I have to find the distribution of the r.v. $-\sum_i^{10} log(x_i)$ and present it in the form of $\chi^2$ distribution. Given that: $-log(x_i) \sim Exp(\theta) $.
Then I know that the sum of exponantially distributed r.v is a Gamma distribution $\Gamma(n,\theta) = \Gamma(10,\theta)$. I also know that $$\Gamma(n,2) <=> \chi^2(2n).$$
I don't know how to make the above transformation for the case when in Gamma distribution we have $\theta$ and not $2$.
PS I have seen these to eqalities in literature, used alternately: $\sum Exp(\theta) <=> \Gamma(n,\theta)$ and $\sum Exp(\theta) <=> \Gamma(n,\frac{1}{\theta})$. Which one is correct?
Let $X_i \sim Exp(\theta)$,for $i=1,..,10$, be i.i.d. random variables, then $\sum_{i=1}^{10} X_i \sim Gamma(10, \theta) $, now multiplying it by $2\theta$ you arrive at $$ 2\theta \sum_{i=1}^{10} X_i \sim Gamma(10, 1/2) \equiv\chi^2_{(20)} $$