Let $\mathbf{p}=\{p_{1},\cdots, p_{K}\}$ be a probability distribution with a fixed positive integer $K$, that is, $\mathbf{p}$ satisfies
1) $0< p_{k} < 1$ for every $k$, and
2) $ \sum_{k}p_{k}=1$.
Consider the following equations:
$ \sum_{k=1}^{K}p_{k}=c_{1}=1$
$\sum_{k=1}^{K}p_{k}^{2}=c_{2}$
$\cdots$
$\sum_{k=1}^{K}p_{k}^{K}=c_{K} $.
Obviously the equations don't necessarily have a solution, in general, for an arbitrarily given set of
{$c_{1}=1,c_{2},\cdots,c_{K}$}. But does anyone know a sufficient condition on $c_{k}$, $k=1,,\cdots, K$, to assure the existence of a solution?
For example, if $K=3$, the $p_i$ will be roots of the cubic polynomial $P(x) = 6 x^3 - 6 x^2 + 3 (1-c_2) x + 3 c_2 - 2 c_3 - 1$. You can use Sturm's theorem to count the number of roots between $0$ and $1$. The Sturm sequence is $$\eqalign{P_0(x) &= P = 6 x^3 - 6 x^2 + 3 (1-c_2) x + 3 c_2 - 2 c_3 - 1\cr P_1(x) &= P' = 18 x^2-12 x-3 c_2 +3 \cr P_2(x) &= -\text{rem}(P_0,P_1) = (2 c_2 - 2/3) x - 8 c_2/3 + 2 c_3 + 2/3\cr P_3(x) &= -\text{rem}(P_1,P_2) = \dfrac{9(3\,{c_{{2}}}^{3}-21\,{c_{{2}}}^{2}+36\,c_{{2}}c_{{3}}-18\,{ c_{{3}}}^{2}+9\,c_{{2}}-8\,c_{{3}}-1) }{(3 c_2-1)^2}}$$ You want the difference between the number of sign changes at $x=0$ and at $x=1$ to be $3$. That is, $P_0(0), P_1(0), P_2(0), P_3(0)$ should have alternating signs, while $P_0(1), P_1(1), P_2(1), P_3(1)$ should all have the same sign.