# $\mathbb{E}[X^4]=1$, $X,Y$ iid, what's the best upper bound of $\mathbb{E}[(X-Y)^4]$?

Let $X,Y$ be i.i.d. random variables, $\mathbb{E}[X^4]=1$, what's the best upper bound for $\mathbb{E}[(X-Y)^4]$ ?

1. A trivial upper bound is $16$, since $(X-Y)^4 \leq 8 (X^4+Y^4)$ then take expectation on both sides. However, equality cannot be achieved.

2. My guess of the best upper bound will be $8$, achieved when $X$ is uniform at random from $\{-1, +1\}$.

## 1 Answer


Your guess is correct. Indeed, it is well known (see e.g. Bertsimas--Popesku, page 781) that real numbers $m_0=1,m_1,\dots,m_{2\ell}$ are the moments of orders $0,1,\dots,2\ell$ of a real-valued random variable $X$ iff the matrix $M:=(m_{i+j})_{i,j=0}^\ell$ is nonnegative-definite, that is, iff all the principal minors of $M$ are $\ge0$; here $\ell$ is a natural number; in our case, $\ell=2$. Also, given $\E X^4=1$, we have $$\E(X-Y)^4=2-8m_3m_1+6m_2^2.$$ Thus, the problem is a simple problem of real algebraic geometry, which can be solved algorithmically. Using the Mathematica command Maximize[], we get the result:

Added: Here is an elementary solution, without using Mathematica: Since $m_4=1$, the condition $\det M\ge0$ implies $$2m_3m_1\ge m_2^2-m_4+(m_3^2+m_4m_1^4)/m_2\ge m_2^2-1,$$ whence $$\E(X-Y)^4=2-8m_3m_1+6m_2^2\le2-4(m_2^2-1)+6m_2^2=2m_2^2+6\le8,$$ since $m_2^2\le m_4=1$. The equality in the inequality in question is attained only if $m_2^2=m_4=1$ and $2m_3m_1=m_2^2-1=0$, that is, only if $\PP(X=1)=\PP(X=-1)=1/2$.

• Note that your formula is $\operatorname{trace}(M D MD)$ where $D$ is a matrix with alternating plus and minus signs. The same works for the analogous problem with $(X-Y)^{2\ell}$. However, I wasn't able to use this idea to solve the analogous problem. Apr 29, 2018 at 6:01
• I have added an elementary solution, without using Mathematica. Apr 29, 2018 at 12:07