Distribution of max of ratio of a sequence of uniform variables

Question

Consider a sequence of uniform random variables $X_i,i=1,2,\cdots,n$ in the interval $(0,1)$.
It can be said that the density of ratio of two iid uniform variables in this interval is \begin{equation} f(z)= \begin{cases} \frac{1}{2},& 0<z<1 ;\\ \frac{1}{2z^2},z \geq 1 \end{cases} \end{equation} We know that the distribution of max of $n$ iid random variables is given by $(F(x))^n.$But here as the CDF is a piecewise function ,I am not able to figure out how it can be done.I would be highly obliged for any help/hints/suggestions

Answer 1

The function you call $F(z)$ is actually the density, not a cumulative distribution function. Usually we denote a probability density with a lowercase $f$ so as to distinguish it from the cumulative distribution function, for which we use $F$, and is how you have used it when writing the CDF of the maximum order statistic $(F(x))^n$.

Since the density is piecewise, we integrate to compute the CDF:

$$F_Z(z) = \Pr[Z \le z] = \int_{t=0}^z f_Z(t) \, dt. \tag{1}$$ If $z \le 1$, then this is just $$F_Z(z) = \int_{t=0}^z \frac{1}{2} \, dt = \frac{z}{2}, \quad 0 \le z \le 1. \tag{2}$$ If $z > 1$, then $$F_Z(z) = \int_{t=0}^1 \frac{1}{2} \, dt + \int_{t=1}^z \frac{1}{2t^2} \, dt = \frac{1}{2} + \frac{z-1}{2z} = 1 - \frac{1}{2z}, \quad z > 1. \tag{3}$$ So the CDF is the piecewise function $$F_Z(z) = \begin{cases} 0, & z < 0 \\ \frac{z}{2}, & 0 \le z \le 1 \\ 1 - \frac{1}{2z}, & z > 1. \end{cases} \tag{4}$$

The maximum order statistic $Z_{(n)}$ of an iid sample $(Z_1, Z_2, \ldots, Z_n)$ where each $Z_i$ is the ratio of two uniform random variables on $(0,1)$, therefore has the CDF $$\begin{align} F_{Z_{(n)}}(z) = \Pr[Z_{(n)} \le z] &= \Pr[\max(Z_1, Z_2, \ldots, Z_n) \le z] \\ &= \Pr\left[\bigcap_{i=1}^n (Z_i \le z) \right] \\ &\stackrel{\text{ind}}{=} \prod_{i=1}^n \Pr[Z_i \le z] \\ &\stackrel{\text{id}}{=} (\Pr[Z \le z])^n \\ &= (F_Z(z))^n \\ &= \begin{cases} 0, & z < 0 \\ (\frac{z}{2})^n, & 0 \le z \le 1 \\ (1 - \frac{1}{2z})^n, & z > 1. \end{cases} \tag{5} \end{align}$$

The density of $Z_{(n)}$ is simply the derivative: $$f_{Z_{(n)}}(z) = \frac{d}{dz}\left[F_{Z_{(n)}}(z)\right] = \begin{cases} \frac{nz^{n-1}}{2^n}, & 0 \le z \le 1 \\ \frac{n}{2z^2} (1 - \frac{1}{2z})^{n-1}, & z > 1. \end{cases} \tag{6}$$