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CDF of absolute value...

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Joe Bloe...
Posted: Thu Nov 05, 2009 11:40 am
Guest
I have a bivariate simmetric distribution (X,Y) and want Pr(|X| < a, |Y| < b).
By simply drawing rectangles, it's apparent that in terms of the (X,Y) CDF
that probability is

Pr(|X| < a, |Y| < b) = Pr(X < a, Y < b)
- Pr(X < -a, Y < b) - Pr(X < a, Y < -b)
+ Pr(X < -a, Y < -b)

Is there a similar general formula for many dimensions that gives
Pr(|X1| < x1, ..., |Xk| < xk) in terms of the CDF of (X1,...Xk)?
 
Jack Tomsky...
Posted: Thu Nov 05, 2009 12:12 pm
Guest
[quote]I have a bivariate simmetric distribution (X,Y) and
want Pr(|X| < a, |Y| < b).
By simply drawing rectangles, it's apparent that in
terms of the (X,Y) CDF
that probability is

Pr(|X| < a, |Y| < b) = Pr(X < a, Y < b)
- Pr(X < -a, Y < b) - Pr(X < a, Y < -b)
+ Pr(X < -a, Y < -b)

Is there a similar general formula for many
dimensions that gives
Pr(|X1| < x1, ..., |Xk| < xk) in terms of the CDF of
(X1,...Xk)?
[/quote]


The generalization is

Pr(|X1| < x1, ..., |Xk| < xk) =
Sum[sgn(Prod(xi))*Pr(X1 < +/-x1, ..., Xk < +/-xk)],

where the sum is over all 2^k combinations of k-tuples of +/- and sgn is the sign function. That is, sgn is +1 if the number of negative xi is even and -1 if the number of negative xi is odd.

Jack
www.tomskystatistics.com
 
Joe Bloe...
Posted: Thu Nov 05, 2009 12:41 pm
Guest
Jack Tomsky <jtomsky at (no spam) ix.netcom.com> wrote:
[quote]I have a bivariate simmetric distribution (X,Y) and
want Pr(|X| < a, |Y| < b).
By simply drawing rectangles, it's apparent that in
terms of the (X,Y) CDF
that probability is

Pr(|X| < a, |Y| < b) = Pr(X < a, Y < b)
- Pr(X < -a, Y < b) - Pr(X < a, Y < -b)
+ Pr(X < -a, Y < -b)

Is there a similar general formula for many
dimensions that gives
Pr(|X1| < x1, ..., |Xk| < xk) in terms of the CDF of
(X1,...Xk)?

The generalization is

Pr(|X1| < x1, ..., |Xk| < xk) =
Sum[sgn(Prod(xi))*Pr(X1 < +/-x1, ..., Xk < +/-xk)],

where the sum is over all 2^k combinations of k-tuples of +/- and
sgn is the sign function. That is, sgn is +1 if the number of
negative xi is even and -1 if the number of negative xi is odd.

Jack
www.tomskystatistics.com
[/quote]
Thank you, Jack.
Is there a reference to this?
 
Jack Tomsky...
Posted: Thu Nov 05, 2009 7:24 pm
Guest
[quote]Jack Tomsky <jtomsky at (no spam) ix.netcom.com> wrote:
I have a bivariate simmetric distribution (X,Y)
and
want Pr(|X| < a, |Y| < b).
By simply drawing rectangles, it's apparent that
in
terms of the (X,Y) CDF
that probability is

Pr(|X| < a, |Y| < b) = Pr(X < a, Y < b)
- Pr(X < -a, Y < b) - Pr(X < a, Y < -b)
+ Pr(X < -a, Y < -b)

Is there a similar general formula for many
dimensions that gives
Pr(|X1| < x1, ..., |Xk| < xk) in terms of the CDF
of
(X1,...Xk)?

The generalization is

Pr(|X1| < x1, ..., |Xk| < xk) =
Sum[sgn(Prod(xi))*Pr(X1 < +/-x1, ..., Xk < +/-xk)],

where the sum is over all 2^k combinations of
k-tuples of +/- and
sgn is the sign function. That is, sgn is +1 if
the number of
negative xi is even and -1 if the number of
negative xi is odd.

Jack
www.tomskystatistics.com

Thank you, Jack.
Is there a reference to this?
[/quote]


Joe, the way I looked at it was as follows. The identity is true for k=1, since Pr(|X|<a) = Pr(X<a)-Pr(X<-a).

For k=2, you came up with your result geometrically by constructing rectangles.

For k=3, I worked it out in terms of a right-angled parallelepiped centered at the origin.

The results for k=1,2, and 3 suggested, heuristically, a generalization to k dimensions which are the sums and differences of 2^k probabilities, half of them (2^(k-1)) being plus and half of them being minus.

Since the identity is rather simple, I'm sure that there must be a neat formal derivation, possibly by mathematical induction.

Jack
www.tomskystatistics.com
 
Joe Bloe...
Posted: Fri Nov 06, 2009 10:38 am
Guest
Jack Tomsky <jtomsky at (no spam) ix.netcom.com> wrote:
[quote]I have a bivariate simmetric distribution (X,Y) and
want Pr(|X| < a, |Y| < b).
By simply drawing rectangles, it's apparent that in
terms of the (X,Y) CDF
that probability is

Pr(|X| < a, |Y| < b) = Pr(X < a, Y < b)
- Pr(X < -a, Y < b) - Pr(X < a, Y < -b)
+ Pr(X < -a, Y < -b)

Is there a similar general formula for many
dimensions that gives
Pr(|X1| < x1, ..., |Xk| < xk) in terms of the CDF of
(X1,...Xk)?

The generalization is

Pr(|X1| < x1, ..., |Xk| < xk) =
Sum[sgn(Prod(xi))*Pr(X1 < +/-x1, ..., Xk < +/-xk)],

where the sum is over all 2^k combinations of k-tuples of +/- and
sgn is the sign function. That is, sgn is +1 if the number of
negative xi is even and -1 if the number of negative xi is odd.

Thank you, Jack.
Is there a reference to this?

Joe, the way I looked at it was as follows. The identity is true
for k=1, since Pr(|X|<a) = Pr(X<a)-Pr(X<-a).

For k=2, you came up with your result geometrically by constructing
rectangles.

For k=3, I worked it out in terms of a right-angled parallelepiped
centered at the origin.

The results for k=1,2, and 3 suggested, heuristically, a
generalization to k dimensions which are the sums and differences of
2^k probabilities, half of them (2^(k-1)) being plus and half of them
being minus.

Since the identity is rather simple, I'm sure that there must be a
neat formal derivation, possibly by mathematical induction.
[/quote]
Once you pointed it out, I can see that the formula is correct. I just
wanted to know if that is a some kind of textbook result. As a new
dimension is added, the terms that involve its positive bound should
be added with the same sign as before. Therefore, for the remaining,
negative bound terms, the sign must be reversed. This gives your
formula.
 
 
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