It is common to use weights in applications like mixture modeling and to linearly combine basis functions. Weights $w_i$ must often obey $w_i ≥$ 0 and $\sum_{i} w_i=1$. I'd like to randomly choose a weight vector $\mathbf{w} = (w_1, w_2, …)$ from a uniform distribution of such vectors.

It may be tempting to use $w_i = \frac{\omega_i}{\sum_{j} \omega_j}$ where $\omega_i \sim$ U(0, 1), however as discussed in the comments below, the distribution of $\mathbf{w}$ is not uniform.

However, given the constraint $\sum_{i} w_i=1$, it seems that the underlying dimensionality of the problem is $n-1$, and that it should be possible to choose a $\mathbf{w}$ by choosing $n-1$ parameters according to some distribution and then computing the corresponding $\mathbf{w}$ from those parameters (because once $n-1$ of the weights are specified, the remaining weight is fully determined).

The problem appears to be similar to the sphere point picking problem (but, rather than picking 3-vectors whose $ℓ_2$ norm is unity, I want to pick $n$-vectors whose $ℓ_1$ norm is unity).


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    $\begingroup$ Your method does not generate a uniformly distributed vector on the simplex. To do what you want correctly, the most straightforward way is to generate $n$ iid $\mathrm{Exp}(1)$ random variables and then normalize them by their sum. You could try to do it by finding some other method to draw only $n-1$ variates directly, but I have my doubts regarding the efficiency tradeoff since $\mathrm{Exp}(1)$ variates can be very efficiently generated from $U(0,1)$ variates. $\endgroup$ – cardinal Aug 9 '11 at 22:24

Choose $\mathbf{x} \in [0,1]^{n-1}$ uniformly (by means of $n-1$ uniform reals in the interval $[0,1]$). Sort the coefficients so that $0 \le x_1 \le \cdots \le x_{n-1}$. Set

$$\mathbf{w} = (x_1, x_2-x_1, x_3 - x_2, \ldots, x_{n-1} - x_{n-2}, 1 - x_{n-1}).$$

Because we can recover the sorted $x_i$ by means of the partial sums of the $w_i$, the mapping $\mathbf{x} \to \mathbf{w}$ is $(n-1)!$ to 1; in particular, its image is the $n-1$ simplex in $\mathbb{R}^n$. Because (a) each swap in a sort is a linear transformation, (b) the preceding formula is linear, and (c) linear transformations preserve uniformity of distributions, the uniformity of $\mathbf{x}$ implies the uniformity of $\mathbf{w}$ on the $n-1$ simplex. In particular, note that the marginals of $\mathbf{w}$ are not necessarily independent.

3D point plot

This 3D point plot shows the results of 2000 iterations of this algorithm for $n=3$. The points are confined to the simplex and are approximately uniformly distributed over it.

Because the execution time of this algorithm is $O(n \log(n)) \gg O(n)$, it is inefficient for large $n$. But this does answer the question! A better way (in general) to generate uniformly distributed values on the $n-1$-simplex is to draw $n$ uniform reals $(x_1, \ldots, x_n)$ on the interval $[0,1]$, compute

$$y_i = -\log(x_i)$$

(which makes each $y_i$ positive with probability $1$, whence their sum is almost surely nonzero) and set

$$\mathbf w = (y_1, y_2, \ldots, y_n) / (y_1 + y_2 + \cdots + y_n).$$

This works because each $y_i$ has a $\Gamma(1)$ distribution, which implies $\mathbf w$ has a Dirichlet$(1,1,1)$ distribution--and that is uniform.

[3D point plot 2]

