From: Ala' Jaouni <ajaouni_at_gmail.com>

Date: Wed, 26 Mar 2008 14:26:59 -0700

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Wed 26 Mar 2008 - 21:29:56 GMT

Date: Wed, 26 Mar 2008 14:26:59 -0700

X1,X2,X3,X4 should have independent distributions. They should be between 0 and 1 and all add up to 1. Is this still possible with Robert's method?

Thanks

On Wed, Mar 26, 2008 at 12:52 PM, Ted Harding
<Ted.Harding_at_manchester.ac.uk> wrote:

*> On 26-Mar-08 20:13:50, Robert A LaBudde wrote:
*

> > At 01:13 PM 3/26/2008, Ala' Jaouni wrote:

*> >>I am trying to generate a set of random numbers that fulfill
**> >>the following constraints:
**> >>
**> >>X1 + X2 + X3 + X4 = 1
**> >>
**> >>aX1 + bX2 + cX3 + dX4 = n
**> >>
**> >>where a, b, c, d, and n are known.
**> >>
**> >>Any function to do this?
**> >
**> > 1. Generate random variates for X1, X2, based upon whatever
**> > unspecified distribution you wish.
**> >
**> > 2. Solve the two equations for X3 and X4.
**>
**> The trouble is that the original problem is not well
**> specified. Your suggestion, Robert, gives a solution
**> to one version of the problem -- enabling Ala' Jaouni
**> to say "I have generated 4 random numbers X1,X2,X3,X4
**> such that X1 and X2 have specified distributions,
**> and X1,X2,X3,X4 satisfy the two equations ... ".
**>
**> However, suppose the real problem was: let X2,X2,X3,X4
**> have independent distributions F1,F2,F3,F4. Now sample
**> X1,X2,X3,X4 conditional on the two equations (i.e. from
**> the coditional density). That is a different problem.
**>
**> As a slightly simpler example, suppose we have just X1,X2,X3
**> and they are independently uniform on (0,1). Now sample
**> from the conditional distribution, conditional on
**> X1 + X2 + X3 = 1.
**>
**> The result is a random point uniformly distributed on the
**> planar triangle whose vertices are at (1,0,0),(0,1,0),(0,0,1).
**>
**> Then none of X1,X2,X3 is uniformly distributed (in fact
**> the marginal density of each is 2*(1-x)).
**>
**> However, your solution would work from either point of
**> view if the distributions were Normal.
**>
**> If X1,X2,X3,X4 were neither Normally nor uniformly
**> distributed, then finding or simulating the conditional
**> distribution would in general be difficult.
**>
**> Ala' Jaouni needs to tell us whether what he precisely
**> wants is as you stated the problem, Robert, or whether
**> he wants a conditional distribution for given distributions
**> if X1,X2,X3,X4, or whether he wants something else.
**>
**> Best wishes to all,
**> Ted.
**>
**> --------------------------------------------------------------------
**> E-Mail: (Ted Harding) <Ted.Harding_at_manchester.ac.uk>
**> Fax-to-email: +44 (0)870 094 0861
**> Date: 26-Mar-08 Time: 19:52:16
**> ------------------------------ XFMail ------------------------------
**>
*

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