Suppose that X is a geometric random variable with parameter (probability of success) p. Show that Pr(X>a+b∣X>a)=Pr(X>b).

madeeha1d8 2022-09-19 Answered
Geometric Distribution Probability Problem
Suppose that X is a geometric random variable with parameter (probability of success) p.
Show that Pr ( X > a + b X > a ) = Pr ( X > b ).
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Answers (2)

ruinsraidy4
Answered 2022-09-20 Author has 11 answers
Step 1
There are two ways to get Pr ( X > n ), the easy way and the hard way. Since there is something to be learned from both, we do it both ways.
The hard way: The probability that X = n + 1 is the probability of n failures in a row, then success. This is ( 1 p ) n p
Similarly, the probability that X = n + 2 is the probability of n + 1 failures in a row, then success. This has probability ( 1 p ) n + 1 p.
Continue, and add up. There is a ( 1 p ) n p "in" each term. Take it out as a common factor. So we get ( 1 p ) n p ( 1 + ( 1 p ) + ( 1 p ) 2 + ) .
The geometric series inside the brackets has sum 1 1 ( 1 p ) = 1 p
Step 2
The easy way: We have X > n precisely if we get n failures in a row, probability ( 1 p ) n .
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koraby2bc
Answered 2022-09-21 Author has 2 answers
Step 1
Here's how to handle an infinite geometric series when the index starts at n instead of 0:
k = n a x k = a x n + a x n + 1 + a x n + 2 + a x n + 3 + = ( a x n ) + ( a x n ) x + ( a x n ) x 2 + ( a x n ) x 3 + = b + b x + b x 2 + b x 3 +
Now it starts at index 0. And of course b is a x n .
It looks as if you can do the rest.
Second method: You mentioned that the probability of "success" is p. That means the probability of success on each trial is p.
If X is defined as the number of trials needed to get one success, then the event X > n is the same as the event of failure on all of the first n trials, so that probability of that is ( 1 p ) n .
If X is defined as the number of trials needed to get one failure, then the event X > n is the same as the event of success on all of the first n trials, so the probability of that is p n .
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