Unbiased estimator of $X^k$ given independent unbiased estimators of $X$ Suppose we are interested estimating $X^k$, and we have access to independent unbiased estimators $Y_i$ for $i = \{1,2,\dots, k\}$, i.e., $\mathbb{E}[Y_i] = X$.
A straightforward way to construct an unbiased estimator of $X^k$ is to define $Y^{(k)} := \prod_{i=1}^kY_i$, then $\mathbb{E}[Y^{(k)}]=X^k$ by independence.
What are other methods for estimating $X^k$ using only $c < k$ number of estimators? Is it even possible in this general setup? Is there a name to this type of problem?
 A: The obvious analogy here is to use independent estimators $Y_1,...,Y_c$ with expectations:
$$\mathbb{E}(Y_i) = X^{p_i}
\quad \quad \quad \sum p_i = k.$$
In theory this is possible so long as you have a method to construct the required estimators for this problem.  The independence requirement will generally make it impractical, since in most cases it will require you to partition your dataset to form the estimators seperately.  Consequently, while it is possible to get an unbiased estimator by this method, it will tend to be a poor estimator (with high variance), owing to the fact that it uses a partition of the data where only a small amount of the data is used for each part of the estimator.

An example using IID data with zero mean: Suppose you have IID data $X_1,...,X_n$ from some distribution with zero mean and finite variance $\sigma^2 < \infty$, and you want to estimate $\mathbb{E}(X^k)$.  In this case the sample mean and sample variance give unbiased estimators for the first and second raw moments:
$$\mathbb{E}(\bar{X}) = \mathbb{E}(X) = 0
\quad \quad \quad 
\mathbb{E}(S^2) = \mathbb{E}(X^2) = \sigma^2.$$
Now, suppose we partition our data into $c < k$ parts where we have at least two data points in each part (so that we can form the sample variance).  Denote these partition parts by $\boldsymbol{X}_{(1)},...,\boldsymbol{X}_{(c)}$ and denote statistics using these samples with the corresponding subscripts.  Now, choose some values $p_1,...,p_c \in \{ 1,2 \}$ with $p_1 + \cdots + p_c = k$ and form the estimator:
$$\text{Est} \equiv \prod_{i:p_i = 1} \bar{X}_{(i)} \times \prod_{i:p_i = 2} S_{(i)}^2 .$$
The partition of the data means that the parts are independent, so we have:
$$\begin{align}
\mathbb{E}(\text{Est}) 
&= \prod_{i:p_i = 1} \mathbb{E}(\bar{X}_{(i)}) \times \prod_{i:p_i = 2} \mathbb{E}(S_{(i)}^2) \\[6pt]
&= \prod_{i:p_i = 1} \mathbb{E}(X) \times \prod_{i:p_i = 2} \mathbb{E}(X^2) \\[6pt]
&= \prod_{i:p_i = 1} \mathbb{E}(X^{p_i}) \times \prod_{i:p_i = 2} \mathbb{E}(X^{p_i}) \\[6pt]
&= \prod_{i} \mathbb{E}(X^{p_i}) \\[6pt]
&= \mathbb{E}(X^{\sum p_i}) \\[12pt]
&= \mathbb{E}(X^k). \\[12pt]
\end{align}$$
Note that although this is an unbiased estimator, it will be a terrible estimator when the number of partition pieces is large.
A: To illustrate the point that the answer depends on the underlying statistical model: If $Y\sim\mathcal E(1/X)$, an exponential variable, then
$$\mathbb E[Y^k]= X^k\Gamma(k+1)$$
meaning that $Y^k/\Gamma(k+1)=Y^k/k!$ is an unbiased estimator of $X^k$, based on a single observation. This extends to Gamma variables, obviously.
