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Suppose I have a vector of positive weights $a=(a_1, a_2, a_3, a_4)'$ such that $a_2=a_3$ and $a_1+a_2+a_3+a_4=1$. Is there any way to construct a joint sampling distribution for $a$ with a compact functional form ?

P.S. The context of this problem follows from a StackOverflow
question. There @josilber gave a very nice and easy sampling mechanism to sample this kind of weights.

Community
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Dey
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1 Answers1

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If you rewrite the expression into ($b_1$ + $b_2$ + $b_3$ = 1) and assume that $b_i$ is $U(0,1)$, you could clearly see it is Dirichlet distribution. It could be sampled from gammavariates with parameter vector $\vec{\alpha}=1$. After sampling reassign $a_1$ = $b_1$, $a_2$ = $b_2$/2, $a_3$ = $b_2$/2, $a_4$ = $b_3$

Severin Pappadeux
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  • Thanks. I think I got your point - we can assume $(b_1, b_2, b_3)$ to be $Dirichlet(\alpha_1, \alpha_2, \alpha_3)$ and $(b_1, b_2, b_3)$ will have the same distribution as $(a_1, a_2, a_3, a_4)$ with $a_2=a_3$. By the way why you are telling each A, B, C follows U(0,1) ? They do not. Only after two transformations ($-ln(x_i)$ and $x_i/\sum x_i$) on U(0,1) variate we can get Dirichlet. – Dey Sep 18 '15 at 04:53
  • @SoumenDey By the way why you are telling each A, B, C follows U(0,1) ? Well, PDF is a product of $x^{\alpha_i-1}_i$, and lies within $0...1$ range. In case of $\vec{\alpha}=1$ PDF is constant thus making it $U(0,1)$, isn't it? https://en.wikipedia.org/wiki/Dirichlet_distribution – Severin Pappadeux Sep 18 '15 at 16:03
  • @SoumenDey No, I'm most likely wrong. While $X_i$ is within $[0...1]$ interval and PDF is constant (for $\alpha=1$), mean and variance is different from $U(0,1)$. – Severin Pappadeux Sep 18 '15 at 19:32