Let x, y be i.i.d. random variables with zero mean. Let the
distribution of each of them be D. Let A be the area in the
2-dimensional plane that consists of the fourth and the eigth octants;
i.e., (x, y) is in A iff either y>x & x>0 & y>0, or y<x & x<0 & y<0.
I'd like to know a sufficient condition for E[x - y| (x,y) in A] > 0;
i.e., for the conditional mean of x-y given that the point (x,y) is in
A to be positive.

I think I have to clarify what sort of condition I look for. (Clearly,
it will impose some restriction on D.)

Of course, I don't want the condition that trivially rephrases my
required result in the form of integrals of the p.d.f. of D. Neither
do I want such a restrictive condition as a specific distribution
family that works - one would be quite easy to construct by
straightforward numerical computation (say, if D is truncated normal,
it is sufficient that the truncation is done to the right of a certain
percentile point).

What I do want is a non-trivial condition, that has some generality
(say, the set of distributions allowed by this condition should be
larger than continuum). Perhaps, something referring to the skewness
of D, or its other moments, or its hazard rate, etc.

Dear mm1234,

I have a condition that is fairly general. While it is not very far
from just a restatement of your original inequality, I do feel that it
elucidates the nature of the inequality, and it does enable one to
write down quite easily large families of distributions that satisfy
it. Still, if you are not satisfied please ask for a clarification and
I will see if I can still simplify it further.

Broadly speaking, your inequality concerns comparing the negative part
of the distribution D with the positive part of it. It will be
satisfied if the negative part of D is "less concentrated around its
mean value" (see below) than the positive part.

More precisely, define two distributions D+ and D- as:

D+ is the distribution of x conditioned on x>0
D- is the distribution of -x conditioned on x<0
(let's assume for convenience that D does not have an atom at 0)
(also let's assume that P(x > 0) = 1/2 - this simplifies the
inequalities below and still leaves us with a fairly broad class of
distributions; I will indicate below how to treat the more general
case)

Introduce two i.i.d. variables x' and y' having distribution D+,
and two i.i.d. variables x'' and y'' having distribution D-.

Also, write the set A you defined as the (disjoint) union of two sets
A1 and A2, where

A1 = { 0 > x > y },        A2 = { y > x > 0 }

Now your original inequality     E[ x - y | A ] > 0
can be restated as

E[ (x-y)1_A1 + (x-y)1_A2 ] > 0

(1_A1 is the indicator function of A1 etc.), or

E[ (x-y)1_A1 ] > E[ (y-x)1_A2 ]

This can be written in terms of the conditioned r.v.s x',y',x'',y'':

E[ (x''-y'') | x''>y'' ] > E[ (y'-x') | y'>x' ]

(here we use the assumption that P(x>0)=1/2, otherwise there will be a
factor
alpha on one side of the inequality and a factor 1-alpha on the other
side,
where alpha = P(x>0))

Since x'',y'' are i.i.d. and x',y' are i.i.d. this can be rewritten as

E|x''-y''| > E|x'-y'|

Which is (for now) my final form of the inequality, in other words the
sufficient condition you were asking for. For two i.i.d. random
variables u,v, the quantity E|u-v| is known as the DIVERGENCE of the
distribution of u. It is a bit like the variance in that it measures
how much u is concentrated around its mean value, except it is much
less convenient to work with than variance because it is not additive
with respect to independent sums. Still, you can now construct large
families of distributions D satisfying your original inequalities by
fixing their negative and positive parts separately and then giving
each of them probability 1/2 of happening to make up D. (Note that
under this assumption, since you specified that D has mean 0, the mean
of D+ must equal the mean of D-) If you fix them in such a way that
the divergence of D- is bigger than the divergence of D-, then your
inequality will be satisfied.

I hope this is the kind of answer you were looking for, please do not
hesitate to ask for clarification. If this is not exactly what you
were looking for, it would help me to know in more detail the
application for which you need this, so I can understand better the
kind of condition you are looking for.

Regards,
dannidin

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Clarification of Answer by dannidin-ga on 01 Nov 2002 03:24 PST

I forgot to mention as a simple (trivial) example that a very large
family of distributions D satisfying your condition is the
distributions which are supported on the interval (-infinity,0), in
other words where x is a negative r.v.

-dannidin

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Request for Answer Clarification by mm1234-ga on 04 Nov 2002 02:35 PST

Thanks, this is helpful. And this is exactly the type of answer I was
looking for. However, I have a couple of concerns about the specific
condition you suggest.

