Probability That the Maximum Falls in an Interval
Suppose $X_1, \dots, X_n$ are i.i.d. $\text{Uniform}(0, a)$ random variables, and define $M_n = \max(X_1, \dots, X_n)$.
For a subinterval $(b, c] \subseteq (0, a)$, derive a formula for $P(b < M_n \leq c)$.
Then compute the answer for $n = 4$, $a = 5$, $b = 2$, $c = 4$.
Hints
- The key property of the maximum: $\{M_n \leq x\}$ happens exactly when every observation is at most $x$. Use independence to factor the joint probability.
- For uniform random variables on $(0, a)$, the CDF is simply $F(x) = x/a$. So $P(M_n \leq x) = (x/a)^n$.
- An interval probability is a difference of CDF values: $P(b < M_n \leq c) = F_{M_n}(c) - F_{M_n}(b)$. Plug in the specific numbers.
Worked Solution
How to Think About It: The maximum of i.i.d. random variables is the simplest order statistic to work with, because the event $\{M_n \leq x\}$ factors cleanly: the max is at most $x$ if and only if every single observation is at most $x$. Since the variables are independent, this turns a joint probability into a product of marginals. That is the entire trick. Once you have the CDF of the max, any interval probability is just a difference of CDF values.
Quick Estimate: With $n = 4$ uniforms on $(0, 5)$, the maximum tends to be close to 5. Its mean is $\frac{n}{n+1} \cdot a = \frac{4}{5} \cdot 5 = 4$. So $P(2 < M_4 \leq 4)$ should be substantial but not huge -- the max is often near 4 but also often above 4. Rough guess: somewhere around 0.3 to 0.5.
Approach: Use the independence factorization for the CDF of the maximum, then take a difference.
Formal Solution:
The CDF of the maximum is:
$P(M_n \leq x) = P(X_1 \leq x, \dots, X_n \leq x) = \prod_{i=1}^n P(X_i \leq x) = \left(\frac{x}{a}\right)^n$
for $0 \leq x \leq a$, since each $X_i \sim \text{Uniform}(0, a)$ has CDF $F(x) = x/a$.
The interval probability follows by subtraction:
$P(b < M_n \leq c) = P(M_n \leq c) - P(M_n \leq b) = \left(\frac{c}{a}\right)^n - \left(\frac{b}{a}\right)^n$
Plugging in $n = 4$, $a = 5$, $b = 2$, $c = 4$:
$P(2 < M_4 \leq 4) = \left(\frac{4}{5}\right)^4 - \left(\frac{2}{5}\right)^4 = \frac{256}{625} - \frac{16}{625} = \frac{240}{625} = \frac{48}{125}$
Answer: $P(b < M_n \leq c) = \left(\dfrac{c}{a}\right)^n - \left(\dfrac{b}{a}\right)^n$. For the given values, the answer is $\dfrac{48}{125} = 0.384$.
Intuition
The maximum of i.i.d. random variables is the gateway to order statistics, and the core insight is always the same: convert a statement about the max into a statement about all the individual variables. The event "the maximum is at most $x
This pattern shows up constantly in quantitative finance. The price of a best-of option (payoff depends on the max of several assets) uses exactly this logic. The distribution of the maximum drawdown, the worst-case loss in a portfolio, and extreme value theory all build on the CDF of order statistics. Whenever you see "max" or "min" of independent quantities, your first instinct should be to convert to individual CDFs and multiply.