A random sample, $X_1, X_2,... X_n$ is drawn from a Pareto population with pdf
$$f(x|\theta)=\frac{\theta}{x^2}I_{[\theta,\infty)}(x)$$
I've been trying to figure out the distribution of
$$T=\log\Big[\frac{\prod_{i=1}^n X_i}{X_{(1)}^n}\Big]$$
where $X_{(1)}$ is the first order statistic, or $X_{(1)}=\min_i X_i$. The problem is boiled down to finding the distribution of $W_i=\log\Big(\frac{X_i}{X_{(1)}}\Big)=\log(X_i)-\log(X_{(1)})=Y_i-Z$ for $i=2,...n$, since $T=\sum_{i=2}^nW_i$
I made some transformation and ended up with the pdf of $Y_i$ and $Z$ are given by
$$f_Z(z)=\frac{n\theta^n}{e^{zn}}$$
$$f_Y(y)=\theta e^{-y}$$
$Y_i$ and $Z$ are independent so the next thing is to find the distribution of $Y_i-Z$ by taking the integral of the product of those 2 pdf and the result should yield to $Y_i-Z$ following exponential(1). However, from what I've derived, I can't get the exponential(1) and I don't know where I went wrong in any calculations or reasoning, can you please help me have a look? Many thanks.
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diidoobiib
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1$\theta$ merely determines the units in which $x$ is measured, so set it to a convenient value like $1$ to do all your initial calculations. With this in mind, can you find the common distribution of the $\log X_i$? You will find it easy to work with. – whuber May 20 '19 at 22:47
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2See https://stats.stackexchange.com/q/345895/119261. – StubbornAtom May 20 '19 at 23:01