Assume $x_i$ is a series of independent samples from a identical distribution , $1\leq i\leq n$ .
Define $\mu=\frac{1}{n}\sum_{i=1}^n x_i$ is the estimated mean.
From the estimation of sample variance, it illustrated $E\left[\frac{\sum_{i=1}^n(x_i-\mu)^2}{n-1}\right]=E[(x-E(x))^2]$.
Therefore, is it possible to prove that $E\left[\frac{\sum_{i=1}^n(x_i-\mu)^k}{n-1}\right]=E[(x-E(x))^k]$ for any $k\in N^+$ ?
Just want to ask if it is possible, and if yes, may I have the proof.
Thanks very much!
In the following I'll use the conventional notation $\bar x:=\frac1n\sum x_i$ for the sample mean, instead of $\mu$, which is often the symbol used for $E(x)$.
The assertion $$E\left[\frac{\sum_{i=1}^n(x_i-\bar x)^k}{n-1}\right]=E[(x-E(x))^k]\tag1 $$ is true for $k=1$ and $k=2$, but that's all. For $k=3$ here is a counterexample with sample size $n=2$:
Let $x_1$ and $x_2$ be independently drawn from a distribution such that $E(x)=0$ but $E(x^3)\ne0$. (For example, you could let $x$ take value $2$ with probability $\frac13$, and value $-1$ with probability $\frac23$, but the detail doesn't matter.)
To evaluate (1) for $k=3$ with the given $x_1$, $x_2$, calculate $\Delta_1:= x_1-\bar x$ and $\Delta_2:=x_2-\bar x$ and find that $\Delta_1 + \Delta_2=0$. This means that $\sum_{i=1}^2(x_i-\bar x)^3 = (\Delta_1)^3 + (-\Delta_1)^3=0$, so the LHS of (1) equals zero. However, the RHS of (1) equals $E[(x-0)^3]=E(x^3)$, which is nonzero by construction.
ADDED: A variant of assertion (1) is true when $k=3$: $$ E\left[\sum_{i=1}^n(x_i-\bar x)^3\right] =\frac{(n-1)(n-2)}nE[(x-E(x))^3].\tag2 $$ To prove this, wlog $E(x)=0$. Write $x_i-\bar x=\sum_{j=1}^n c_{ij}x_j$, where $$ c_{ij}=\begin{cases} 1-\frac1n&\text{if $i=j$}\\ -\frac1n&\text{if $i\ne j$} \end{cases}. $$ Expanding $(x_i-\bar x)^3=(\sum_{j=1}^n c_{ij}x_j)^3$ gives a linear combination of terms $x_{j_1}x_{j_2}x_{j_3}$, where the three indices $j_1, j_2,j_3$ are chosen from $1,\ldots, n$ with replacement. To compute $E(x_i-\bar x)^3$, we need consider only terms where these three indices are the same, since all other terms will have at least one index appearing exactly once and will therefore have zero expectation. So $$ E(x_i-\bar x)^3=E\sum_{j=1}^n (c_{ij}x_j)^3=\left( \sum_{j=1}^n c_{ij}^3\right)E(x^3) =\left(\left(1-\frac1n\right)^3+(n-1)\left(-\frac1n\right)^3\right)E(x^3).\tag3 $$ Summing (3) over all $i$ and rearranging gives (2) for the case $E(x)=0$.
For $k>3$ the above approach gives a formula for $E\sum (x_i-\bar x)^k$; but note there will be additional terms involving second or higher central moments of $x$.