# $X_i \sim \text{Uniform}(0, \theta)$ iid; $Y = \max{(X_1,..,X_n)}$. Why is $\theta$ necessarily larger than $y$?

I'm going through Statistical Inference by Casella & Berger, and on page 419, in the intro section of interval estimation there is the following example (note: most of the text was left out as it's irrelevant to my question):

Let $X_1,...,X_n$ be a random sample from a $\text{Uniform}(0, \theta)$ population, and let $Y = \max{\{X_1,...,X_n\}}$, ... Note that $\theta$ is necessarily larger than $y$.

My question is why is $\theta$ necessarily larger than $y$.

• What is your understanding of a "uniform(0, theta) population"?
– whuber
Aug 17 '18 at 15:20
• In a continuous distribution, such as $\text{Uniform}(0,\theta)$ what is $P(X)=c$ for some specific real value $c$? And for that distribution what is $P(X)>\theta$? Aug 17 '18 at 15:29
• @Alexis My current thinking is because X is distributed as a continuous uniform, then $P(X=x)=0$, so $P(X_{(n)}=\theta)=0$. Perhaps I'm still somewhat confused when thinking about zero probability events and realized values of a continuous random variable.
– Zian
Aug 17 '18 at 16:05
• The relevant probability is $\Pr(Y \gt \theta).$ Can you compute that?
– whuber
Aug 17 '18 at 16:31
• I think it is is better to say '$Y$ is smaller than or equal to $\theta$' instead of '$Y$ is smaller than $\theta$' or equivalent '$\theta$ is bigger than $Y$'. Do you get why each of the $X_i$ should be smaller than or equal to $\theta$? Aug 17 '18 at 18:09

Your thinking is correct. Since the probability if $\theta$ being bigger than $Y$ is equal to 1:
$$\begin{split} P(Y < \theta) &= P(X_1 < \theta \ \cap \ ... \ \cap \ X_n < \theta)\\ &= P(X_1 < \theta)\ ...\ P(X_n < \theta)\\ &= [F(\theta) - P(X_1 = \theta)] \ ... \ [F(\theta) - P(X_n = \theta)] \\ &= [F(\theta) - 0] \ ... \ [F(\theta) - 0]\\ &= 1 \cdot 1 \cdot \ ... \ \cdot 1\\ &= 1 \end{split}$$
Where $F(x)$ is the CDF of the $\text{Uniform}(0, \theta)$ population and we are assuming the $X_1 ... X_n$ are independent.
$F(x)= \begin{cases} 0 & \text{for }x < 0 \\ \frac{x}{\theta} & \text{for }0 \le x \le \theta \\ 1 & \text{for }x > \theta \end{cases}$