former proof has been deleted
@Boby The last answer didn't hold due to Lemma $1$ failing, so I have attempted with a different approach. Apologies for the lengthy delay.
We want to show that no non-negative random variable $X$ satisfies the system of equations:
$$\mathbb{E}\left[\text{sign}\left(X-\frac{k+1}{2}\right)X^k\right]=0, \forall k \in \mathbb{N}_0$$
Given that $X\ge0$, the sign function $\text{sign}\left(X-\frac{k+1}{2}\right)$ behaves as follows:
- $\text{sign}\left(X-\frac{k+1}{2}\right)=-1$ when $X<\frac{k+1}{2}$,
- $\text{sign}\left(X-\frac{k+1}{2}\right)=0$ when $X=\frac{k+1}{2}$,
- $\text{sign}\left(X-\frac{k+1}{2}\right)=1$ when $X>\frac{k+1}{2}$.
So the expectation can be written as
$$\mathbb{E}\left[\text{sign}\left(X-\frac{k+1}{2}\right)X^k\right]=\int_0^\infty\left(x-\frac{k+1}{2}\right)x^kdF_X(x)$$
$$=\left(-\int_0^{\frac{k+1}{2}}x^kdF_X(x)\right)+\left(\int^\infty_{\frac{k+1}{2}}x^kdF_X(x)\right)$$
$$=0.$$
This simplifies to
$$\int_0^{\frac{k+1}{2}}x^kdF_X(x)=\int^\infty_{\frac{k+1}{2}}x^kdF_X(x)$$
Let us denote
$$I_k=\int_0^{\frac{k+1}{2}}x^kdF_X(x), \ J_k=\int^\infty_{\frac{k+1}{2}}x^kdF_X(x)$$
So
$$I_k=J_k$$
Moreover the $k$-th moment of $X$ is
$$\mathbb{E}\left[X^k\right]=I_k+J_k=2I_k=2J_k$$
So we have
$$\mathbb{E}\left[X^k\right]=2\int^\infty_{\frac{k+1}{2}}x^kdF_X(x)$$
We will relate the tail probability $P\left(X\ge\frac{k+1}{2}\right)$ to the moment $\mathbb{E}\left[X^k\right]$. Since $x\ge\frac{k+1}{2}$ in the integral we have
$$x^k\ge\left(\frac{k+1}{2}\right)^k, \ \forall x\ge\frac{k+1}{2}$$
Therefore:
$$J_k=\int^\infty_{\frac{k+1}{2}}x^kdF_X(x)\ge\left(\frac{k+1}{2}\right)^kP\left(X\ge\frac{k+1}{2}\right)$$
From $\mathbb{E}\left[X^k\right]=2J_k$,
$$\frac{1}{2}\mathbb{E}\left[X^k\right]=J_k\ge\left(\frac{k+1}{2}\right)^kP\left(X\ge\frac{k+1}{2}\right)$$$$P\left(X\ge\frac{k+1}{2}\right)\le\frac{\frac{1}{2}\mathbb{E}\left[X^k\right]}{\left(\frac{k+1}{2}\right)^k}$$
We now see how $P\left(X\ge\frac{k+1}{2}\right)$ behaves as $k\rightarrow\infty$.
Case 1: $\mathbb{E}\left[X^k\right]$ is bounded
Suppose $\mathbb{E}\left[X^k\right]\le C$ for all $k$, where $C$ is a constant. This would happen if $X$ is bounded above. Then:
$$P\left(X\ge\frac{k+1}{2}\right)\le\frac{\frac{1}{2}C}{\left(\frac{k+1}{2}\right)^k}$$
As $k\rightarrow\infty$, $\left(\frac{k+1}{2}\right)^k\rightarrow\infty$ faster than any exponential function, so $P\left(X\ge\frac{k+1}{2}\right)\rightarrow0$.
