!} + \sum_{n=0}^{\infty} \frac{|\xi|^{2n} \mathbb{E}(|X|^{2n})}{(2n)!} \\ &\stackrel{(3)}{\leq} \sum_{n=0}^{\infty} \frac{|\xi|^{2n+1}}{(2n+1)!} +\frac{1}{|\xi|} \sum_{n=0}^{\infty} \underbrace{(2n+2)}_{\leq 2^{(2n+2)}} \frac{|\xi|^{2n+2} \mathbb{E}(|X|^{2n+2})}{(2n+2)!} + \sum_{n=0}^{\infty} \frac{|\xi|^{2n} \mathbb{E}(|X|^{2n})}{(2n)!} \\ &\stackrel{(2)}{\leq} \exp(|\xi|) + \frac{1}{|\xi|} \mathbb{E}\exp(2|\xi| \cdot |U|) + \mathbb{E}\exp(|\xi| \cdot |U|)<\infty \end{align*} for all $\xi \neq 0$. Now take some $\xi> 0$. Because of the above estimate, the dominated convergence theorem gives

$$\mathbb{E}\exp(-\xi X) =\mathbb{E}\exp \left( \sum_{k=0}^{\infty} \frac{(-\xi)^k}{k!} X^k \right) \stackrel{\text{dom. conv.}}{=} \sum_{k=0}^{\infty} \frac{(-\xi)^k}{k!} \mathbb{E}(X^k).$$

Hence, by $(1)$,

$$\mathbb{E}\exp(-\xi X) = \sum_{k=0}^{\infty} \frac{(-\xi)^k}{k!} \mathbb{E}(U^k) = \mathbb{E}\exp(-\xi U) = \exp(\xi^2/2).$$

Consequently, $\mathbb{E}\exp(\xi X)=\exp(\xi^2/2)$ for all $\xi \in \mathbb{R}$, which implies $X \sim N(0,1)$.

saz
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    There's a bit of work to be done here to justify differentiating under the integral in the first step. The usual domination argument fails because you no longer have access to a negative power of the exponential to control the contribution from negative values of the random variable. – Chris Janjigian Nov 20 '19 at 19:52
  • @ChrisJanjigian You are right; I was too quick about that. I don't have a quick fix (do you?) but I will think about it. – saz Nov 21 '19 at 07:25