Let $X,Y$ be real random variables.
I know that if $\mathbb{E}[X^n] =\mathbb{E}[Y^n] $ and $\mathbb{E}[X^n] \leq \frac{Mn!}{r^n} \hspace{0.1cm} \forall \hspace{0.1cm} n \in \mathbb{N}$, then $\varphi_X = \varphi_Y$.
I was wondering how can we weaken the hypothesis on $Y$, for example, is it possible to require that $\mathbb{E}[Y^n] =\mathbb{E}[X^n]$ and the bound be true only for even $n \in \mathbb{N}$ ? I think the answer is afirmative, but if so why should $Y$ verify $\mathbb{E}[Y^n] \leq \frac{Mn!}{r^n}$ for odd $n$ ?
I try something the inequality $\lvert y \rvert \leq \frac{y^2}{2} + \frac{1}{2}$ to use that $\lvert y^n \rvert \leq \lvert y^{n-1} \rvert \cdot (\frac{y^2}{2} + \frac{1}{2}) \leq \frac{\lvert y^{n-1} \rvert}{2} + \frac{\lvert y^{n+1} \rvert}{2}$ in order to use that $n+1$ and $n-1$ are both even but didn't menage to conclude from here.
Any help, hint or solution would be appreciated.