Let $X_1, X_2, . . . , X_n$ be a random sample from a uniform distribution on $[0, \theta]$. Suppose results $x_1, x_2, . . . , x_n$ are observed. Since $f(x) = 1/\theta$ for $0 \leq x \leq \theta$, the likelihood function is:
$L(\theta) = 1/\theta^n$ $0 \leq x_1 \leq \theta$, $. . .$ , $0 \leq x_n \leq \theta$ and zero otherwise
As long as $\theta \geq \max(x_i)$, the likelihood is $1/\theta^n$, which is positive, but as soon as $\theta < \max(x_i)$, the likelihood drops to zero.
QUESTION: Let $Y = \max(X_i)$. Derive the CDF of $Y$ and then show that the PDF of $Y$ is $fy(y) = ny^{n-1}/\theta^n$ $0\leq y\leq \theta$ and zero otherwise
My answer: CDF of $Y = (Y/\theta)^n$. Is this correct?
If $$Y = X_{(n)} = \max_i X_i,$$ then $$\Pr[Y \le y] = \Pr\left[\max_i X_i \le y\right] = \Pr\left[\bigcap_{i=1}^n (X_i \le y) \right] \overset{\text{ind}}{=} \prod_{i=1}^n \Pr[X_i \le y].$$ This is because the largest observation is at most $y$ if and only if each observation is at most $y$. Since $$\Pr[X_i \le y] = F_X(y) = \frac{y}{\theta}$$ for $0 \le y \le 1$, we obtain $$F_Y(y) = \begin{cases} 0, & y \le 0 \\ \left(\frac{y}{\theta}\right)^n, & 0 < y \le \theta \\ 1, & \theta < y. \end{cases}$$ Differentiating, we obtain the density $$f_Y(y) = \begin{cases} \frac{ny^{n-1}}{\theta^n}, & 0 < y \le \theta, \\ 0, & \text{otherwise},\end{cases}$$ as claimed.