First, I am not too happy with the assumption P[x>0] = 1/2. Not
because it's restrictive (I don't mind that), but because it's
counterintuitive. Really, suppose that D- has bigger "divergence" than
D+. What can we say about skewness of D? Of course, there are numerous
conflicting definitions of skewness, so I am not rigorous here; but
this seems a clear case of a left-skewed distribution if there ever
was one. Unfortunately, one of the main measures of skewness is
Pearson's 2nd coefficient (the difference between the mean and the
median, scaled by the st. dev.). A "typical" left-skewed distribution
should have negative Pearson's 2nd coefficient, implying P[x>0] > 1/2
(since the median is to the right of the mean, which is zero).

Sadly, if we try to consider the case P[x>0] > 1/2, things will get
even worse. This implies that the coefficient alpha to which you
referred is less than 1/2, and so the overall conclusion about D now
becomes ambiguous: the higher divergence of the left side is offset by
the fact that the left side is multiplied by a smaller number (alpha <
1-alpha).

I am also concerned, though to a lesser extent, with the use of
E|x'-y'| to measure the divergence (or spread) of a distribution. I
would have no problem with using E|x - Ex|, or E|x - Median[x]|, or
Median|x - Median[x]| (respectively, mean absolute deviation from
mean, mean abs. dev. from median, and median abs. deviation from
median). But E|x'-y'| does not directly relate to either of those
measures. Of course, E[(x'-y')^2] does equal to 2 E[(x-Ex)^2] = 2
var[x]. This provides some justification for using E|x'-y'| as a
measure of spread.

And a very minor point -- the trivial example yuo mention wouldn't
work since I require E x = 0.

Let me know what you think.

Thanks again!

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Clarification of Answer by dannidin-ga on 04 Nov 2002 04:04 PST

Hi mm1234,

Regarding your concern about skewness and how it conflicts with my
condition that P(x>0)=1/2: I want to emphasize that my form of the
inequality is not directly related to skewness, which is a measure of
how the distribution is biased more towards the left side relative to
its mean than towards the right side. What we are doing here is rather
comparing the amount of "spread" of the left side of the distribution
relative to the amount of spread of the right side. So this is more a
"second-moment" version of what you call skewness (which is a
"first-moment" kind of concept). Indeed, for a distribution with mean
zero that satisfies P(x>0)=1/2, skewness as you define it is equal to
zero. However, this does not rule out the left side of the
distribution being more "spread out" (in the sense of divergence) than
the right side of the distribution. And again, if you allow arbitrary
values of P(x>0), you just need to give the correct weights (i.e.
P(x>0) and 1-P(x>0)) to the divergence of the left- and right- sides
when you compare them. This does not conflict with the concept of
skewness.

About your unease regarding my use of divergence as opposed to the
more traditional ways of measuring "spread": I was uneasy about this
myself, so over the weekend I checked an old paper I remembered
reading once that discussed comparisons of the different measures of
spread. As you yourself noticed, putting two independent copies of the
original random variable can be related to just one copy if we were
talking about variance rather than divergence. But in fact using the
Cauchy-Schwarz inequality (or alternatively the Holder inequality) can
bring us to divergence, since:

divergence = E|x-y| < or = sqrt(E(x-y)^2) = sqrt(2 var(x)) = sqrt(2) *
s.d(x)

(s.d(x) = standard deviation). In other words the divergence of x is
bounded
from above by sqrt(2) times the standard deviation of x which is a far
more
convenient measure of spread. As for a lower bound (which you need
since you
are comparing two divergences), this is more difficult. Some
computations using Fourier transforms show the following inequality

divergence = E|x-y| > or = E|x-Ex|, which relates the divergence to
the more familiar "mean absolute deviation from mean". The paper I
found this in is

von Bahr, B., Esseen, C.G. Inequalities for the r-th absolute moment
of a sum of random variables, 1<=r<=2.
Ann. Math. Statist. 36 (1965) 299-303.

Using these two inequalities you can rewrite my condition in terms of
these simpler measures of spread.

I hope this makes things clearer, and again if you have any more
questions please ask.

Regards,
dannidin

Thanks, that looks really good. I really liked your justification for
the E|x-y| as a measure of divergence. And I guess I am comfortable
that you can increase the spread of D- even while keeping its mean and
its p.d.f. at 0 constant (so that the transition from D- to D+ would
result in smooth p.d.f. of D at zero).