I am wondering if the following inequality is correct and can be shown?
Let $A$ and $B$ be random vectors of dimension $n$. Then for $ p \ge 1$ \begin{align} E^{\frac{1}{2p}} \left[ \left| Tr \left\{(A-B)(A-B)^T \right\} \right|^p \right] \le E^{\frac{1}{2p}} \left[\left| Tr \left\{AA^T \right\} \right|^{p} \right]+ E^{\frac{1}{2p}} \left[\left| Tr \left\{BB^T \right\} \right|^{p}\right] \end{align}
For $n=1$ the inequality becomes
\begin{align} E^{\frac{1}{2p}} \left[ |A-B|^{2p} \right] \le E^{\frac{1}{2p}} \left[|A|^{2p}\right]+ E^{\frac{1}{2p}} \left[ |B|^{2p}\right] \end{align} which simply follows from Minkowski inequality.
How would one even start proving an inequality like this? If you can also point me to some similar result it will be great. The only related discussion on this cite that I found is here
Any help would be greatly appreciated. Thank you
So we have some probability space $\{\Omega,\mathbb{P}\}$, and vector-valued random variables: $A,B:\Omega\to\mathbb{R}^n$. In general, for every $x\in\mathbb{R}^n$,
$$\hbox{Tr}(xx^T)=\langle x,x\rangle = ||x||^2$$ where $\langle,\cdot,\rangle, ||\cdot||$ denote the Euclidean inner-product and the Euclidean norm, respectively. So we can write:
$$\begin{align} E^{\frac{1}{2p}}\left(\left |\hbox{Tr}\{(A-B)(A-B)^T\}\right|^p\right)&=\left(\int_{\Omega}||A(\omega)-B(\omega)||^{2p}\,d\mathbb{P}(\omega)\right)^{\frac{1}{2p}}\\ &\leq^{(1)} \left(\int_{\Omega}(||A(\omega)||+||B(\omega)||)^{2p}\,d\mathbb{P}(\omega)\right)^{\frac{1}{2p}}\\ &\leq^{(2)} \left(\int_{\Omega}||A(\omega)||^{2p}\,d\mathbb{P}(\omega)\right)^{\frac{1}{2p}}+\left(\int_{\Omega}||B(\omega)||^{2p}\,d\mathbb{P}(\omega)\right)^{\frac{1}{2p}}\\ &=E^{\frac{1}{2p}}\left|\hbox{Tr}(AA^T)\right|^p+E^{\frac{1}{2p}}\left|\hbox{Tr}(BB^T)\right|^p \end{align}$$
Inequality (1) follows from the triangle inequality of the Euclidean norm.
Inequality (2) follows from the triangle inequality (Minkowski's inequality) for the Banach space $L^{2p}(\Omega,\mathbb{P})$, for the functions $f(\omega)=||A(\omega)||$ and $g(\omega)=||B(\omega)||$.
Note that everything is non-negative, so the absolute values in the last row are redundant.