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From: Robert Israel on 7 Jul 2010 11:07
Ray Vickson <RGVickson(a)shaw.ca> writes:

> On Jul 6, 7:02=A0pm, Robert Israel
> <isr...(a)math.MyUniversitysInitials.ca> wrote:
> > Robert Israel <isr...(a)math.MyUniversitysInitials.ca> writes:
> > > WizardOfOzzz <eric_pen...(a)hotmail.com> writes:
> >
> > > > Hey all,
> >
> > > > I have a problem where I have two 1D distributions represented with
> > > > mean/variance. =A0I need to calculated a third distribution that is
> > > > t=
> he
> > > > minimum of a random choice from both.
> >
> > > If X and Y are independent random variables with CDF F_X and F_Y
> > > respectively,
> > > Z =3D min(X,Y) has CDF F_Z(z) =3D 1 - (1 - F_X(z))(1 - F_Y(z)). =A0If
> > > X=
> and Y
> > > have
> > > densities f_X and f_Y, then Z has density
> > > f_Z(z) =3D (F_Z)'(z) =3D f_X(z) (1 - F_Y(z)) + f_Y(z) (1 - F_X(z)).
> >
> > > > I also need to represent the resulting distribution using only
> > > > mean/variance, even if that isn't the best approximation.
> >
> > > > Any ideas? =A0I figure it there should be a closed form solution
> > > > from=
> the
> > > > original means/variances to calculate the minimum mean/variance.
> >
> > > No, the mean and variance of Z do not depend only on the means and
> > > variances
> > > of X and Y.
> >
> > > But by "represented with mean/variance" I suppose you're assuming
> > > distributions
> > > from a specific two-parameter family, e.g. normal distributions. =A0In
> > > =
> that
> > > case
> > > the mean m_z and variance v_z of Z are rather complicated functions of
> > > the means and variances m_x, v_x, m_y, v_y of X and Y, involving some
> > > non-elementary integrals.
> >
> > For simplicity, suppose m_x =3D 0 and v_x =3D 1 (to which the general
> > cas=
> e can
> > be reduced by linear transformation). =A0Then a series expansion around
> > m=
> _y =3D 0,
> > v_y =3D 1 is
> >
> > m_z =3D -1/sqrt(pi) - (v_y - 1)/(4 sqrt(pi)) + m_y/2 + (v_y - 1)^2/(32
> > sq=
> rt(pi))
> > =A0 =A0 - m_y^2/(4 sqrt(pi)) + m_y^2 (v_y - 1)/(16 sqrt(pi))
> > =A0 =A0 - (v_y - 1)^3/(128 sqrt(pi)) - 3 m_y^2 (v_y - 1)^2/(128 sqrt(pi))
> > =A0 =A0 + 5 (v_y - 1)^4/(2048 sqrt(pi)) + m_y^4/(96 sqrt(pi)) + ...
> >
> > v_z =3D 1 - 1/pi + (pi - 1)(v_y - 1)/(2 pi) + (pi-2) m_y/(2 sqrt(pi))
> > =A0 =A0 + m_y (v_y - 1)/(2 sqrt(pi)) + m_y (v_y - 1)^2/(8 sqrt(pi))
> > =A0 =A0 - m_y^4/(24 pi) + m_y^3 (v_y - 1)/(24 sqrt(pi))
> > =A0 =A0 - 3 m_y (v_y - 1)^3/(64 sqrt(pi)) + ...
> >
> > and this should provide a fairly good approximation if m_y is close to 0
> > =
> and
> > v_y close to 1.
> > --
>
> How did you get these expressions?

