Probability of sum of random variables: pivotal quantity I'm trying to find a confidence interval for a parameter $\theta$, yet I'm not quite sure I'm following the right steps.
The probability function is such that:
$$f(x|\theta) = \frac{1}{\theta} x^{-\left(\frac{1}{\theta}+1\right)}$$
I have shown that for a sample of size $n$, $ \sum_{i=1}^{n} \ln(X_i)$ is a sufficient statistic. I am trying to prove that $Q(\theta,x) = \frac{1}{\theta} \sum_{i=1}^{n} \ln(X_i)$ is a pivotal quantity. 
To do so, I have to prove that its density is independent of $\theta$. So I did: 
$$ P\left(\frac{1}{\theta} \sum_{i=1}^{n} \ln(X_i) \leq a \right) = \sum_{i=1}^{n} P\left( \ln(X_i) \leq \theta a\right) = \sum_{i=1}^{n} P\left(X_i \leq e^{a\theta}\right) $$ 
Evaluating the integral, it's not hard to show that the density for $Q$ does not depend on $\theta$. Problem is, I considered that: $$ P\left(\frac{1}{\theta} \sum_{i=1}^{n} \ln(X_i) \leq a \right) = \sum_{i=1}^{n} P\left( \ln(X_i) \leq \theta a\right)$$ because it's a sample and then all $X_i$ must be iid. Can I do this step? It seems wrong to me, because I feel I'd have to use a convolution. But the solution interestingly is independent of $\theta$. Can I get any tips?
Thanks! 
 A: Since you do not mention the support of the distribution, I am assuming $X$ has a Pareto density with shape parameter $1/\theta$, i.e. the pdf is
$$f(x\mid\theta)=\frac{1}{\theta x^{1/\theta+1}}\mathbf1_{x>1}\quad,\,\theta>0$$
The distribution function of $X$ is
\begin{align}
F(x\mid\theta)&=\int_1^x f(t\mid\theta)\,dt
\\&=1-x^{-1/\theta}\qquad,\quad x>1
\end{align}
Now we know that
\begin{align}
&\qquad 1-X^{-1/\theta}\sim U(0,1)
\\\\&\implies X^{-1/\theta} \sim U(0,1)
\end{align}
So for a random sample $(X_1,X_2,\ldots,X_n)$ drawn from the above distribution, we have
\begin{align}
X_i^{-1/\theta}\stackrel{\text{i.i.d}}\sim U(0,1)&\implies -\ln\left(X_i^{-1/\theta}\right)\stackrel{\text{i.i.d}}\sim \text{Exp with mean }1
\\&\implies -\sum_{i=1}^n\ln\left(X_i^{-1/\theta}\right)\sim \text{Gamma}(1,n)
\end{align}
You can of course directly derive the distribution of $-\ln \left(X^{-1/\theta}\right)$ starting from the distribution of $X$. Hence $T=\frac{1}{\theta}\sum\limits_{i=1}^n\ln X_i$ has a $\text{Gamma}(1,n)$ distribution with density
$$f_T(t)=\frac{e^{-t}t^{n-1}}{\Gamma(n)}\mathbf1_{t>0}$$
