We have a random variable $x$ with p.d.f. $\sqrt{\dfrac{\theta}{\pi x}}\exp(-x\theta)$, $x>0$ and $\theta$ a positive parameter.
We are required to show that $2\theta x$ has a $\chi^2$ distribution with $1$ degree of freedom and deduce that, if $x_1,\dots,x_n$ are independent r.v. with this p.d.f., then $2\theta\sum_{i=1}^n x_i$ has a $\chi^2$ distribution with $n$ degrees of freedom.
Using transformation $y=2\theta x$ I found the pdf of $$y=\frac{1}{\sqrt{2\pi}}y^{-1/2}e^{-y/2}.$$ How do I find the distribution of $2\theta\sum_{i=1}^n x_i$? Do I need to find the likelihood function (which contains $\sum_{i=1}^n x_i$) first? How do I recognise the degrees of freedom of this distribution (Is it $n$ because it involves $x_1,\dots,x_n$, i.e. $n$ random variables?
The probability density function of a random variable $Z$ which follows the chi-square distribution with $n$ degrees of freedom is given by $$f_Z(z)=\frac{1}{2^{\frac{1}{2}}\Gamma{(\frac{n}{2})}}z^{\frac{n}{2}-1}e^{-\frac{z}{2}}$$ for $z>0$.
You already derived the $pdf$ of the transformed random variable $Y=2\theta X$. By matching that $pdf$ of $Y$ with the $pdf$ of $Z$, you can notice that $Y$ is actually following chi-square distribution with $d.f.$ n (where n=1).
Again $X_1,\ldots,X_n$ are independent random variable implies that $Y_1,\ldots,Y_n$ also independent random variable.
Note that, sum of independent chi- square distribution is also a chi-square distribution with $d.f.$ equals to sum of the corresponding degree of freedoms.
Using the above result you can say that the distribution of $\sum_{i=1}^n Y_i$ is chi-square distribution with $d.f.$ n.