What is the reasoning behind the string of equality $P(x_{(n)} \le t) = P(X_i \le t, \ \text{for each $i$}) = \{\Phi(t - \theta)\}^n$?

Question

I am currently studying the textbook In All Likelihood by Yudi Pawitan. Example 2.4 of chapter 2.2 Examples says the following:

Example 2.4: Suppose $x$ is a sample from $N(\theta, 1)$; the likelihood of $\theta$ is $$L(\theta) = \phi(x - \theta) \equiv \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} (x - \theta)^2}.$$ The dashed curve in Figure 2.3(d) is the likelihood based on observing $x = 2.45$.
Suppose it is known only that $0.9 < x < 4$; then the likelihood of $\theta$ is $$L(\theta) = P(0.9 < X < 4) = \Phi(4 - \theta) - \Phi(0.9 - \theta),$$ where $\Phi(x)$ is the standard normal distribution function. The likelihood is shown in solid line in Figure 2.3(d).
Suppose $x_1, \dots, x_n$ are an identically and independently distributed (iid) sample from $N(\theta, 1)$, and only the maximum $x_{(n)}$ is reported, while the others are missing. The distribution function of $x_{(n)}$ is $$\begin{align} F(t) &= P(x_{(n)} \le t) \\ &= P(X_i \le t, \ \text{for each $i$}) \\ &= \{\Phi(t - \theta)\}^n. \end{align}$$ So, the likelihood based on observing $x_{(n)}$ is $$L(\theta) = p_\theta (x_{(n)}) = n\{ \Phi(x_{(n)} - \theta)\}^{n - 1} \phi(x_{(n)} - \theta).$$

What is the reasoning behind the string of equality $P(x_{(n)} \le t) = P(X_i \le t, \ \text{for each $i$}) = \{\Phi(t - \theta)\}^n$?

Answer 1

Preliminary note: There is inconsistency in the use of lower/upper case notation for random variables in the question. I have chosen to correct this in the answer so my capitalisation is different to the question.

The first equation comes from the fact that $X_{(n)} = \max \{ X_1,...,X_n \}$, which gives the logical equivalence:

$$X_{(n)} \leqslant t \quad \quad \iff \quad \quad \begin{matrix} X_1 \leqslant t, \\ X_2 \leqslant t, \\ \vdots \\ X_n \leqslant t. \\ \end{matrix}$$

In simple terms, this is saying that if the maximum of $X_1,...,X_n$ is no greater than $t$ then there cannot be any value in $X_1,...,X_n$ that is greater than $t$, and likewise, if there is no value in $X_1,...,X_n$ that is greater than $t$ then the maximum of $X_1,...,X_n$ is no greater than $t$. Once you impose this logical equivalence for the underlying events, you then have:

$$\mathbb{P}(X_{(n)} \leqslant t) = \mathbb{P}(X_1 \leqslant t,X_2 \leqslant t,...,X_n \leqslant t).$$

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The second equation comes from the fact that the random variables $X_1,...,X_n$ were stipulated to be independent and identically distributed (IID) normal random variables with mean $\theta$ and unit variance, which means that:

$$\begin{align} \mathbb{P}(X_1 \leqslant t,X_2 \leqslant t,...,X_n \leqslant t) &= \prod_{i=1}^n \mathbb{P}(X_i \leqslant t) \\[6pt] &= \prod_{i=1}^n \mathbb{P}(X_i-\theta \leqslant t-\theta) \\[6pt] &= \prod_{i=1}^n \Phi(t-\theta) \\[12pt] &= [\Phi(t-\theta)]^n. \\[6pt] \end{align}$$

(The penultimate step here comes from the fact that $X_1-\theta,...,X_n-\theta \sim \text{IID N}(0,1)$ and $\Phi$ is the notation used for the CDF of the standard normal distribution.)

Answer 2

The more general relation is

$$ \mathbb P[X_{i;n}\leq t]=\sum_{k=i}^n{{n}\choose {k}}F^k(t)[1-F(t)]^{n-k}$$

which has been derived in this CV post.

Breaking it down for $i=n,$\begin{align}\mathbb P[X_{n;n}\leq t]&=\mathbb P[X_i\leq t; ~\forall i=1,2,\ldots,n]\\&=\mathbb P[(X_1\leq t) \cap~(X_2\leq t) \cap\cdots\cap (X_n\leq t) ]\\&=\mathbb P[X_1\leq t] \times \mathbb P[X_2\leq t]\times\cdots\times\mathbb P[X_n\leq t]\\&=[F(t)]^n.\end{align}