# A Minkowski-like inequality

Assume that $X$ and $Y$ are two arbitrary non-negative random variables. Is the following inequality true for $1\leq\alpha\leq 2$?

\begin{align} \left(\mathbb{E}\left[X^\alpha\right]-\mathbb{E}\left[X\right]^\alpha\right)^{\frac{1}{\alpha}}+\left(\mathbb{E}\left[Y^\alpha\right]-\mathbb{E}\left[Y\right]^\alpha\right)^{\frac{1}{\alpha}}\geq \left(\mathbb{E}\left[(X+Y)^\alpha\right]-\left(\mathbb{E}\left[X+Y\right]\right)^\alpha\right)^{\frac{1}{\alpha}} \end{align}

• you mean that $X,Y$ take non-negative real values? Feb 6, 2018 at 19:01
• Yes, $X$ and $Y$ take non-negative real values. Feb 6, 2018 at 19:23
• I have restored the original condition $\alpha\ge1$ on $\alpha$. Feb 6, 2018 at 19:54

The answer to this question is yes.

Indeed, let $a:=\alpha$, and suppose $1\le a\le2$. Without loss of generality (wlog), $1<a\le2$, $EX^a<\infty$, and $Y$ takes only values in a fixed finite set $S\subset[0,\infty)$. The inequality in question is equivalent to the following: $$h(t):=h_{X,Y}(t):=[E(X+tY)^a-E^a(X+tY)]^{1/a}-(EX^a-E^a X)^{1/a}-t(EY^a-E^a Y)^{1/a}\le0$$ for all $t\ge0$, and hence to the condition $h'(0)\le0$ (because $\frac d{dt}h_{X,Y}(t)=\frac d{du}h_{X+tY,Y}(u)|_{u=0}$). In turn, the condition $h'(0)\le0$ can be rewritten as $$(EX^{a-1}Y-E^{a-1}X\,EY)(EX^a-E^a X)^{1/a-1}\le(EY^a-E^a Y)^{1/a},$$ and then in this H\"older-like form: $$F(\mu):=(EX^{a-1}Y-E^{a-1}X\,EY)_+^a-(EX^a-E^a X)^{a-1}(EY^a-E^a Y)\le0,$$ where $u_+^a:=\max(0,u)^a$ and $\mu=\mu_Y$ is the probability distribution of $Y$, with the random variable (r.v.) $X$ considered fixed.

Obviously, the function $F$ is convex on the simplex $M_S$ of all probability distributions $\mu_Y$ on $S$. So, the maximum of $F$ on $M_S$ is attained when $\mu$ is a Dirac measure -- that is, when the r.v. $Y$ is a nonnegative constant $c$, so that
$F(\mu)=c^a(EX^{a-1}-E^{a-1}X)_+^a=0$, because $1<a\le2$ and hence $EX^{a-1}\le E^{a-1}X$. So, the maximum of $F$ on the simplex $M_S$ is $0$, which completes the proof for the case $1\le a\le2$.

Moreover, following the lines of the above proof, it should be clear that the inequality in question fails to hold in general for any $a>2$. Indeed, then, for instance, $F(\mu)=\frac12-(\frac12)^{a-1}>0$ if $Y=1$ and $P(X=0)=P(X=1)=1/2$, which implies $h'(0)>0$, which implies $h(t)>0$ for small enough $t>0$.

• Sorry, I have forgotten to mention that $\alpha\leq 2$. Feb 6, 2018 at 19:56
• Then you should probably ask that as a separate question, with that additional condition. Feb 6, 2018 at 19:57
• If $1\le\alpha\le2$, then the inequality holds. I have now added the corresponding proof. Feb 6, 2018 at 21:19
• This proof is mistaken, since $EX^{a-1}Y$ and hence $F(\mu)$ are not actually functions of $\mu$, even with $X$ fixed. A correct (but much involved) proof can be found at arxiv.org/abs/1807.11108 . Aug 1, 2018 at 14:50