Maple 14.

with(Statistics): assume(v_y > 0):
X:= RandomVariable(Normal(0,1));
Y:= RandomVariable(Normal(mu_y, sqrt(v_y)));
Z:= piecewise(X<Y,X,Y);
mu_z:= Mean(Z);
mtaylor(mu_z, [mu_y=0, v_y = 1]);
v_z:= Mean(Z^2) - mu_z^2;
mtaylor(v_z, [mu_y=0, v_y=1]);

Actually, you can also do a one-variable series expansion in powers of mu_y:

map(simplify,series(mu_z,mu_y));

1/2 1/2 1/2
2 (1 + v_y~) 2 2
- ------------------ + 1/2 mu_y - --------------------- mu_y +
1/2 1/2 1/2
2 Pi 4 Pi (1 + v_y~)

1/2
2 4 6
---------------------- mu_y + O(mu_y )
3/2 1/2
48 (1 + v_y~) Pi


map(simplify,series(v_z,mu_y));

1/2
Pi v_y~ - v_y~ + Pi - 1 (v_y~ - 1) 2 -2 + Pi
----------------------- - --------------------- mu_y + -------
2 Pi 1/2 1/2 4 Pi
2 (1 + v_y~) Pi

1/2
2 (v_y~ - 1) 2 3 1 4
mu_y + ---------------------- mu_y - ---------------- mu_y
3/2 1/2 12 (1 + v_y~) Pi
12 (1 + v_y~) Pi

