2

I am trying to get the distribution of a weighted sum when the weights are uncertain:

$S = \sum\limits_{i=1}^N w_iC_i = \mathbf{w}\cdot \mathbf{C}$ where vector $\mathbf{w}$ is random with components having an N-dimensional Dirichlet distribution,: $\mathbf{w} \sim \mathcal{D}_{\theta_1,\theta_2...\theta_n}$ such that $\sum\limits_{i=1}^N w_i = 1$

The vector $\mathbf{C}$ is an N-dimensional fixed vector, whose components are the terms being randomly weighted.

I think that I can approximate the dirichlet by a multivariate gaussian, with variance-covaraince matrix determined from the dirichlet variances-covariances. Then the weighted sum would could be modeled as the a truncated normal distribution.

However, is there any theory out there about the actual distribution of the above operation?

2 Answers2

2

I'm very interested in this distribution as well, unfortunately I believe TenaliRaman's answer is only an approximation. Perhaps it only holds when the distribution is sufficiently concentrated to be approximately Gaussian..?

I tried the following in R:

require(gtools)
library(MASS)

n <- 7000
C <- c(0,1,0.2,0.5)
alpha <- rep(0.2,length(C))
x <- rdirichlet(n, alpha)%*%C


fits <- fitdistr(x,"beta",list(shape1=1,shape2=1))

y <- rbeta(n,shape1=fits$estimate[1],shape2=fits$estimate[2])

plot(sort(x),sort(y))
abline(a=0,b=1)

Here's the result:

QQ plot of Dirichlet dot product and samples from Beta fit

Clearly not a match.

martin
  • 161
1

If you shift the value of $C$ as $\hat{C} = \frac{C - \min(C)\textbf{1}}{\max(C) - \min(C)}$, then I believe the distribution of $S = w^{\top}\hat{C}$ is a beta distribution. I don't have a proof for this but if we do the following:

  1. Take a random $C$, compute $\hat{C}$.
  2. Take a random $\alpha$.
  3. Sample several random $w$'s from $Dir(\alpha)$, call it $W = [w_1 w_2 \ldots w_n]$.
  4. Compute $\hat{S} = W^{\top}\hat{C}$
  5. Fit a beta distribution over $\hat{S}$.
  6. Do a qqplot of $\hat{S}$ vs. random values samples from the beta distribution.

If we do the above, the qqplot looks like this: qqplot

TenaliRaman
  • 3,916
  • This is VERY cool. Wow. I'll need some sort of proof, but your numerical results are very convincing. –  Mar 28 '14 at 17:59
  • I changed my mind about the proof...that chart, and the approach you used to produce it, is sufficient for my purposes. Thanks again! –  Mar 31 '14 at 14:38
  • Do you have the code for your generated plot? I'm more convinced by the answer from https://math.stackexchange.com/users/134328/martin below. – sqrt Feb 21 '18 at 21:47