# An inequality for two independent identically distributed random vectors in a normed space

Suppose that $X$ and $Y$ are independent identically distributed random vectors in a separable Banach space $B$. Does it always follow that $E\|X-Y\|\le E\|X+Y\|$?

Some background information on this question can be found at the end of the note posted on arXiv at additive decomposition of norms. In particular, the inequality in question holds if $B$ is two-dimensional or Euclidean.

• OK, it is true if $B$ is $2$-dimensional or $L_p$, $1\le p\le 2$, because it is true when $B$ is the real line. But do you know whether it is true for $B=L_4$? – Bill Johnson Jun 2 '15 at 1:46
• I think it's a good question. I haven't considered that. – Iosif Pinelis Jun 2 '15 at 2:49

To get dimension 4, I think the norm \begin{equation} \|a\| = \max_{\{i,j\}} |a_i-a_j| \vee \|a\|_\infty, \end{equation} where $a=(a_1,a_2,a_3,a_4)$, works. Again $X$ samples the unit vector basis uniformly.

EDIT: It looks like a variation takes care of dimension 3. Use again \begin{equation} \|a\| = \max_{\{i,j\}} |a_i-a_j| \vee \|a\|_\infty, \end{equation} where $a=(a_1,a_2,a_3)$. This time, let $X$ sample $e_1, e_2, e_3$ and $-(e_1+e_2+e_3)/2$ uniformly. The first three vectors have norm 1 and the last one norm $1/2$. In the (unlikely) event that my arithmetic is correct, the expectation of $\|X+Y\|$ is $19/16$ and the the expectation of $\|X-Y\|$ is $21/16$.

If this is correct, the only remaining thing is whether the inequality is true for random variables that take on three non zero values.

• This is very nice, thank you Bill! Now only dimension 3 remains. In fact, your construction works for any dimension $\ge4$, and it seems to be the simplest construction so far! – Iosif Pinelis Jun 2 '15 at 20:14
• Alas, for dimension $3$ my calculations (with Mathematica) suggest that $E\|X-Y\|=\frac{18}{16}$ and $E\|X+Y\|=\frac{22}{16}>E\|X-Y\|$. The Mathematica notebook and its pdf copy can be found at dropbox.com/s/z69emx0royqdjjt/2iidVectors.nb?dl=0 and dropbox.com/s/6fcl07z3evdpjw6/2iidVectors.pdf?dl=0 , respectively. – Iosif Pinelis Jun 3 '15 at 0:21
• Oh, there should be a minus sign in front of the fourth sample. Sorry for the typo. – Bill Johnson Jun 3 '15 at 1:01
• This is great! As far as I am concerned, this completes the answer. Thank you! – Iosif Pinelis Jun 3 '15 at 3:10
• I added the missing minus sign. – Bill Johnson Jun 3 '15 at 11:17

I'm not in my best shape at the moment, so, please, check thoroughly what is written below.

The distribution does not matter much, but the norm does. So, we want to take $N\ge 3$ standard basis vectors $e_n$ and see if there is a chance to create a permutation invariant norm in $\mathbb R^N$ such that

$$(*)\qquad N(N-1)\|e_1+e_2\|+2N\|e_1\|< N(N-1)\|e_1-e_2\|.$$

Now, just define the norm of $v$ as the infimum of $\sum_{i< j}|a_{i,j}|$ with $v=\sum_{i< j}a_{ij}(e_i+e_j)$. Then, obviously, $\|e_1\|\le 3/2$ ($2e_1=(e_1+e_2)+(e_1+e_3)-(e_2+e_3)$), $\|e_1+e_2\|\le 1$. However, if $$e_1-e_2=\sum_{i< j}a_{i,j}(e_i+e_j)\,,$$ then $2=\sum_{k>2}(a_{1k}-a_{2k})\le\sum_{i<j}|a_{ij}|$, so $\|e_1-e_2\|\ge 2$ and if $N$ is large, the inequality holds.

• I really wish I could accept both your answer and Bill Johnson's, but the system seems to force me to choose just one answer to accept. This is a difficult choice for me. On the one hand, your answer was first, and it was great, and you did answer the question just as it was stated. On the other hand, Bill's answer is more complete (in fact, quite complete, as far as I am concerned), and his construction is simpler and more transparent. – Iosif Pinelis Jun 3 '15 at 3:22
• No problem whatsoever. I said everything I have to say on this issue long ago (see mathoverflow.net/questions/66162/… ) and it takes way more than 4 years to change my core opinions :-). My only request to the people who upvote my answer is to upvote Bill's one as well. He really deserves it, IMHO :-). Now I'll take care of some other fish to fry for a while... – fedja Jun 3 '15 at 14:32
• Thanks, fedja; with a few more points and $3.95 I can get an espresso. More seriously, you homed in on the essential points and I just added some windrow dressing. What attracted me (other than it being proposed by Pinelis) was that the answer violates Kwapien's dictum that probabilistic inequalities that are true for the real line are also true for Banach space random variables. – Bill Johnson Jun 3 '15 at 18:07 • @BillJohnson -with a few more points and 3.95 I can get an espresso- Actually, I would rather redeem them for getting more people who read other posts of mine (and that does not even require$3.95 !). Believe it or not, there are many souls out there who would benefit from learning from you (even if they disagree to you to the extent that they would rather challenge you to a shooting duel at 20 yards than recognize that they owe you something). More seriously, the reputation points are just a numerical quantity reflecting something very real: your professional reputation... – fedja Jun 4 '15 at 10:11

fedja: This is a very nice and elegant answer, thank you! (Have you seen/used such a norm ever before?) Before accepting your answer, I'd like to wait a bit, so that other participants feel more encouraged to present other answers.

In fact, let me give here a modification of your answer. I noticed that for $N=3$ your norm of $v=(v_1,\dots,v_N)$ is $\sum_i|v_i-\frac12\,\sum_1^N v_j|$. So, I thought such a modified norm will work as well, for all $N\ge3$, perhaps with some other factor in place of $\frac12$. This does not seem to work, though. [Addendum: Of course, since the conjectured inequality in my question holds when $B$ is one-dimensional, it was silly for me to think even for a moment that such an $\ell_1$ norm as $\|v\|:=\sum_i|v_i-\frac12\,\sum_1^N v_j|$ could possibly provide a counterexample.]

However, if I am not mistaken, if $\|v\|:=\max_i|v_i-\frac14\,\sum_1^N v_j|$ for $v=(v_1,\dots,v_N)$, then $\|e_1\|=3/4$, $\|e_1+e_2\|=1/2$, and $\|e_1-e_2\|=1$, so that your main inequality $(*)$ holds for all $N\ge5$.

Only the dimensions $N=3,4$ now remain to be considered.

• Oops! You are right, Iosif; my arithmetic was bad, not for the first time. :) – Bill Johnson Jun 2 '15 at 16:49
• I've struggled with the arithmetic for quite some time. :-) It looks like fedja's original construction works for $N\ge5$, but I don't know about $N=3,4$. – Iosif Pinelis Jun 2 '15 at 17:19