Re: X+Y | -w<=X-Y<=w
- From: "bokdan.hermicz@xxxxxxxxxxxxxx" <bokdan.hermicz@xxxxxxxxxxxxxx>
- Date: Sun, 29 Jul 2007 16:20:28 -0000
On 29 Jul., 17:19, "Stephen J. Herschkorn" <sjhersc...@xxxxxxxxxxxx>
wrote:
In sci.stat.math, bokdan hermicz wrote:If X and Y follow a uniform distribution with support (0,1) then Z=X+Y
if X and Y are Gaussian and independent and w=Inf P(X+Y |-w<=X-Y<=w)
is equal to the unconditional probability P(X+Y). This implies that
Z=X+Y follows a Gaussian with mean E(X)+E(Y) and Var(Z)=Var(X)+Var(Y).
I do not undertstand what you mean by "This implies that." The fact
that has normal distrbution with parameters EX + EY and Var(X) +
Var(Y) has nothing to do with your conditioning event.
(for X and Y independent) follows a triangular distribution. But the
density of P(Z | |X-Y| <= w) has a trapezoidal shape with a flat part
in the middle, if the condition really binds (w <1). Therefore the
resulting variance of Z depends on the conditioning event. Moreover,
the shape of the density depends on the conditioning event. And I
think this should also be the case, if X and Y are Gaussian. If X and
Y are independently uniformly distributed the variance of P(Z | |X-Y|
<= w) is smaller for larger values if w, because the shape of the
resulting variable Z approaches the shape of the triangular
distribution.
If w is smaller than Inf:
Intuition suggests that Var(Z) > Var(X)+Var(Y) but does anyone know a
formula / approximation for this?
Let A = {|X - Y| <= w}. Are you asking if Var(Z | A) > Var(X | A) +
Var(Y | A) when w is finite? That depends on the sign of Cov(X, Y |
A). Do you have empirical evidence to suggest that this conditional
covariance is always negative?
I am asking, if Var( Z | A) > Var(Z). As written above I do think that
the variance (and the distribution) depends on the conditioning event.
And this is the link to the second question, if the resulting Z can be
approximated by a Gaussian and what is the approximated variance?
And is there a approximation for the case that X and Y are dependent.
Second issue:
Can the resulting distribution of Z approximated by a Gaussian
distribution?
--
Stephen J. Herschkorn sjhersc...@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey and Manhattan
.
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