Prove / Disprove that $E\big((X−a)^4\big)$ is minimized when $a=E(X)$.

Question

Let $X$ be a random variable with moment up to order $4$. Prove / disprove that $E\big((X−a)^4\big)$ is minimized when $a=E(X)$.

I've already proved that $E((X−a)^2)$ is minimized as such, but I'm lost with the 4 power. I've tried things from differentiating to opening up the parenthesis and trying to work from there but nothing worked.

plus I couldn't come up with any counter-example.

any help is appreciated! thanks

Answer 1

Expanding $\mathbb{E}\left[(X-a)^4\right]$ yields $$f(a):=\mathbb{E}\left[(X-a)^4\right]=\mathbb{E}[X^4]-4\,\mathbb{E}[X^3]\,a+6\,\mathbb{E}[X^2]\,a^2-4\,\mathbb{E}[X]\,a^3+a^4\,.$$ Thus, $$f'(a)=-4\,\mathbb{E}[X^3]+12\,\mathbb{E}[X^2]\,a-12\,\mathbb{E}[X]\,a^2+4\,a^3\,.$$ Therefore, if $a=\mathbb{E}[X]$ minimizes $f$, then $f'\big(\mathbb{E}[X]\big)=0$, and so $$-\mathbb{E}[X^3]+3\,\mathbb{E}[X^2]\,\mathbb{E}[X]-2\,\big(\mathbb{E}[X]\big)^3=0\,.\tag{*}$$ Hence, if (*) does not hold, then you have found a counterexample, say, a random variable $X$ with $\mathbb{E}[X]=0$ but $\mathbb{E}[X^3]\neq 0$.

Take $X$ to be the discrete random variable with values $-1$ and $2$ such that $$\mathbb{P}[X=-1]=\frac{2}{3}\text{ and }\mathbb{P}[X=2]=\frac13\,.$$ Then, $$\mathbb{E}[X]=0\text{ whereas }\mathbb{E}[X^3]=2\neq 0\,.$$ Indeed, $f(a)=6-8a+12a^2+a^4$ so $f(a)$ is minimized when $$a=18\left(\sqrt[3]{2}-1\right)\neq 0=\mathbb{E}[X]\,.$$

Answer 2

For a counter example take a look at a random variable with $X\in\{1,-2\}$ with probabilities $$P(X=+1) =\frac{2}{3}, \qquad P(X=-2) = \frac{1}{3}\,.$$

Now, we can see that $$E(X) =\frac{2}{3} \cdot 1 + \frac{1}{3}\cdot(-2) = 0\,.$$

However, $$E((X-a)^4) = \frac{2}{3} \cdot (1-a)^4 + \frac{1}{3}\cdot(-2-a)^4\,.$$

We have $$ \frac{81}{16}= E((X+\tfrac12)^4) < E(X^4) = 6$$ providing a counterexample.