I am looking for a (somewhat intuitive) sufficient condition for a
nonnegative random variable X to satisfy the following property:

f(x) + (x - c) f'(x) >= 0,

where f(x) is the p.d.f. of X, and c is a given constant strictly
greater than the expecation E[X] of X. (If it helps, c is actually
equal to E[X|X>Y], where X, Y i.i.d.) Note: f'(x) is the derivative of f(x).

If this question requires more research time, let me know.

By intuitive, I mean something that has to do with the moments of X,
or its hazard rate, or other properties of X that are "higher level"
than the p.d.f. Also, I don't want a specific example of (a family of)
r.v. that satisfies the given property -- I want something more
general. See the answer to my other question (ID 95175) for an idea of
what I mean.

Thanks!

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Request for Question Clarification by mathtalk-ga on 20 Nov 2002 05:58 PST

Hi, mm1234-ga:

Are you sure you mean that f(x) is the probability density function
(of random variable X), and that f'(x) is the derivative of f(x)?  I
would have expected your question, from a probability point of view,
to involve a cumulative distribution function and _its_ derivative
(the probability density function).

Note that the derivative is a very local construct.  Given a p.d.f.
f(x) which satisfies the property, one can make an arbitrarily small
perturbation (in the L2 sense) that causes the property to fail. 
Therefore very strong conditions are needed to imply the inequality.

Would unimodality be the sort of property (in combination with other
conditions) that might be used in answering your question?

thanks, mathtalk-ga

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Clarification of Question by mm1234-ga on 20 Nov 2002 12:28 PST

Yes, f(x) is the p.d.f., not c.d.f., and f'(x) is the (very
unintuitive) derivative of the p.d.f. function. The reason the
condition is so strange is that it's actually a sufficient condition
for the more complex property to be satisfied. Unfortunately, the
original property is probably too complicated to work with, but what
the hell, here it goes:

Integral from 0 to infinity of

x |x - c| f(x) (f(x) + (x - c) f'(x)) dx

is positive, where |x - c| is the absolute value of x - c. So you can
see, a small perturbation would change nothing. I just (stupidly?)
tried to achieve this property by forcing the term in the parenthesis
to be always positive.

Unimodality as a property is ok (unfortunately, if you look at the
signs, it's not going to make things work by itself).

Hi, mm1234-ga:

Thanks for the clarification.  It puts your question in a much better
context for me.

I think I understand your injunction that you are not interested in a
simple mathematical restatement of the condition or in examples of
distribution families that satisfy the condition.  However it might be
helpful to all to provide some of those facts as comments (not as an
answer).

The condition:

f(x) + (x - c) f'(x) >= 0

is equivalent to bounds on d(ln(f(x))/dx as follows:

for x < c, d(ln(f(x)))/dx <= 1/(c - x)

for x > c, d(ln(f(x)))/dx >= 1/(c - x)

A natural candidate distribution is the simple exponential on [0,+oo):

f(x) = exp(-x)

for which d(ln(f(x)))/dx = -1 identically.  We can thus see
immediately that the first half of the bounds (x < c) is easily
satisfied, and that the second half will _fail_ for all sufficiently
large x, e.g. x > c+1.

Other candidates could be checked, e.g. gamma distributions and
censored normal distributions (X > 0).  If you have already done so
and would be willing to share the results, I'd be interested to read
them.  My impression is that the pointwise version of the condition
requires a very "sparse" tail to the distribution, as for example the
exponential distribution fails by having a "thick" tail.

regards, mathtalk-ga

Hi, mm1234-ga:

Just to follow up on my previous comment, where I should that the
standard exponential distribution fails the "pointwise" version of
your condition, by mentioning that it satisfies your original
"integral" form of the condition.

I computed c = E[X|X>Y] for this distribution to be 3/2.  Then:

INTEGRAL x |x - c| f(x) (f(x) + (x - c) f'(x)) dx OVER [0,+oo)

works out to be slightly more than 0.3, hence positive.  Perhaps this
gives a little concrete insight into the degree to which the pointwise
condition is stronger than the integral version.

regards, mathtalk-ga

thanks.. yeah, I guess the pointwise condition is very restrictive,
but i just don't see how to approach the integral condition. the
latter seems to be even harder to find intuition for.