My question can be thought as a direct continuation of this important question and its first answer (but see this question also), which gives us the joint distribution of the max and min of an iid random sample. In my question, I just wanted to make sure if this formula for the ratio of maximum to minimum of an iid random sample is correct, given that first answer to the abovementioned question, which is indeed correct.

Proposition: Let $\{N_1 \dots N_n\}$ be iid random sample generated by a positive random variable $N > 0$ with CDF $F_N$ and PDF $f_N.$ Let $U_n, L_n$ denote the maximum and minimum of the random sample. Let us denote the PDF of their joint distribution by $f_{U_n, L_n}$ and that of the ratio by $f_{\frac{U_n}{L_n}}. $ Then we have the following expression of the PDF of the ratio:

$$ f_{\frac{U_n}{L_n}}(z) = n(n-1)\int_{s=0}^{\infty}sf_N(s)f_N(sz)(F_N(sz)-F_N(s))^{n-2}, z \in [1, \infty) $$

Note that: $\frac{U_n}{L_n} \ge 1.$ So let $ z \ge 1.$ Then: \begin{align*} \\& f_{\frac{U_n}{L_n}}(z) \\& = \frac{d}{dz} P[\frac{U_n}{L_n} \le z] \\&= \frac{d}{dz} P[{U_n} \le z L_n] \\&= \frac{d}{dz} \int_{ \{ t \le z s\} } f_{L_n, U_n}(s,t) ds dt \\&= \frac{d}{dz} \int_{s=0}^{\infty} \left[ \int_{t=0}^{zs} f_{L_n, U_n}(s,t) ds \right] dt \\&= \int_{s=0}^{\infty} \frac{d}{dz} \left[ \int_{t=0}^{zs} f_{L_n, U_n}(s,t) ds \right] dt \\&= \int_{s=0}^{\infty} s f_{L_n, U_n}(s,sz) ds \\&= n(n-1) \int_{s=0}^{\infty} s f_N(s)f_N(sz)(F_N(sz)-F_N(s))^{n-2} ds \\& \text{ (The next step is only for Uniform distributions) } \\&= n(n-1) \int_{s=0}^{1/z} s f_N(s)f_N(sz)(F_N(sz)-F_N(s))^{n-2} ds (\text{ because when } sz > 1, f_N(sz)=0. ) \end{align*}

Note that in the second to last line of the above, proof, I indeed used the general (not for uniform distribution) expression for $f_{L_n, U_n}$ provided by the first answer to the question I mentioned in the very first line. In the last line, we specialized for uniform distributions.

EDIT: After a few first comments, I'm including a check for $n=2, N \sim \mathcal{U}(0,1)$ below:

Plug in $n=2, $ and assume $N \sim U(0,1),$ then the last line of my calculation becomes $2 \int_{0}^{1/z}sds= 1/z^2,$ which is the PDF of the inverse uniform distribution. Now note that, in this question, they treat the distribution of $L_2/U_2$ and obtain it as a uniform distribution $\mathcal{U}(0,1),$ implying that at least in this simple case, my calculations are correct.

Indeed a similar calculation also shows that when $X_i \sim_{i.i.d.} \mathcal{U}(0,1), U_n/L_n$ has PDF $f_{U_n/L_n}(z)= \frac{n-1}{z^n}, z \geq 1.$

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    $\begingroup$ Does you general result matches the $n=2$ particular case treated here with an especialy simple expression? $\endgroup$
    – Jean Marie
    Sep 4, 2020 at 15:27
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    $\begingroup$ @JeanMarie I'll check - but in the link you cited, the author is asking for the reciprocal of what I'm asking: (s)he's asking for the distribution of $\frac{L_2}{U_2},$ not that of $\frac{U_2}{L_2},$ which is what I'm asking for. $\endgroup$ Sep 4, 2020 at 15:32
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    $\begingroup$ @LearningMath easy enough to adapt the same argument to the inverse of the fraction, would it match? $\endgroup$
    – gt6989b
    Sep 4, 2020 at 15:39
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    $\begingroup$ @gt6989b It seems to match. If I plug in $n=2, $ and assume $N \sim U(0,1),$ then the last line of my calculation becomes $2 \int_{0}^{1/z}sds= 1/z^2,$ which is the PDF of the inverse uniform distribution - en.wikipedia.org/wiki/…. $\endgroup$ Sep 4, 2020 at 15:57
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    $\begingroup$ Distribution of $U_2/L_2$ for $U(0,1)$ is in fact derived here. $\endgroup$ Sep 4, 2020 at 16:14


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