Natural Science Forum / Physics / General Physics / July 2008

From Osher Doctorow

Consider:

1) (x o y) = xy + F(x, y) - yFX(x),  or for short xy + F - FX

We can factor this as:

2) (x o y) = y(x - FX(x)) + F(x, y)

For X, Y nonnegative continuous random variables, we know that x > = 0
and y > = 0, and it is known from mathematical probability-statistics
that FX(x) > = and F(x, y) > = 0 always.

When x = FX(x), which only happens for all x for the uniform
probability distribution function (pdf) on [0, 1], we have:

3) (x o y) = 0 + F(x, y) = F(x, y) when x= FX(x) for all x in [0, 1]
(so X uniform on [0, 1])

I will prove here that for the Exponential cumulative distribution
function FX(x):

4) FX(x) = 1 - exp(-x)

we have:

5) (x o y) > = xy for X an exponentially distributed random variable,
with = only at x = 0.

Simply use the second derivative test on:

6) exp(-x) + x - 1

We have:

7) Dx[exp(-x) + x - 1] = -exp(-x) + 1 = 0 iff 1 = 1/exp(x) iff exp(x)
= 1 iff x = 0

8) Dxx[exp(-x) + x - 1] = Dx[-exp(-x) + 1] = exp(-x) > 0 always

So by the second derivative test, we have:

9) exp(-x) + x - 1 has a relative minimum at x = 0, at which point
exp(-x) + x - 1 = 0, and therefore exp(-x) + x - 1 > 0 if x is not 0,
which is to say FX(x) = 1 - exp(-x) < x if x is not 0.

Q.E.D.

The integral mean value theorem proves that FX(x) < = x for X a
continuous nonnegative random variable provided that the probability
density function of X, fX(x), does not exceed 1.  Although this does
not always hold, for example for a very sharply peaked Gaussian/normal
distribution close to its mean (with very small variance), it holds
for a very large variety of families of random variables.

Osher Doctorow