Let $X_{(1)}\geq X_{(2)}\geq\cdots X_{(K)}$ be the order statistics of the random variables $X_1,\,X_2,\,\ldots,X_K$, which are independent and identically distributed exponential random variables with parameter 1.

Now let $Z_k=\frac{Y_k}{X_{(k)}}$, where $Y_1,\,Y_2,\,\ldots,Y_K$ are also independent and identically distributed exponential random variables with parameter 1.

I wish to find the cumulative distribution function (CDF) of $\max\limits_{k}\{Z_k\}$, which is given by

$$\text{Pr}\left[\max\limits_{k}Z_k\leq z\right]$$

How can I find the above CDF?

The problem is that $\{Z_k\}_{k=1}^K$ are not independent, because the denominator is an order statistics. I tried to condition on the denominator and then average as

$$\begin{split} &\underbrace{\int\int\cdots\int}_{K-\text{Fold}}\text{Pr}\left[\frac{Y_1}{\alpha_1}\leq z, \frac{Y_2}{\alpha_2}\leq z,\cdots, \frac{Y_K}{\alpha_K}\leq z\right]f_{X_{(1)}, X_{(2)}, \ldots, X_{(K)}}(\alpha_1, \alpha_2, \ldots, \alpha_K)\,d\alpha_1\cdots d\alpha_K\\ =& \underbrace{\int\int\cdots\int}_{K-\text{Fold}} \prod_{k=1}^K \text{Pr}\left[\frac{Y_k}{\alpha_k}\leq z\right]f_{X_{(1)}, X_{(2)}, \ldots, X_{(K)}}(\alpha_1, \alpha_2, \ldots, \alpha_K)\,d\alpha_1\cdots d\alpha_K \end{split}$$

but got stuck in evaluating it.

Thanks in advance for any tips.


1 Answer 1


Define $\tilde{Z}_{k} = \frac{Y_{k}}{X_{k}}$ and $Z_{k} = \frac{Y_{k}}{X_{(k)}}$ with $Y_{k}$ and $X_{k}$ as in the question.

The random vectors $[\tilde{Z}_{1},\tilde{Z}_{2},...,\tilde{Z}_{K}]$ and $[Z_{1},Z_{2},...,Z_{K}]$ will have different distributions, however...

The distribution of the order statistics $\tilde{Z}_{(K)} = \max\limits_{k}\frac{Y_{k}}{X_{k}}$ and $Z_{(K)} = \max\limits_{k}\frac{Y_{k}}{X_{(k)}}$ will be equivalent since the ordering in the denominator is "undone" by introducing new random variables and taking the maximum.

So, to find the density of $Z_{(K)}$, it suffices to find the density of $\tilde{Z}_{(K)}$.

Distribution of $\tilde{Z}_{k}$

The CDF of the ratio of two exponential distributions with rate 1 is $F(z) = (1+z^{-1})^{-1}$, for $z>0$. See relevant question here.

Distribution of $\tilde{Z}_{(K)}$

$\tilde{Z}_{(K)}$ has CDF: $Pr(\tilde{Z}_{(K)} < z) =F(z)^{K}$ by rules for order statistics. So $Pr(\tilde{Z}_{(K)} < z) = (1+z^{-1})^{-K}$.

This is also the CDF of the maximum you are after.

Note: I'm using the standard definition of order statistics where $X_{(i)} \leq X_{(j)}$ for $i < j$.

  • $\begingroup$ Why the distribution of $\tilde{Z}_{(K)}$ and $Z_{(K)}$ are the same again? $\tilde{Z}_{(K)} = \max_k \tilde{Z}_{k}$, while $Z_{(K)} = \frac{Y_{i_K}}{X_{(K)}}$, where $i_K$ is the index of $X_{(K)}$. These are two different random variables, aren't they? I think your definition of $Z_k$ is not correct (I just realized that my definition isn't correct, too). It should be $Z_k=\frac{Y_{i_k}}{X_{(k)}}$. $\endgroup$
    – BlackMath
    Dec 6, 2018 at 7:17
  • $\begingroup$ I was defining $\tilde{Z}_{(K)} = \max_k \tilde{Z}_{k}$ and $Z_{(K)} = \max_k Z_{k}$ (to find what you asked for: $\text{Pr}\left[\max\limits_{k}Z_k\leq z\right]$). Right? $\endgroup$
    – BonStats
    Dec 6, 2018 at 11:17
  • $\begingroup$ $\tilde{Z}_{(K)}$ is correct, but $Z_{(K)}$ is not. $Z_{(K)}$ is defined as $$Z_{(K)} = \frac{Y_{i_K}}{X_{(K)}}$$ where $$i_K = \text{arg} \max_k X_k$$. $\endgroup$
    – BlackMath
    Dec 6, 2018 at 19:56
  • $\begingroup$ Maybe ask a new question then? Regardless of how you index the $Y_{i}$ they are still iid, hence I think that if $S = \frac{Y_{i_K}}{X_{(K)}}$ and $T = \frac{Y}{X_{(K)}}$, with $Y \sim \text{Exp}(1)$, $S$ and $T$ will have the same distribution. $\endgroup$
    – BonStats
    Dec 7, 2018 at 4:56
  • $\begingroup$ That's right, but $\frac{Y_{i_{K}}}{X_{(K)}}\neq \max_k \frac{Y_k}{X_{k}}$, does it? $\endgroup$
    – BlackMath
    Dec 7, 2018 at 6:15

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