Bound on third moment when the variable is bounded and has finite expectation If I have a random variable $X$ such that $|X| \leq 1$ and $\mathbb{E}(X) = c < \infty$, I want to find the tightest upper bound for $\mathbb{E}(X^3)$. What would be the most optimal bound to use?
 A: Let $f(x)=x^3$ and let $S\subset \mathbb R^2$ be the convex hull of the points $\{(x,f(x)): x\in[-1,1]\}$. The set $S$ is the set of all possible $(EX,EX^3)$ values for random variables $X$ with $P(|X|\le1)=1.$ The intersection of $S$ with the vertical line cut out by $x=c$ is the interval from $(c,L(c))$ to $(c,U(c))$, where $L$ and $U$ are the lower convex  and upper concave envelopes  of $f$ on $[-1,1]$.  Then the desired upper bound is $U(c)$.
More explicitly: Draw a line through $(1,1)$, tangent at $(-1/2, -1/8)$ to the graph of $f$. The graph of $U$ follows that of $f$ for  $x\in[-1,-1/2]$ and that of the tangent line for $x\in[-1/2,1]$.  The formula for $U$ is $$U(x)=\begin{cases}-x^3&x\le -1/2\\ -\frac 1 8 +\frac 3 4 (x+1/2)& x\gt-\frac 1 2 \end{cases}.$$
If $c=0$, for example, $U(0)=1/4$, which is attained by a random variable $X$ taking the value $X=1$ with probability $1/3$ and the value $X=-1/2$ with probability $2/3$. Then $$EX^3=\frac 2 3 (-\frac 1 8 )+\frac 1 
 3(1)=\frac 1 4 .$$ 
