just a manipulation problem for statistics

Question

I am a little bit confused when i was asked to

Suppose $X_1, X_2,\ldots,X_n$ is a simple random sample from a continuous distribution with density function $f(x)$. Consider the new random variable $T$, defined as:

$$T=\max(X_1, X_2,\ldots,X_n)$$

Find an expression for $g(t)$, the density function of the random variable $T$.

What does this max actually mean? It seems to me it is similar to an MLE problem where I times all the likelihood. Am I right?

because the answer is $P(T\le t)=\prod_i P(X_i\le t)$.

Answer 1

Let $F_X$ be the CDF of the $X_i$ (which are i.i.d), and let $$ T_n = \max \{X_1,\ldots,X_n\} \text{.} $$ In plain english, what happens here is that you draw $n$ samples from some population, and then pick that largest one. That's your sample $T_n$.

Obviously $P(T_1 \leq x) = P(X_1 \leq x) = F_X(x)$. And since the $X_i$ are all independent, $$ P(T_2 \leq x) = P(X_1 \leq x, X_2 \leq x) = P(X_1 \leq x)P(X_2 \leq x) = \left(F_X(x)\right)^2 \text{.} $$ In the general case you get $$ F_{T_n}(x) = P(T_n \leq x) = P(X_1,\ldots,X_n \leq x) = \prod_{k=1}^n P(X_i \leq x) = \left(F_X(x)\right)^n \text{.} $$ and from that you find that density by differentiating, i.e. $$ f_{T_n}(x) = \frac{d}{dx} F_{T_n}(x) = n\left(F_X(x)\right)^{n-1}F_X'(x) = n\left(F_X(x)\right)^{n-1}f_X(x) \text{.} $$

If we throw caution to the wind for a second, and treat densities as if they were probabilities, we can easily interpret that result. Let's look at the factors of $f_{T_n}(x)$, i.e. of the "probability" that $T_n = x$, individually.

1. $f_X(x)$. This is simply the "probability" that some $X_i=x$ (for a fixed $i$)
2. $\left(F_X(x)\right)^{n-1}$. This is the probability we get a value less than $x$ exactly (n-1) times.
3. $n$. This accounts for the fact that (1) assumed $i$ is fixed. Since we're taking the maximum of $n$ values, there are $n$ possible choices as to which of the values is the maximum.