I found a question given by my teacher which went like this:
Suppose $X_0,X_1,...,X_n$ are i.i.d. random variable following $\chi_1^2$ distribution. Then by proving necessary result, find the distribution of $Z=\sum_{i=1}^n c_i \sqrt{X_i}$ where $c=\left(c_1,c_2,...c_n\right)'$ is a vector of known constants.
So here's what I tried till now:
Let $X\sim\chi^2_n$. Then $$f_X\left(x\right)=\begin{cases} \frac{x^{n/2-1}e^{-x/2}}{2^{n/2}\Gamma\left(\frac{n}{2}\right)},& x\geq0\\ 0,&\text{otherwise} \end{cases}$$ Then, to find the distribution of $Y'=\sqrt{X}$, I used the transformation technique and proceeded as follows: $$f_{Y'}\left(y'\right)=f_X\left(y'^2\right)\left|\frac{dy'^2}{dy'}\right|=\frac{\left(y'^2\right)^{n/2-1}e^{-y'/2}}{2^{n/2}\Gamma\left(\frac{n}{2}\right)}2y'=\frac{y'^{n-1}e^{-b^2/2}}{2^{n/2-1}\Gamma\left(\frac{n}{2}\right)} ,y'\geq0$$ As, $X_1,X_2,...,X_n\sim\chi^2_1$, I replaced the $n$ in the above pdf by $1$ and thus got the pdf of $\sqrt{X_1},\sqrt{X_2},...,\sqrt{X_n}$ as $$f_{X_i}\left(x_i\right)=\frac{\sqrt{2}e^{-x_i^2/2}}{\sqrt{\pi}},x_i\geq0$$ This seems like the pdf of a Half Normal Distribution.
This is all I did and I could go no further. Can anyone suggest me some hints as to how to go ahead in this problem? Also, did I do any mistake till now, like approaching the problem in this manner cause I don't know how to proceed any further. Alternative solution suggestions are very much welcome.
Hint:
Let's denote $Y_i = \sqrt{X_i}$. As $Y_i$ are i.i.d and follows the half-normal distribution, perhaps you could find the distribution of $Z$ by using the characteristic function
\begin{align} E(e^{itZ}) &= E\left(e^{it\sum_{i=1}^n c_i\sqrt{X_i}}\right) = \prod_{i=1}^n E\left(e^{itc_i Y_i}\right) =\prod_{i=1}^n \phi_{Y_i}(c_it) \end{align}
Using Mathematica, the characteristic function of half normal distribution is $$\phi_{Y_i}(t) = e^{-\frac{\pi t^2}{4}}\left(1+\mathbf{i} \text{ Erfi}\left( \frac{\pi t}{2} \right) \right)$$ where $\text{Erfi}(t)$ is the imaginary error function.