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We are given statistical sample of $X=(X_1,X_2,X_3)$, where $X_i\sim\Gamma(\alpha_i,1)$ and independent. Let $Z=X_1+X_2+X_3$ and $T$ three dimensional statistic $T:=(\frac{X_1}{Z},\frac{X_2}{Z},\frac{X_3}{Z})$.

We are asked to check are $T$ and $Z$ independent.

Hint: Find joint density function of $T$.

So far, I manage to find that:

$$Z\sim\Gamma(\alpha_1+\alpha_2+\alpha_3,1)$$ and that:

$$\frac{X_i}{Z}\sim\beta'(\alpha_i,\alpha_1+\alpha_2+\alpha_3)$$.

So:

$$f_T(y_1,y_2,y_3)=\prod_{i=1}^{3}f_{\frac{X_i}{Z}}(y_i).$$

However, now when I have to prove the independence of $T$ and $Z$ I get lost. How should I use this joint distribution function? Should I try to deduce something special about this distribution? Please, help.

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  • $\begingroup$ Your expression for the PDF of $T$ is wrong, actually $T$ has no PDF since $T_1+T_2+T_3=1$. To check that $T$ and $Z$ are independent using PDFs, one needs to use random variables with PDFs, for example, it suffices to show that $(Z,T_1,T_2)$ is independent by computing its joint PDF, using the Jacobian change of variable formula. Any trouble doing this? $\endgroup$ – Did Jun 25 '17 at 21:23
  • $\begingroup$ Can you explain why $T_1+T_2+T_3=1$ means that there is no PDF? $\endgroup$ – Aidas Jun 26 '17 at 6:29
  • $\begingroup$ One has $P(T\in H)=1$ where the hyperplane $H=\{(t_1,t_2,t_3)\in\mathbb R^3\mid t_1+t_2+t_3=1\}$ has Lebesgue measure zero, hence no PDF can exist. $\endgroup$ – Did Jun 26 '17 at 8:21

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