Find the sampling distribution of the MLE of the uniform distribution

Question

The MLE is $ \theta = max [x1,...,xn] $ And $ P(max [Xi] < t) = P(Xi < t)^n = P(t/\theta) $

But the question asks me to show that $ P(max[Xi]< t) = (min[\theta, t]/ \theta)^n * I[t>0] $

Where I is the indicator function. How do I show this please?

Answer 1

Suppose $X_1, X_2, \ldots, X_n \ \sim \ U(0, \theta)$. The CDF is $$F_{X_i}(t)=\begin{cases} 0 \quad & \text{if} \quad t \leq 0 \\ \frac{t}{\theta} \quad & \text{if} \quad 0 < t < \theta \\ 1 \quad & \text{if} \quad t \geq \theta\end{cases}$$

Denote $X_{(n)} = max[Xi]$ be the largest order statistic, then \begin{align*} P(max[Xi]<t) & = P(X_{(n)}<t) \\ &=P(X_1<t)P(X_2<t)\cdots P(X_n<t) \\ & = F_{X_1}(t)F_{X_2}(t)\cdots F_{X_n}(t) \\&= \begin{cases} 0 \quad & \text{if} \quad t \leq 0 \\ \frac{t^n}{\theta^n} \quad & \text{if} \quad 0 < t < \theta \\ 1 \quad & \text{if} \quad t \geq \theta\end{cases} \\ &=\left(min[\theta, t]/\theta \right)^n \cdot I[t>0] \end{align*}