# Evaluating min/max probability with Geometric random variables

This is an exercise problem that I am solving from the book Probability and Random Processes for Electrical and Computer Engineers (Gubner, 2002) that I'm having trouble understanding. For anyone curious, the problem is Exercise 18 from Chapter 2.

Suppose that $$X_1,\dots,X_n$$ are independent, $$\text{geometric}_1(p)$$ random variables. Evaluate $$P(\text{min}(X_1,\dots,X_n) \gt l)$$ and $$P(\text{max}(X_1,\dots,X_n) \le l)$$.

The textbook solution is as follows:

\begin{align} P(\text{min}(X_1,\dots,X_n) \gt l) & = P\left( \bigcap_{k = 1}^n \{X_k \gt l\} \right) \\\ & = \prod_{k = 1}^n P(X_k \gt l) \\\ & = \prod_{k = 1}^n p^l \\\ & = p^{ln} \end{align}

and

\begin{align} P(\text{max}(X_1,\dots,X_n) \le l) & = P\left( \bigcap_{k = 1}^n \{X_k \le l\} \right) \\\ & = \prod_{k = 1}^n P(X_k \le l) \\\ & = \prod_{k = 1}^n (1 - p)^l \\\ & = (1 - p^l)^n \end{align}

I'm having trouble understanding pretty much the entire solution. If I were to summarize what's particularly puzzling me:

1. How was the first line of each solution derived (i.e. $$P(\text{min}(X_1, \dots, X_n \gt l) = P(\cap_{k = 1}^n \{X_k \gt l \})$$ and the corresponding equation for $$\text{max}$$)?

2. The PMF for $$\text{Geometric}_1(p)$$ is given as $$p_X(k) = (1 - p)p^{k - 1}$$. How is $$P(X_k \gt l) = p^l$$? The same applies for the $$\text{max}$$ case as well.

• If the smallest of $n$ numbers is greater than $l$ than each number must be greater than $l$—otherwise the smallest could not be! Applied to RVs and in set notation $\{X>l\}=\cap_{k=1}^n \{X_k>l\}$ where $X=\min\{X_1,\dotsc, X_n\}$. Is that clearer? – Nap D. Lover Sep 22 '19 at 2:28

1) This is because the minimum of a set of numbers is greater than $$l$$ if and only if all the numbers are greater than $$l$$. So $$\min(X_1,\ldots,X_n)>l$$ if and only if $$X_1 > l$$ and $$X_2 > l$$ and ... and $$X_n > l$$. Note that the event "$$X_1 > l$$ and $$X_2 > l$$ and ... and $$X_n > l$$" is written as $$\bigcap\limits_{k=1}^{n}\{ X_k > l\}$$. Thus $$P(\min(X_1,\ldots,X_n) > l) = P\left(\bigcap\limits_{k=1}^{n}\{ X_k > l\}\right).$$
The $$\max$$ case is similar reasoning; it's because the maximum of a set of numbers is less than or equal to $$l$$ if and only if all the numbers are less than or equal to $$l$$.
Note that what I said above holds true regardless of the distribution of the $$X_k$$'s.
2) Note that if $$X$$ has PMF $$(1-p)p^{k-1}$$ for $$k\ge 1$$, then for any integer $$l\ge 1$$, we have \begin{align*}P(X \le l) &= \sum\limits_{k=1}^{l} P(X=k) \\ &= \sum\limits_{k=1}^{l} (1-p)p^{k-1} \\ &= (1-p)\times \frac{1-p^{l}}{1-p} \quad(\text{using the }\textbf{geometric series formula})\\ &= 1-p^l.\end{align*}
(And so $$P(X > l) = 1-P(X\le l) = p^l$$.)