Boundedness of Moments of Random Variables I'm sure this must be well-known so hints and/or a reference are fine. Let $\epsilon>0$. Must there exist a random variable $X$ such that: 
(i) $X \geq 0$ a.s. 
(ii) $E[X] \leq \epsilon$ 
(iii) $E[X^2] \geq 1$ 
(iv) $E[X^3] \leq 10^{100}$?
If we just want the first 3 conditions we can construct a solution along the lines of $X=\epsilon n$ with probability $\frac{1}{n\sqrt{n}}$. But this won't satisfy the fourth property. 
 A: If $\epsilon>0$ is very small, then we cannot find such $X$. It is because, informally speaking, if we have a large $E[X^2]/E[X]$, then we should also have a large $E[X^3]/E[X^2]$. As $\epsilon \to 0^+$, the former diverges to $\infty$, while the latter is bounded by $10^{100}$. Hence it cannot happen.
To see this in a formal argument note first that
$$
X(X-1/\epsilon)^2=X^3-(2/\epsilon)X^2+X/\epsilon^2\ge 0.
$$ By taking expectation, we have
$$
10^{100}+1/\epsilon \ge E[X^3]+E[X]/\epsilon^2\ge 2E[X^2]/\epsilon\ge 2/\epsilon.
$$ This gives
$$
\epsilon \ge 10^{-100},
$$ hence for $\epsilon \in (0,10^{-100})$ $X$ satisfying the given conditions does not exist.
If $\epsilon \in [ 10^{-100},1]$, we may define
$$
P(X=1/\epsilon)=\epsilon^2,\quad P(X=0)=1-\epsilon^2.
$$ By this, we get $E[X]=\epsilon, $ $E[X^2]=1$ and
$E[X^3]=1/\epsilon \le 10^{100}$. For $\epsilon\ge 1$, we can choose $X\equiv 1$. 
To conclude, a random variable satisfying all the given conditions exists if and only if $\epsilon\ge 10^{-100}$.
