TITLE: Bound third moment and Fourth moment
QUESTION [7 upvotes]: 1) I have seen it mentioned that if X is a zero mean random variable then $\frac{E[X^4]}{\sigma^4} \geq \left\{\frac{E[X^3]}{\sigma^3}\right\}^2 + 1$ , where $E[X^2]=\sigma^2$ .
Is this true, if yes ,what is the proof for this?
(I am able to prove that $\frac{E[X^4]}{\sigma^4} \geq \left\{\frac{E[X^3]}{\sigma^3} \right\}^2 $)
2) Similar to the above statement, is it possible to find an upper bound or lower bound for the third moment $E[X^3]$ in terms of the second moment $E[X^2]$ , for a zero mean $X$?
I had tried using Holder's and Jensen's inequality to prove these, but it didn't give me the right direction for the bounds. I got bounds like $E[X^3] \leq \sqrt{E[X^2]\cdot E[X^4]} )$. This gives me that $E[X^4] \geq \{E[X^3]/\sigma\}^2$

REPLY [4 votes]: This answer proves that if $X$ is a random variable that satisfies $E[X]=0$ and $E[X^2]=1$, $E[X^3] \in \mathbb{R}$, then
$$ E[X^4] \geq 1 + E[X^3]^2$$
As in my other answer, the scaling $X=Y/\sigma$ proves $E[(Y/\sigma)^4] \geq 1 + E[(Y/\sigma)^3]^2$. 

Fix $m \geq 0$.  In my other answer I constructed the following random variable to show tightness of the conjectured inequality:
\begin{align}
X^* &= \left\{ \begin{array}{ll}
a &\mbox{ with prob $1/(1+a^2)$} \\
(-1/a)  & \mbox{ with prob $a^2/(1+a^2)$} 
\end{array}
\right. \\
a &= (1/2)\left(m + \sqrt{m^2+4}\right)
\end{align}
Here I show that this random variable solves the problem of minimizing $E[X^4]$ subject to $E[X]=0$, $E[X^2]=1$, $E[X^3]=m$, and that the minimizing value is: 
$$ E[X^4] = c = 1 + m^2 $$
Once we show this, it follows that for any random variable $X$ with $E[X]=0, E[X^2]=1$, $E[X^3]\geq 0$ we get: 
$$ E[X^4] \geq 1+E[X^3]^2 $$
Now if $X$ is a random variable that satisfies $E[X]=0, E[X^2]=1, E[X^3]<0$, we can define $Y=-X$ to conclude 
$$ E[X^4]=E[Y^4] \geq 1 + E[Y^3]^2 = 1 + E[X^2]^2 $$
So the desired result would be true regardless of whether $E[X^3]\geq 0$ or $E[X^3]<0$.  

Fix $m \geq 0$. Define $\mathcal{S}$ as the space of all random variables $X$ that satisfy $E[X^4]<\infty$. Define the set $\mathcal{A} \subseteq \mathbb{R}^4$ by: 
$$ \mathcal{A} = \{\left(E[X], E[X^2], E[X^3], E[X^4]\right) : X \in \mathcal{S}\} $$
Consider the following two optimization problems: 

Problem 1 (Constrained problem):
\begin{align}
\mbox{Minimize:} \quad & y_4 \\
\mbox{Subect to:} \quad & y_1 = 0 \\
& y_2 = 1\\
& y_3 = m\\
& (y_1, y_2, y_3, y_4) \in \mathcal{A}
\end{align}
Problem 2 (Unconstrained problem): Given real numbers $\lambda_1, \lambda_2, \lambda_3$, we have
\begin{align}
\mbox{Minimize:} \quad & y_4 + \lambda_1 y_1 + \lambda_2 (y_2-1) + \lambda_3 (y_3-m) \\
\mbox{Subect to:} \quad & (y_1, y_2, y_3, y_4) \in \mathcal{A}
\end{align}

Lemma:  (Lagrange multiplier theorem) If $(y_1, y_2, y_3, y_4)$ is a solution to the unconstrained problem that satisfies $y_1=0, y_2=1, y_3=m$, then it is also a solution to the constrained problem.
Proof: Clearly $(y_1, y_2, y_3, y_4)$ satisfies the constraints of the constrained problem.  Let  $(w_1, w_2,w_3,w_4)$ be another vector that satisfies the constraints of the constrained problem. We want to show that $y_4 \leq w_4$.  Since $(w_1, w_2,w_3,w_4) \in \mathcal{A}$ and  $(y_1, y_2, y_3, y_4)$ solves the unconstrained problem over all vectors in $\mathcal{A}$ we get
$$ y_4 + \lambda_1y_1 + \lambda_2(y_2-1) + \lambda_3(y_3-m) \leq w_4 + \lambda_1w_1 + \lambda_2(w_2-1) + \lambda_3(w_3-m) $$
However, since $y_1=w_1=0$, $y_2=w_2=1$, $y_3=w_3=m$, the above inequality reduces to 
$$ y_4 \leq w_4$$
$\Box$
Now define $a= (1/2)\left(m + \sqrt{m^2+4}\right)$ (this is consistent with the definition of $a$ for the 2-valued random variable $X^*$ above). Note that $a \geq 1$. Define the following particular real numbers: 
\begin{align}
\lambda_1 &= -2/a + 2a\\
\lambda_2 &= 1/a^2 - 4 + a^2\\
\lambda_3 &= -2a+2/a\\
c &= 1+m^2
\end{align}
Define the function $f:\mathbb{R}\rightarrow\mathbb{R}$ by
$$ f(x) = x^4 + \lambda_1x + \lambda_2(x^2-1) + \lambda_3(x^3-m)$$
I designed these $\lambda_i$ values to get the following nice factorization: For all $x \in \mathbb{R}$ we have: 
$$ f(x) = (x-a)^2(x+1/a)^2+c \geq c $$
Thus for all random variables $X \in \mathcal{S}$ we have: 
$$ f(X) \geq c $$
and so 
$$ E[f(X)] \geq c$$
That is,
$$E[X^4] + \lambda_1E[X] + \lambda_2(E[X^2]-1) + \lambda_3(E[X^3]-m) \geq c $$
On the other hand, the 2-valued random variable $X^*$ defined above takes values either $a$ or $-1/a$, and so for all realizations of that random variable we get
$$ f(X^*) = (X^*-a)^2(X^*+1/a)^2+c = c $$
And so $E[f(X^*)]=c$. 
Thus, that 2-valued random variable minimizes the expression 
$$ E[X^4] + \lambda_1E[X] + \lambda_2(E[X^2]-1) + \lambda_3(E[X^3]-m)$$
over all random variables $X \in \mathcal{S}$.  In particular, its moments $$(y_1^*,y_2^*,y_3^*,y_4^*) = (E[X^*], E[(X^*)^2], E[(X^*)^3], E[(X^*)^4])$$
solve the unconstrained optimization problem. Further, from my previous answer, we know those moments for $X^*$ satisfy $y_1^*=0, y_2^*=1, y_3^*=m$, and so by the Lagrange multplier lemma we conclude that $(y_1^*,y_2^*,y_3^*,y_4^*)$ is also optimal for the constrained problem. In particular, the value $y_4^*$ achieved by this 2-valued random variable is the minimum possible value of $E[X^4]$ over all random variables that satisfy $E[X]=0, E[X^2]=1, E[X^3]=m$.  In fact, we know: 
$$ y_4^* = E[(X^*)^4] = 1+m^2 =c$$
$\Box$