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In a related question, it is noted that two functions of independent random variables are themselves independent.

Does this result extend to three or more functions of independent random variables? Also, are the functions necessarily mutually independent?

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In the case of three or more variables you can use the same lines of reasoning in that thread you linked to.

To address your other question, if I understand it correctly, you can define transformations of independent random variables that use both "old" random variables to get each of the "new" random variabes, and still end up with independent things.

Consider the following pretty well known example. Assume $X_1,X_2 \sim \text{Gamma}(\alpha, \beta)$. That means the original density is $$ f_{X_1,X_2}(x_1,x_2) = \frac{1}{\Gamma(\alpha)^2 \beta^{2\alpha}}x_1^{\alpha-1}x_2^{\alpha-1}\exp\left[-\frac{x_1+x_2}{\beta}\right]. $$ Then define $Y_1 = X_1/(X_1 + X_2)$ and $Y_2 = X_1 + X_2$. Then the new joint density is $$ f_{Y_1,Y_2}(y_1,y_2) = \left[\frac{\Gamma(2\alpha)}{\Gamma(\alpha)^2 }y_1^{\alpha-1}(1-y_1)^{\alpha-1} \right]\left[\frac{1}{\beta^{2\alpha}\Gamma(2\alpha)}y_2^{2\alpha-1}\exp\left[-\frac{y_2}{\beta}\right] \right]. $$ $Y_1 \sim \text{Beta}(\alpha,\alpha)$, $Y_2 \sim \text{Gamma}(2\alpha,\beta)$, and they are independent.

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  • $\begingroup$ So in other words, we can avoid thinking about mutual independence for $n$ random variables by making a function of $n-1$ random variables and realizing that it must be independent of a function of the other random variable (provided that the $n$ random variables are mutually independent to begin with)? $\endgroup$ – Taliant Mar 5 '18 at 5:59
  • $\begingroup$ Then the two resulting random variables are independent, yes. That falls under the lines of reasoning of that other post. $\endgroup$ – Taylor Mar 5 '18 at 6:07

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