Let $X$ be a real random variable, with standard normal distribution.
Is the distribution of $X^2$ determined by its moments?
In general, if $n \in \mathbb N$, is the distribution of $X^n$ determined by its moments?
Let $X$ be a real random variable, with standard normal distribution.
Is the distribution of $X^2$ determined by its moments?
In general, if $n \in \mathbb N$, is the distribution of $X^n$ determined by its moments?
On
The most the moments could possibly determine is the distribution, not the random variable itself. I assume that's what you meant.
$X$ has moments $M_n(X) = 0$ for $n$ odd, $2^{n/2} \Gamma((n+1)/2)/\sqrt{\pi}$ for $n$ even. Thus the moments of $X^k$ are $M_n(X^k) = M_{nk}(X)$.
Carleman's condition says if a measure $\mu$ has moments $M_n$ of all orders and these satisfy $\sum_n M_{2n}^{-1/(2n)} = \infty$ then the moments determine the distribution. In the case of $X^k$ we have
$$ M_{2n}(X^k)^{1/(2n)} = M_{2nk}(X)^{1/(2n)} \sim Const \cdot n^{-k/2} $$
so the case $k=2$ does satisfy Carleman's condition, but any $k > 2$ does not. But this is only a sufficient condition, so I don't know whether the moments determine the distribution for $k > 2$.
Thus: