Show that for $0The distribution function for a power family distribution is given by $$F(y)=\begin{cases} 0, & y<0\\ \left(\frac{y}{\theta}\right)^\alpha, &0\le y \le \theta \\ 1, ,&y>\theta,\end{cases}$$ where $\alpha$, $\theta$ > 0. Assume that a sample of size $n$ is taken from a population with a power family distribution and that $\alpha = c$ where $c > 0$ is known.
Show that for $0<k<1$ $$P(k < \frac{Y_{(n)}}{\theta} \le 1) = 1 - k^{cn}.$$
I am told that $$P(k\theta < Y_{(n)} < \theta) = F_{Y_{(n)}}(\theta) - F_{Y_{(n)}} (k\theta).$$
I don't see why. Could someone help me understand it by providing the intermediate steps?
 A: You are asked $\Pr\left[k<\dfrac{Y_{(n)}}{\theta}\le 1\right]$. Now, take a look the part: $k<\dfrac{Y_{(n)}}{\theta}\le 1$. Multiply each side by $\theta$, you will obtain: $k\theta<Y_{(n)}\le\theta$. 
Let $Y_1,\cdots, Y_n$ be a random variable from a given power family distribution. Here, $Y_{(n)}$ is $n$-th order statistics. Therefore, $Y_{(n)}=\max[Y_1,\cdots, Y_n]$. Note that $Y_{(n)}\le y$ equivalence to $Y_i\le y$ for $i=1,2,\cdots,n$. Hence, for $0\le y\le\theta$, the fact that $Y_1,Y_2,\cdots, Y_n$ are i.i.d. implies
$$
F_{Y_{(n)}}(y)=\Pr[Y_{(n)}\le y]=\Pr[Y_1\le y,Y_2\le y,\cdots, Y_n\le y]=(\Pr[Y_i\le y])^n=\left(\frac{y}{\theta}\right)^{cn}.
$$
Thus,
$$
\begin{align}
\Pr\left[k<\dfrac{Y_{(n)}}{\theta}\le 1\right]&=\Pr[Y_{(n)}\le\theta]-\Pr[Y_{(n)}\le k\theta]\\
\Pr[k\theta<Y_{(n)}\le\theta]&=F_{Y_{(n)}}(\theta)-F_{Y_{(n)}}(k\theta)\\
&=\left(\frac{\theta}{\theta}\right)^{cn}-\left(\frac{k\theta}{\theta}\right)^{cn}\\
&=1-k^{cn}.
\end{align}
$$
$$\\$$

$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
