Statistic of functions of multiple random variables If there are $n$ independent random variables $x_1,\, x_2,\cdots, x_n$, where $x_i$ follow the p.d.f. $f_i(x_i)$. Now if there is a function $y=f(x_1,\, x_2,\cdots, x_n)$, is there a simple way to find the p.d.f. of random variable $y$.
I know the standard way of doing this is to calculate $P(y\leq y_0)$ and then differentiates it. But given a huge number of $x_i$, this task seems intractable. There is some theorem where we transform $x_1,\, x_2,\cdots, x_n$ to $y_1,\, y_2,\cdots, y_n$ such as theorem 3.9.5 of DeGroot & Schervish's book. Is there any similar result when I am only interested in one function $y=f(x_1,\, x_2,\cdots, x_n)$.
I notice the it is important to make sure the transformation is one-to-one in the similar theorem, which may not be true in a general $y=f(x_1,\, x_2,\cdots, x_n)$. I am not sure whether a general theorem exists.
What about the more specific question below:
If $x_1,\, x_2,\cdots, x_n$ follow uniform distribution on $[0,1/n]$, $y=\sqrt{\sum_i a_i \, x_i^2}$, where $a_i$ is a real constant. Is that possible to find the p.d.f of $y$? What if we can also assume $n\rightarrow \infty$?
 A: Unfortunately there is no general method like this.
The method given in the book you've mentioned is likely the well-known  variable transormation method, however that requires $f$ to be bijective, and such, invertable. That's also why the dimensionality of the input and output spaces were equal.
The method is the following:
If $f:\mathbb{R}^n \to \mathbb{R}^n$, $f(x_1,\dots,x_n)=(y_1,\dots,y_n)$, with $x := (x_1,\dots,x_n)$ and $y := (y_1,\dots,y_n)$, and $x$ having joint p.d.f. $f_x$ and joint c.d.f. $F_x$, then $\forall i \in \{1,\dots,n\}$:
$$\frac{\partial}{\partial a_i}\mathbb{P}(y_1 \le a_1,\dots,y_n \le a_n) = \frac{\partial}{\partial a_i}\mathbb{P}(f(x) \le (a_1,\dots,a_n)) = \\ = \frac{\partial}{\partial a_i}\mathbb{P}(x \le f^{-1}(a_1,\dots,a_n)) = \frac{\partial}{\partial a_i} F_x(f^{-1}(a_1,\dots,a_n)) = \\ = f_x(f^{-1}(a_1,\dots,a_n))\cdot \left(\frac{\partial f^{-1}}{\partial a_i}\right)(a_1,\dots,a_n)$$
where the derivative of the inverse of $f$ is often called the Hessian matrix with elements
$$H_{i,j}(a_1,\dots,a_n) = \left(\left(\frac{\partial f^{-1}}{\partial a_i}\right)(a_1,\dots,a_n)\right)_j$$
So the p.d.f. of $y$ in this case is
$$f_y(a) = f_x(f^{-1}(a))[H_1(a) \cdots H_n(a)]$$
where $H_k(a)$ is the $k$'th row of the Hessian matrix evaluated at $a$.
