23. