4. Let Y denote a geometric random variable with probability of success p. a. Sh

Question

4. Let Y denote a geometric random variable with probability of success p. a. Show that for a positive integer a, P(Y > a) -qa. b. Show that for positive integers a and b, P(Y> a + blY> a) qb P(Y>b). This result implies that, for example, (Y>71Y> 2)= P(Y > 5). This is called the memoryless property of the geometric distribution. c. In the development of the distribution of the geometric random variable, we assumed that the experiment consisted of conducting identical and independent trials until the first success was observed. In light of these assumptions, why if the result in part (b) "obvious"

Explanation / Answer

Question 4

(a) P(Y > a) = qa

Here Y is a geometric random variable with probability of success p.

so, first we will analyse the question here. Here it is asked that what is the probability that first success will appear after a failures, that is P(Y > a)

so that means we get a failures in initial a attempt. As all events are independent.

Pr(a failurs in a event) = Pr(Intial a attempts fail) = (1-p)a = qa = P(Y > a)

(b) Here Pr(Y > a + b l Y > a)

Here verbally the question means that if in a attempt there is no success, what is the probability that there be no success in (a + b) attempt. As Y > a + b is a subset of Y > a

so, we can say that

Pr(Y l a + b l Y >a) = Pr(Y l a + b) / Pr(Y > a) = qa+b /qa = qb

(c) Here we can intrepret the problem in (b) i.e. that as all events are independent here. so as we know that there is no success in intitial a attempts, so now we have to find that there is no success in a + b attempts. so, that means we want to find the probability that there will be no success in next b attempts. NOw as all events are independet,

Pr(No success in any b number of attempts) = qb

SO hence proved that the result in part(b) is obvious.

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