# Bound central moments of even order with raw moments of same order

Let $$(\Omega, \mathcal A, P)$$ be a probability space and consider a real-valued random variable $$X \colon \Omega \to \mathbb{R}$$. It holds $$\mathrm{Var}(X) = \mathbb{E}[(X - \mathbb{E}[X])^2] = \mathbb{E}[X^2] - \mathbb{E}[X]^2 \leq \mathbb{E}[X^2].$$ Is this true for higher moments of even order? Precisely, does it hold for $$p = 1, 2, \ldots$$ $$\mathbb{E}[(X - \mathbb{E}[X])^{2p}] \leq C(p)\mathbb{E}[X^{2p}],$$ for some positive constant $$C(p)$$ depending only on $$p$$ and such that $$C(1) = 1$$?

Edit:

In this thread it is proved that the inequality above does not hold for $$C(p) = 1$$ for all $$p = 1,2,\ldots$$ Still, could it hold with a $$p$$-dependent proportionality constant?

Edit 2:

It is possible to prove that the inequality holds with $$C(p) = 2^{2p}$$ using Hölder's and Jensen's inequality. In particular, Hölder's inequality yields for any real numbers $$a$$ and $$b$$ $$(a-b)^{2p} \leq 2^{2p-1}(a^{2p}+b^{2p}),$$ so that by linearity of $$\mathbb{E}$$ $$\mathbb{E}[(X - \mathbb{E}[X])^{2p}] \leq 2^{2p-1}\left(\mathbb{E}[X^{2p}]+\mathbb{E}[X]^{2p}\right).$$ Noew Jensen's inequality yields $$\mathbb{E}[X]^{2p} \leq \mathbb{E}[X^{2p}]$$ and we can conclude that $$\mathbb{E}[(X - \mathbb{E}[X])^{2p}] \leq 2^{2p-1} \cdot 2 \mathbb{E}[X^{2p}] =2^{2p}\mathbb{E}[X^{2p}].$$ Is there any sharper bound? In this way, for $$p = 1$$ we obtain $$C(1) = 4$$ instead of $$C(1) = 1$$.

$$C(p)=2^{2p-1}$$ (gaining a factor 2) is possible. This answer https://math.stackexchange.com/a/3040487/484640 shows that inequality $$E[(Y-E[Y])^{m}]\le E[Y^m]$$ for nonnegative $$Y$$. Admitting this result, write $$X=X_+-X_-$$ for the positive and negative parts. By Jensen's inequality $$\frac{1}{2^m}E[(X-E[X])^m] \le \frac{E[(X_+-E[X_+])^m] + E[(X_--E[X_-])^m]}{2} \le \frac{E[X_+^m]+ E[X_-^m]}{2} = \frac{E[|X|^m]}{2}.$$ With $$m=2p$$, $$C(p)=2^{2p-1}$$ holds. It is the same as taking $$a=X_+-E[X_+]$$ and $$b=X_- - E[X_-]$$ and taking expectation in the inequality $$(a-b)^{2p}\le 2^{2p-1}(a^{2p}+b^{2p})$$ given in the question.
• @G.Gare Yes, since $|X|=X_+-X_-$ and $X_+X_-=0$. Commented Nov 4, 2022 at 8:28
• Very nice. The fact that for $p=1$ we have $C(1) = 1$ is still not reached, though, as in your calculation we have $C(1) = 2$. How do you think this could be explained? Commented Nov 4, 2022 at 8:30
• For example, it would be very good to obtain $2^{p-1}$. Commented Nov 4, 2022 at 8:38