Suppose that X_1,…,X_n are independent, geometric_1(p) random variables. Evaluate P(min(X_1,…,X_n)>l) and P(max(X_1,…,X_n) le l).

limfne2c 2022-10-19 Answered
Evaluating min/max probability with Geometric random variables
Suppose that X 1 , , X n are independent, geometric 1 ( p ) random variables. Evaluate P ( min ( X 1 , , X n ) > l ) and P ( max ( X 1 , , X n ) l )
The textbook solution is as follows:
P ( min ( X 1 , , X n ) > l ) = P ( k = 1 n { X k > l } )   = k = 1 n P ( X k > l )   = k = 1 n p l   = p l n
and P ( max ( X 1 , , X n ) l ) = P ( k = 1 n { X k l } )   = k = 1 n P ( X k l )   = k = 1 n ( 1 p ) l   = ( 1 p l ) n
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 ( min ( X 1 , , X n > l ) = P ( k = 1 n { X k > l } ) and the corresponding equation for max)?
2) The PMF for Geometric 1 ( p ) is given as p X ( k ) = ( 1 p ) p k 1 . How is P ( X k > l ) = p l ? The same applies for the max case as well.
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Answers (1)

encaselatqr
Answered 2022-10-20 Author has 11 answers
Step 1
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 , , 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 k = 1 n { X k > l }. Thus P ( min ( X 1 , , X n ) > l ) = P ( k = 1 n { X k > l } ) ..
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.
Step 2
2) Note that if X has PMF ( 1 p ) p k 1 for k 1, then for any integer l 1, we have
P ( X l ) = k = 1 l P ( X = k ) = k = 1 l ( 1 p ) p k 1 = ( 1 p ) × 1 p l 1 p ( using the  geometric series formula ) = 1 p l .
(And so P ( X > l ) = 1 P ( X l ) = p l .)
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