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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!

StubbornAtom
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YetAnotherUsr
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1 Answers1

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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

$$f_T(t)=\frac{e^{-t}t^{n-1}}{\Gamma(n)}\mathbf1_{t>0}$$

StubbornAtom
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  • Thank you so much! Indeed the support is $x>1$ so it is a Pareto distribution. – YetAnotherUsr Oct 18 '18 at 11:09
  • My point is, even if the $X_i$ are iid I cannot assume that $P\left(\sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} P(X_i) $, right? I say so because I had got to your result doing the step I did. But because $ \frac{1}{\theta}ln(X_i) $ is a $ \Gamma(1,1)$, not because this is a general result. – YetAnotherUsr Oct 18 '18 at 11:28
  • @M.Gonzalez Yes, that equation makes no sense to me. In this context, at least. – StubbornAtom Oct 18 '18 at 17:22
  • @M.Gonzalez If this addressed your queries sufficiently, you might consider accepting the answer. – StubbornAtom Nov 07 '18 at 20:04