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?

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

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.

  • $\begingroup$ 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 $X_{i} | X_1+\cdots+X_{N} = 1$? $\endgroup$ – BlackHat18 Dec 22 '20 at 17:00
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    $\begingroup$ At the outset, this question stipulates that the $X_i$ are independent. $\endgroup$ – whuber Dec 22 '20 at 21:04
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    $\begingroup$ 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 $X_1+⋯+X_N$. $\endgroup$ – Xi'an Dec 23 '20 at 8:37
  • $\begingroup$ Why is $Z$ independent of $X_{1} + \cdots + X_{N}$? $\endgroup$ – BlackHat18 Dec 23 '20 at 17:55
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    $\begingroup$ This is a result that requires a proof, not an obvious fact. See for instance Devroye (1986, XI.4, Theorem 4.1, p.594). $\endgroup$ – Xi'an Dec 23 '20 at 18:18

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