Do p-norms of two discrete probability distributions 'rank' them equivalently? Please, forgive me if this is an elementary question, as well as my the sloppy phrasing and notation.
Suppose we have two discrete probability distributions $p = {\lbrace p_i \rbrace}$ and $q={\lbrace q_i \rbrace}$, $i=1,\dots,n$, where $p_i=P(p=p_i)$ and $q_i=P(q=q_i)$. Let's represent them as vectors $\boldsymbol{p} = [p_i], \boldsymbol{q}= [q_i] \in \mathbb{R}^n$.
If we take the two p-norms $||\cdot||_a$ and $||\cdot||_b$, excluding 1-norm and max-norm then if $||\boldsymbol{p}||_a>||\boldsymbol{q}||_a$ is it the case that also $||\boldsymbol{p}||_b>||\boldsymbol{q}||_b$ holds? In other words, will all the p-norms induce the same 'ranking' of $\boldsymbol{p}$ and $\boldsymbol{q}$?
Would anything change if at least one of the $||\cdot||_a$ and $||\cdot||_b$ were p-quasinorms i.e., $a,b\in(0,1)$ instead?
 A: Neither  of the two versions holds...
Here are counterexamples. Let $a=1.5$, $b=2$. Let
\begin{equation*}
 p=[3,1,4]/8\quad q = [2,2,5]/9;
\end{equation*}
Then, we have
\begin{equation*}
\begin{split}
   \|p\|_a &= 0.7329,\quad \|q\|_a = 0.7299\\
   \|p\|_b &= 0.6374,\quad \|q\|_b = 0.6383.
\end{split}
\end{equation*}
If we use quasinorms, then we can get similar counterexamples.
However, not all is bad. There is a version of the conjecture that does hold. That is, if $p \prec q$, then for all $p$-norms, you'll have the desired monotonicity. That is, $p \prec q$ means the majorization order:
\begin{equation*}
\begin{split}
    &\sum\nolimits_{i=1}^k p_i^\downarrow \le \sum\nolimits_{i=1}^k q_i^\downarrow\quad\text{for}\   k=1,\ldots,n-1\\
     &\sum\nolimits_{i=1}^n p_i^\downarrow = \sum\nolimits_{i=1}^n q_i^\downarrow.
 \end{split}
\end{equation*}
In this case, $\|p\|_a \le \|q\|_a$ for all $a \ge 1$.
A: Nate, first of all thank you for your time. However, I am afraid I am not persuaded by your example.
We have $||\boldsymbol{q}||_a = (q_{2}^a)^{1/a} = q_2 = 1, \forall a$.
On the other hand, we will have $||\boldsymbol{p}||_1 = (1-x) + x = 1 = ||\boldsymbol{q}||_1, \forall x$. The 1-norm case is trivial, since p and q are probability distributions, hence by definition $||\boldsymbol{p}||_1 = ||\boldsymbol{q}||_1=1$, hence I ignore it in my question.
However if $a>1$ then  $||\boldsymbol{p}||_a < 1 = ||\boldsymbol{q}||_a, \forall x \in (0,1)$ and we do get exactly  $||\boldsymbol{p}||_a = 1 = ||\boldsymbol{q}||_a$ for $x \in \lbrace 0,1 \rbrace$, when we actually have $p\equiv q$ if we consider the permutations [1,0,0], [0,1,0], [0,0,1] to correspond to different representations of the same probability distribution (I know I did not mention this in my original question...).
Am I missing something?