Case 2: $\mathbb{E}\left[X^k\right]$ grows exponentially or faster
Suppose $\mathbb{E}\left[X^k\right]\le De^{ak}$ for some constants $D>0$ and $a>0$. Then
$$P\left(X\ge\frac{k+1}{2}\right)\le\frac{\frac{1}{2}De^{ak}}{\left(\frac{k+1}{2}\right)^k}$$
But $\left(\frac{k+1}{2}\right)^k=e^{k\ln\left(\frac{k+1}{2}\right)}$ grows faster than any exponential $e^{ak}$. Therefore, $P\left(X\ge\frac{k+1}{2}\right)\rightarrow0$ as $k\rightarrow\infty$.
In all cases, we have $$\lim_{k\rightarrow\infty}P\left(X\ge\frac{k+1}{2}\right)=0$$
Since $P\left(X\ge\frac{k+1}{2}\right)\rightarrow0$, the tial integral $J_k$ tends to zero unless compensated by very large values of $x^k$. However, from $J_k=\frac{1}{2}\mathbb{E}\left[X^k\right]$, we have
$$\frac{1}{2}\mathbb{E}\left[X^k\right]\ge\left(\frac{k+1}{2}\right)^kP\left(X\ge\frac{k+1}{2}\right)$$
But as $P\left(X\ge\frac{k+1}{2}\right)\rightarrow0$ and $\left(\frac{k+1}{2}\right)^k\rightarrow\infty$, the RHS could still be finite if $\mathbb{E}\left[X^k\right]$ grows accordingly. However, for $\mathbb{E}\left[X^k\right]$ to remain finite, $X$ cannot have heavy tails that compensate for the small probability $P\left(X\ge\frac{k+1}{2}\right)$.
Let us now show that $\mathbb{E}\left[X^k\right]\rightarrow0$ as $k\rightarrow\infty$.
Suppose $X$ is bounded above by some finite $M$. Then for $k$ large enough, $\frac{k+1}{2}>M$ so $P\left(X\ge\frac{k+1}{2}\right)=0$. Thus $J_k=0$ and $\mathbb{E}\left[X^k\right]=2J_k=0$ for large $k$. This implies $\mathbb{E}\left[X^k\right]=0$ for large $k$, which is only possible if $X=0$ almost surely.
If $X$ is unbounded but $\mathbb{E}\left[X^k\right]\rightarrow0$ as $k\rightarrow\infty$, then $X=0$ almost surely.
Why?
Consider for any $\delta>0$,
$$\mathbb{E}\left[X^k\right]\ge\mathbb{E}\left[X^k\mathbb{I}_{\{X\ge\delta\}}\right]\ge\delta^kP(X\ge\delta)$$
If $\mathbb{E}\left[X^k\right]\rightarrow0$ as $k\rightarrow\infty$, then
$$\delta^kP(X\ge\delta)\le\mathbb{E}\left[X^k\right]\rightarrow0$$
Since $\delta^k$ is a constant raised to the $k$-th power, unless $P(X\ge\delta)=0$, the LHS will not tend to zero. Therefore $P(X\ge\delta)=0$ for any $\delta>0$. Thus:
$$P(X>0)=\lim_{\delta\rightarrow0}P(X\ge\delta)=0$$
Therefore $X=0$ almost surely.
Now if $X=0$ almost surely, then for $k=0$;
$$\boxed{\mathbb{E}\left[\text{sign}\left(0-\frac{1}{2}\right)\cdot1\right]=\text{sign}\left(-\frac{1}{2}\right)\cdot1=(-1)\cdot1=-1≠0}$$
Contradicting the given equation
$$\boxed{\mathbb{E}\left[\text{sign}\left(X-\frac{k+1}{2}\right)X^k\right]=0}$$
Therefore $X$ cannot be identically zero. And thus, no non-negative random variable $X$ can satisfy the given system of equations.
$q.e.d$
Notes.
- The key is recognizing the tail probability $P\left(X\ge\frac{k+1}{2}\right)$ must tend to zero as $k\rightarrow\infty$, given growth rate of $\left(\frac{k+1}{2}\right)^k$.
- This leads to $\mathbb{E}\left[X^k\right]\rightarrow0$ as $k\rightarrow\infty$, implying $X=0$ almost surely.
- However $X=0$ almost surely contradicts the original equation for $k=0$ and hence no such $X$ exists.
