# Confusion regarding obtaining a Dirichlet distribution from Beta marginals

Consider $$N$$ independent random variables $$X_{1}, X_{2}, \ldots, X_{N}$$ such that $$\begin{equation} X_{i} \sim \Gamma\left(1, \frac{1}{N}\right), \end{equation}$$ for $$i \in [N]$$. Let $$\begin{equation} Y = \sum_{i=1}^{N} X_{i}. \end{equation}$$

I am trying to find the joint PDF $$f_{\left(X_1, X_2, \ldots, X_{N-1}\right)|Y=1}(x_{1}, x_2, \ldots x_{N-1}).$$

Can we conclude from our given information that the joint PDF is the PDF of a $$\text{Dirichlet}(1, 1, \ldots, 1)$$ distribution?

This answer tackles a similar question, but I am not sure how the answer concludes that the Beta marginals imply a joint distribution which is a Dirichlet (as a comment states, the marginals contain insufficient information about the joint). In other words, my confusion with the question I linked to is this:

Is it true that, for this case at least, the distribution for the random variable $$\frac{X_i}{X_1+\cdots+X_{N}}$$ is the same as the distribution for the random variable $$X_{i} | X_1+\cdots+X_{N} = 1$$?

If so, just out of curiosity, how general is this result? Does it hold for any distribution and any random variable?

• Changed the question slightly. – BlackHat18 Dec 22 '20 at 10:39
• Edited the question so that it is no longer a duplicate. – BlackHat18 Dec 22 '20 at 14:23

## 1 Answer

The joint distribution of $$(X_1,\ldots,X_{N-1},Y)$$ has density $$f_N(x_1,\ldots,x_{N-1},y-x_1-\cdots-x_{N-1})\mathbb I_{y\ge x_1+\cdots+x_{N-1}}$$ where $$f$$ is any density with Exponential margins. (An example is provided by Gaussian copulas.) The conditional distribution of $$(X_1,\ldots,X_{N-1})$$ given $$Y$$ thus has density $$\dfrac{f_N(x_1,\ldots,x_{N-1},y-x_1-\cdots-x_{N-1})\mathbb I_{y\ge x_1+\cdots+x_{N-1}}}{ \int_{[0,1]^N} f_N(x_1,\ldots,x_{N-1},y-x_1-\cdots-x_{N-1})\mathbb I_{y\ge x_1+\cdots+x_{N-1}}\text d x_1\cdots\,\text dx_{N-1}}$$ which can be essentially anything.

For instance, with $$N=3$$ and a Ali–Mikhail–Haq copula, $$f(x_1,x_2,x_3)=e^{-x_1-x_2-x_3}\frac{(1-e^{-x_1})(1-e^{-x_2})(1-e^{-x_3})}{1-\theta e^{-x_1-x_2-x_3}}$$ leads to $$f(x_1,x_2|y)\propto e^{-x_1-x_2-y+x-1+x_2}\frac{(1-e^{-x_1})(1-e^{-x_2})(1-e^{-y+x-1+x_2})}{1-\theta e^{-x_1-x_2-y+x_1+x_2}}$$ i.e. $$f(x_1,x_2|y)\propto (1-e^{-x_1})(1-e^{-x_2})(1-e^{-y+x_1+x_2})\mathbb I_{y\ge x_1+x_2}$$ which does not correspond to a Dirichlet.

• If that is the case, I am confused about why this answer stats.stackexchange.com/questions/252692/distribution-given-sum reaches a particularly nice form for the conditional distribution. Is it true that at least for the case at hand the the distribution for the random variable $\frac{X_i}{X_1+\cdots+X_{N}}$ is the same as the distribution for the random variable $X_{i} | X_1+\cdots+X_{N} = 1$? – BlackHat18 Dec 22 '20 at 17:00
• At the outset, this question stipulates that the $X_i$ are independent. – whuber Dec 22 '20 at 21:04
• The fact that the distribution of the random variable$$Z=X_1/X_1+⋯+X_N$$ is the same as the conditional distribution of the random variable$X_1$ given $X_1+⋯+X_N=1$ is due to $Z$ being independent of $X_1+⋯+X_N$. – Xi'an Dec 23 '20 at 8:37
• Why is $Z$ independent of $X_{1} + \cdots + X_{N}$? – BlackHat18 Dec 23 '20 at 17:55
• This is a result that requires a proof, not an obvious fact. See for instance Devroye (1986, XI.4, Theorem 4.1, p.594). – Xi'an Dec 23 '20 at 18:18