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    $\begingroup$ @Chris If by "Dir(1)" you mean the Dirichlet distribution with parameters $(\alpha_1, \ldots, \alpha_n)$ = $(1,1,\ldots,1)$, then the answer is yes. $\endgroup$ – whuber Aug 10 '11 at 12:04
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    $\begingroup$ (+1) One minor comment: The intuition is excellent. Care in interpreting (a) may need to be taken, as it seems that the "linear transformation" in that part is a random one. However, this is easily worked around at the expense of additional formality by using exchangeability of the generating process and a certain invariance property. $\endgroup$ – cardinal Aug 10 '11 at 12:25
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    $\begingroup$ More explicitly: For distributions with a density $f$, the density of the order statistics of an iid sample of size $n$ is $n! f(x_1)\cdots f(x_n) 1_{(x_1 < x_2 < \cdots < x_n)}$. In the case of $f = 1_{[0,1]}(x)$, the distribution of the order statistics is uniform on a polytope. Taken from this point, the remaining transformations are deterministic and the result follows. $\endgroup$ – cardinal Aug 10 '11 at 12:47
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    $\begingroup$ @cardinal That's an interesting point, but I don't think it matters, although you're right that additional details could help. The swaps (actually reflections, qua linear transformations) are not random: they are predetermined. In effect, $I_{n-1}=[0,1]^{n-1}$ is carved into $(n-1)!$ regions, of which one is distinguished from the others, and there's a predetermined affine bijection between each region and the distinguished one. Whence, the only additional fact we need is that a uniform distribution on a region is uniform on any measurable subset of it, which is a complete triviality. $\endgroup$ – whuber Aug 10 '11 at 12:51
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    $\begingroup$ @whuber: Interesting remarks. Thanks for sharing! I always appreciate your insightful thoughts on such things. Regarding my previous comment on "random linear transformation", my point was that, at least through $\mathbf{x}$, the transformation used depends on the sample point $\omega$. Another way to think of it is there is a fixed, predetermined function $T: \mathbb{R}^{n-1} \to \mathbb{R}^{n-1}$ such that $\mathbf{w} = T(\mathbf{x})$, but I wouldn't call that function linear, though it is linear on subsets that partition the $(n-1)$-cube. :) $\endgroup$ – cardinal Aug 10 '11 at 16:48
    zz <- c(0, log(-log(runif(n-1))))
    ezz <- exp(zz)
    w <- ezz/sum(ezz)

The first entry is put to zero for identification; you would see that done in multinomial logistic models. Of course, in multinomial models, you would also have covariates under the exponents, rather than just the random zzs. The distribution of the zzs is the extreme value distribution; you'd need this to ensure that the resulting weights are i.i.d. I initially put rnormals there, but then had a gut feeling that this ain't gonna work.

  • $\begingroup$ That doesn't work. Did you try looking at a histogram? $\endgroup$ – cardinal Aug 9 '11 at 21:53
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    $\begingroup$ Your answer is now almost correct. If you generate $n$ iid $\mathrm{Exp}(1)$ and divide each by the sum, then you will get the correct distribution. See Dirichlet distribution for more details, though it doesn't discuss this explicitly. $\endgroup$ – cardinal Aug 9 '11 at 22:02
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    $\begingroup$ Given the terminology you are using, you sound a little confused. $\endgroup$ – cardinal Aug 9 '11 at 22:21
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    $\begingroup$ Actually, the Wiki link does discuss this (fairly) explicitly. See the second paragraph under the Support heading. $\endgroup$ – cardinal Aug 9 '11 at 22:28
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    $\begingroup$ This characterization is both too restrictive and too general. It is too general in that the resulting distribution of $\mathbf{w}$ must be "uniform" on the $n-1$ simplex in $\mathbb{R}^n$. It is too restrictive in that the question is worded generally enough to allow that $\mathbf{w}$ be some function of an $n-1$-variate distribution, which in turn presumably, but not necessarily, consists of $n-1$ independent (and perhaps iid) variables. $\endgroup$ – whuber Aug 10 '11 at 2:07

The solution is obvious. The following MathLab code provides the answer for 3 weights.

function [  ] = TESTGEN( )
SZ  = 1000;
V  = zeros (1, 3);
VS = zeros (SZ, 3);
for NIT=1:SZ   
   V(1) = rand (1,1);     % uniform generation on the range 0..1
   V(2) = rand (1,1) * (1 - V(1));
   V(3) = 1 - V(1) - V(2);  
   PERM = randperm (3);    % random permutation of values 1,2,3
   for NID=1:3
         VS (NIT, NID) = V (PERM(NID));
scatter3 (VS(:, 1), VS(:,2), VS (:,3));

enter image description here

  • 1
    $\begingroup$ Your marginals do not have the correct distribution. Judging from the Wikipedia article on the Dirichlet distribution (random number generation section, which has the algorithm you have coded), you should be using a beta(1,2) distribution for V(1), not a uniform[0,1] distribution. $\endgroup$ – soakley Dec 3 '15 at 18:43
  • $\begingroup$ It does appear that the density increases in the corners of this tilted triangle. Nonetheless, it provides a nice geometric display of the problem. $\endgroup$ – DWin Jul 18 '18 at 0:33

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