Same moments and boundedness of one rv implies same distribution If $X$ is a bounded random variable and $Y$ is another random variable (not assumed to be bounded) with $E[X^n] = E[Y^n]$ for $n = 1, 2, 3, . . .,$ . Then , show that $X$ and $Y$ must have same distribution.
 A: Say $X$ is bounded by some $M>0$. Then $E(|X|^n)^{1/n}\leq M$ and $\limsup_n \frac 1n E(|X|^n)^{1/n}=0$, so $\varphi_X$ (the characteristic function of $X$) is analytic at $0$ with infinite radius: $$\forall t\in \mathbb R, \varphi_X(t) = \sum_{n=0}^\infty \frac{(it)^n}{n!}E(X^n)$$
Note that $E(|Y|^{2n})^{1/(2n)} = E(|X|^{2n})^{1/(2n)}\leq M$ and by Lyapunov's inequality, $$E(|Y|^{2n-1})^{1/(2n-1)} \leq E(|Y|^{2n})^{1/(2n)}\leq M$$
hence $\limsup_n \frac 1n E(|Y|^n)^{1/n}=0$ and 
$$\forall t\in \mathbb R, \varphi_Y(t) = \sum_{n=0}^\infty \frac{(it)^n}{n!}E(Y^n)=\sum_{n=0}^\infty \frac{(it)^n}{n!}E(Y^n)=\varphi_X(t)$$
$X$ and $Y$ have the same characteristic distribution, hence the same distribution.

Lemma: If $|X|\leq M$ a.s., $\varphi_X$  is analytic at $0$ with infinite radius.
Proof: By Taylor's theorem with integral remainder $$\begin{aligned}
\forall x\in \mathbb R, e^{ix} 
&= \sum_{k=0}^{n-1} \frac{(ix)^k}{k!} + \int_0^x \frac{(x-t)^{n-1}}{(n-1)!}i^ne^{it} dt 
\end{aligned}$$
hence $$\forall x\in \mathbb R, \left|e^{ix} - \sum_{k=0}^{n-1} \frac{(ix)^k}{k!}\right|\leq \frac{|x|^n}{n!}$$
Let $t\in \mathbb R$. By the reverse triangle inequality, 
$$
\left|E(e^{itX}) - E\left(\sum_{k=0}^{n-1} \frac{(itX)^k}{k!}\right) \right|\leq E\left| e^{itX} - \sum_{k=0}^{n-1} \frac{(itX)^k}{k!} \right|
\leq \frac{|t|^nE(|X|^n)}{n!} \leq \frac{(M|t|)^n}{n!} \to 0
$$
Hence $\displaystyle \varphi_X(t) = \sum_{n=0}^\infty \frac{(it)^n}{n!}E(X^n)$.
A: If $|X| \leq M,$ then $$\Big|E[X^n]\Big| \leq E\big[|X|^n\big] \leq M^n,$$ which implies that $X$ has a moment generating function that converges everywhere.  Since $Y$ shares the same moments, the same is true for its moments, with the same moment generating function.  As any moment generating function with positive radius of convergence characterizes the distribution, $X$ and $Y$ have the same distribution.
