Can we show that $E[|U+V|^k] \ge \max ( E[|U|^k], E[|V|^k] )$

Question

Can we show that $E[|U+V|^k] \ge \max ( E[|U|^k], E[|V|^k] )$ for any $k \ge 1$ and where $V$ and $U$ are independent and in $L^k$.

Here is the proof when $k$ is even and $E[U]=E[V]=0$.

\begin{align} E[|U+V|^k]&=E[(U+V)^k]\\ &= E[ E[(U+V)^k|V]]\\ &= E[ (E[U+V |V])^k] \text{ by Jensen's inequality}\\ &= E[ (V+E[U] )^k] \text{ by independence} \\ &= E[ V ^k] \end{align} repeating the above steps one gets the bound we wanted.

The question is can we show this bound for any $k$ without too many assumptions on $U$ and $V$.

Answer 1

If the random variables are not centered, then we can have a problem for example if $U=1$ and $V=-1$.

If $U$ and $V$ are centered and independent, then the following inequality holds for any $p\geqslant 1$: $$ \mathbb E\left[\left|U+V\right|^p\right]\geqslant\max\left\{\mathbb E\left[\left|U\right|^p\right],\mathbb E\left[\left|V\right|^p\right]\right\}.$$ This follows from Jensen's inequality: denote by $\Pr_U$ and $\Pr_V$ the distributions of $U$ and $V$. Then by independence of $U$ and $V$, $$ \tag{*}\mathbb E\left[\left|U+V\right|^p\right]= \iint_{\mathbb R^2}\left|u+v\right|^p\mathrm{d}{\Pr}_U(u)\mathrm{d}{\Pr}_V(v)$$ and for any fixed $u$, by Jensen's inequality $$\int_{\mathbb R}\left|u+v\right|^p\mathrm{d}{\Pr}_V(v) \geqslant \left|\int_{\mathbb R} \left(u+v\right)\mathrm{d}{\Pr}_V(v)\right|^p=\left|\int_{\mathbb R} u\mathrm{d}{\Pr}_V(v)+\int_{\mathbb R} v\mathrm{d}{\Pr}_V(v)\right|^p.$$ Since $\Pr_V$ is a probability measure, $\int_{\mathbb R} u\mathrm{d}{\Pr}_V(v)=u$ and since $V$ is centered, $\int_{\mathbb R} v\mathrm{d}{\Pr}_V(v)=0$. We thus derive that $$\int_{\mathbb R}\left|u+v\right|^p\mathrm{d}{\Pr}_V(v) \geqslant \left|u\right|^p,$$ which gives, in view of (*), that $$\mathbb E\left[\left|U+V\right|^p\right]\geqslant \mathbb E\left[\left|U\right|^p\right].$$ The wanted inequality follows by changing the roles of $U$ and $V$.