Suppose $X_1, X_2,\ldots, X_N$ are mutually independent and identically distributed. Let $F_X$ denote the cdf of $X_1$.
Would this statement be true or false? I'm totally lost as to how to approach this... Any help would be appreciated. :)
Let $A$ denote the event that the largest $X_i$ is smaller than or equal to $t_1$, and the smallest $X_i$ is bigger than $t_0$. Then $\mathrm{Pr}[A] = [F_X(t_1) − F_X(t_0)]^N .$
On
I'll use the notation $X_{(1)}$ for the smallest $X_i$ and $X_{(N)}$ for the largest $X_i$.
Suppose $X_1 \leq t_0$. The smallest $X_i$, $X_{(1)}$, then must be either $X_1$ or some even smaller value. In any case, $X_{(1)} \leq t_0$ and the event $A$ did not occur. So $A$ can occur only when $X_1 > t_0$.
Suppose $X_1 > t_1$. We can apply an argument to the largest $X_i$, $X_{(N)}$, following the pattern of the argument about the smallest $X_i$, and conclude that $X_{(N)} > t_1$ and that $A$ did not occur. So $A$ can occur only when $X_1 \leq t_1$.
So if $A$ then $t_0 < X_1 \leq t_1$. What is the probability of that event?
Now for the same reasons that $t_0 < X_1 \leq t_1$, if $A$ then we also have $t_0 < X_2 \leq t_1$, $t_0 < X_3 \leq t_1$, and so forth, up to $t_0 < X_N \leq t_1$. Knowing that $X_1, X_2, X_3, \ldots$ are mutually independent, what is the probability that all of these inequalities are satisfied? If all these inequalities are satisfied, is $A$ satisfied? What can we conclude then about $\mathrm{Pr}[A]$?
$$\begin{align*} \Pr[A] &= \Pr\left[\left(t_0 < \min_i X_i\right) \cap \left(\max_i X_i \le t_1\right)\right] \\ &= \Pr[t_0 < X_{(1)} \le X_{(N)} \le t_1] \\ &= \Pr[t_0 < X_1, X_2, \ldots, X_N \le t_1] \\ &= \Pr\left[\bigcap_{i=1}^N t_0 < X_i \le t_1 \right] \\ &\overset{\rm ind}{=} \prod_{i=1}^N \Pr[t_0 < X_i \le t_1] \\ &\overset{\rm i.d.}{=} \Pr[t_0 < X_1 \le t_1]^N \\ &= \left( \Pr[X_1 \le t_1] - \Pr[X_1 \le t_0] \right)^N \\ &= \left(F_X(t_1) - F_X(t_0)\right)^N. \end{align*}$$