# How do you derive the probability distribution for $I$?

I'm reading a book. On pages 105 and 106 it is said that:

For N-look intensity SAR images,
$$I_N=\frac{1}{N}\sum_{i=1}^N I_1(i)=\frac{1}{N}\sum_{i=1}^N (x(i)^2+y(i)^2)$$

Where $x(i)$ and $y(i)$ are the real of the $i$th look (or sample). Since $x(i)$ and $y(i)$ are independently Gaussian distributed, it is well known that $NI_N$ has a Chi-square distribution with $2N$ degrees of freedom. Therefore, the PDF of the N-look intensity is described by $$p_N(I)=\frac{N^N I^{N-1}}{(N-1)!\sigma^{2N}}\exp(-\frac{NI}{\sigma^2}),\quad I\ge 0$$
The mean and the variance are $M_N(I)=\sigma^2$ and $\text{Var}_N(I)=\frac{\sigma^4}{N}$, respectively.

I don't have any problem with $NI_N$ having a chi-square distribution with $2N$ degrees of freedom but as we know the PDF for a chi-square distribution with $f$ degrees of freedom is:
$$h(x) = \begin{cases} \frac{1}{2^{\frac{f}{2}}\Gamma(\frac{f}{2})}x^{\frac{(f-2)}{2}}e^{-\frac{x}{2}}\quad &, x\ge 0\\[2ex] 0\quad &, x\lt 0 \end{cases}$$
so for the distribution of $NI$ we should have:
$$p(NI)=\frac{1}{2^N\Gamma(N)}(NI)^{N-1}\exp(-\frac{NI}{2})=\frac{N^{N-1}I^{N-1}}{2^N(N-1)!}\exp(-\frac{NI}{2})$$
Then if we consider $Y=NI$ as a function of $I$, we have:
$$p(Y)\,dY=p(I)\,dI\;,\;dY=N\,dI\\\frac{N^{N-1}I^{N-1}}{2^N(N-1)!}\exp(-\frac{NI}{2})NdI=p(I)dI\\p(I)=\frac{N^NI^{N-1}}{2^N(N-1)!}\exp(-\frac{NI}{2})$$
Then where does that $\sigma^2$ come from instead of $2$?
As far as I know degree of freedom ($f$) is the only parameter of chi-square distribution. and the mean and the variance of this distribution are $f$ and $2f$ respectively as wiki page says.

I do not know the context, but I would assume that gaussian distribution in the book are not scaled, so their variance is not equal to $1$. Therefore you have another parameter $\sigma^2$.
So what you get is not exactly chi-squared distribution, but chi-squared scaled by a constant. Think of it as Gamma distribution with shape parameter $N$ and scale parameter $\sigma^2/N$.