generate random 6 weights range (-1,1) with sum of absolute value equals 1

[rosiecao2509]

I want to simulate 1000 sets comprising 6 weights (w1, w2, w3, w4, w5, w6). Their values are uniformly distributed ranging from -1 to 1 (but their absolute values are greater than 0.01. The sum of their absolute values is added to 1. 

Can anyone please help?

[mkeintz]

The moment you require the sum of absolute values of w1..w6 to be 1, you can no longer expect each weight w to be uniformly distributed w~uniform(-1,1). 

True, you could distribute w1~uniform(-1,+1), excluding (-0.01,+0.01).  Actually you'd distribute absolute |w1|~uniform(0.01,+1), and then randomly generate a positive or negative sign.  Or to prevent any of w2 through w6 being less than 0.01, you would uniformly distribute |w1|~uniform(0.01,0.95).

Let's say the absolute value of |w1| is 0.4.  Then you could uniformly distribute absolute |w2| over (0.01,0.55), i.e. (0.01, 0.95-0.4), and again randomly generate a sign.

You could similarly generate w3 through w5, and finally directly calculate absolute |w6| from the others, and randomly give it a sign.  

But clearly the only weight distributed uniform(-1,1) is w1.  

So could you clarify?

[FreelanceReinh · Accepted solution]

Hello @rosiecao2509,

As mkeintz has already pointed out, there is no probability distribution satisfying all your requirements. Indeed, note that for |w_i| (i=1, ..., 6) with a uniform distribution on [0.01, 1] you would have the expectation E(|w₁|+...+|w₆|) = 6*(0.01+1)/2 = 3.03, not 1 as per your requirement.

I would probably start with a Dirichlet distribution, Dir(a). For example, consider the special case a=(1,1,1,1,1,1), which yields the uniform distribution over the set of points with positive coordinates in six-dimensional space whose coordinates sum to 1. Once you have generated a random vector (x₁, ..., x₆) from this distribution, you easily get corresponding weights w_i: Just apply the linear transformation x_i → 0.01+0.94*x_i and generate a random sign.

Here is SAS code implementing this approach:

data want(drop=i j);
call streaminit(27182818);
array w[6];
array x[6] _temporary_;
do i=1 to 1000;
  do j=1 to dim(w);
    x[j]=rand('expo');
  end;
  do j=1 to dim(w);
    w[j]=(2*rand('bern',0.5)-1)*(0.01+0.94*x[j]/sum(of x[*]));
  end;
  output;
end;
run;

(Note that the exponential distribution with default parameter 1 is the same as the Gamma(1,1) distribution mentioned in the Wikipedia article.)

In particular, this approach ensures that the six weights have the same probability distribution.

[rosiecao2509]

Thanks a lot Reinhard,
It works perfectly well. Thank you so much for your help.

Wish you all the best.