If $\mathbb E[e^{\xi X}]=e^{\frac{\xi^2}{2}}$ for all $\xi >0$ does $X\sim \mathcal N(0,1)$? 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 ? 
 A: 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 $X \sim N(0,1)$.