1/2
(v_y~ - 1) 2 5 6
- ---------------------- mu_y + O(mu_y )
5/2 1/2
80 (1 + v_y~) Pi
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Ray Vickson on 7 Jul 2010 14:59
On Jul 7, 8:07 am, Robert Israel
<isr...(a)math.MyUniversitysInitials.ca> wrote:
> Ray Vickson <RGVick...(a)shaw.ca> writes:
> > On Jul 6, 7:02=A0pm, Robert Israel
> > <isr...(a)math.MyUniversitysInitials.ca> wrote:
> > > Robert Israel <isr...(a)math.MyUniversitysInitials.ca> writes:
> > > > WizardOfOzzz <eric_pen...(a)hotmail.com> writes:
>
> > > > > Hey all,
>
> > > > > I have a problem where I have two 1D distributions represented with
> > > > > mean/variance. =A0I need to calculated a third distribution that is
> > > > > t=
> > he
> > > > > minimum of a random choice from both.
>
> > > > If X and Y are independent random variables with CDF F_X and F_Y
> > > > respectively,
> > > > Z =3D min(X,Y) has CDF F_Z(z) =3D 1 - (1 - F_X(z))(1 - F_Y(z)). =A0If
> > > > X=
> >  and Y
> > > > have
> > > > densities f_X and f_Y, then Z has density
> > > > f_Z(z) =3D (F_Z)'(z) =3D f_X(z) (1 - F_Y(z)) + f_Y(z) (1 - F_X(z)).
>
> > > > > I also need to represent the resulting distribution using only
> > > > > mean/variance, even if that isn't the best approximation.
>
> > > > > Any ideas? =A0I figure it there should be a closed form solution
> > > > > from=
> >  the
> > > > > original means/variances to calculate the minimum mean/variance.
>
> > > > No, the mean and variance of Z do not depend only on the means and
> > > > variances
> > > > of X and Y.
>
> > > > But by "represented with mean/variance" I suppose you're assuming
> > > > distributions
> > > > from a specific two-parameter family, e.g. normal distributions. =A0In
> > > > =
> > that
> > > > case
> > > > the mean m_z and variance v_z of Z are rather complicated functions of
> > > > the means and variances m_x, v_x, m_y, v_y of X and Y, involving some
> > > > non-elementary integrals.
>
> > > For simplicity, suppose m_x =3D 0 and v_x =3D 1 (to which the general
> > > cas=
> > e can
> > > be reduced by linear transformation). =A0Then a series expansion around
> > > m=
> > _y =3D 0,
> > > v_y =3D 1 is
>
> > > m_z =3D -1/sqrt(pi) - (v_y - 1)/(4 sqrt(pi)) + m_y/2 + (v_y - 1)^2/(32
> > > sq=
> > rt(pi))
> > > =A0 =A0 - m_y^2/(4 sqrt(pi)) + m_y^2 (v_y - 1)/(16 sqrt(pi))
> > > =A0 =A0 - (v_y - 1)^3/(128 sqrt(pi)) - 3 m_y^2 (v_y - 1)^2/(128 sqrt(pi))
> > > =A0 =A0 + 5 (v_y - 1)^4/(2048 sqrt(pi)) + m_y^4/(96 sqrt(pi)) + ....
>
> > > v_z =3D 1 - 1/pi + (pi - 1)(v_y - 1)/(2 pi) + (pi-2) m_y/(2 sqrt(pi))
> > > =A0 =A0 + m_y (v_y - 1)/(2 sqrt(pi)) + m_y (v_y - 1)^2/(8 sqrt(pi))
> > > =A0 =A0 - m_y^4/(24 pi) + m_y^3 (v_y - 1)/(24 sqrt(pi))
> > > =A0 =A0 - 3 m_y (v_y - 1)^3/(64 sqrt(pi)) + ...
>
> > > and this should provide a fairly good approximation if m_y is close to 0
> > > =
> > and
> > > v_y close to 1.
> > > --
>
> > How did you get these expressions?
>
> Maple 14.
>
> with(Statistics): assume(v_y > 0):
> X:= RandomVariable(Normal(0,1));
> Y:= RandomVariable(Normal(mu_y, sqrt(v_y)));
> Z:= piecewise(X<Y,X,Y);
> mu_z:= Mean(Z);
> mtaylor(mu_z, [mu_y=0, v_y = 1]);
> v_z:= Mean(Z^2) - mu_z^2;
> mtaylor(v_z, [mu_y=0, v_y=1]);
>
> Actually, you can also do a one-variable series expansion in powers of mu_y:
>
> map(simplify,series(mu_z,mu_y));
>
>       1/2           1/2                       1/2
>      2    (1 + v_y~)                         2                 2
>    - ------------------ + 1/2 mu_y - --------------------- mu_y  +
>               1/2                        1/2           1/2
>           2 Pi                       4 Pi    (1 + v_y~)
>
>                   1/2
>                  2                 4         6
>         ---------------------- mu_y  + O(mu_y )
>                      3/2   1/2
>         48 (1 + v_y~)    Pi
>
> map(simplify,series(v_z,mu_y));
>
>                                            1/2
>   Pi v_y~ - v_y~ + Pi - 1      (v_y~ - 1) 2              -2 + Pi
>   ----------------------- - --------------------- mu_y + -------
>            2 Pi                         1/2   1/2         4 Pi
>                             2 (1 + v_y~)    Pi
>
>                                1/2
>             2      (v_y~ - 1) 2            3          1             4
>         mu_y  + ---------------------- mu_y  - ---------------- mu_y
>                              3/2   1/2         12 (1 + v_y~) Pi
>                 12 (1 + v_y~)    Pi
>
>                           1/2
>               (v_y~ - 1) 2            5         6
>          - ---------------------- mu_y  + O(mu_y )
>                         5/2   1/2
>            80 (1 + v_y~)    Pi
> --
> Robert Israel              isr...(a)math.MyUniversitysInitials.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel
> University of British Columbia            Vancouver, BC, Canada

OK: for normally-distributed r.v.s we can do it. However, the OP did
not specify any particular distributions (although he/she maybe meant
to have normal r.vs without saying so).

R.G. Vickson
From: Robert Israel on 7 Jul 2010 15:18
Ray Vickson <RGVickson(a)shaw.ca> writes:

| OK: for normally-distributed r.v.s we can do it. However, the OP did
| not specify any particular distributions (although he/she maybe meant
| to have normal r.vs without saying so).

Yes, I did mention "normal distributions", but maybe I didn't make it clear
enough that this was what I was calculating it for.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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Prev: NEWSFLASH! Meami.org finds: As it turns out P.NP-complete problems are a set of problems which any other NP-problem can be reduced to in polynomial time, but which retain the ability to have their solution verified
Next: Collatz-like conjecture using integer square root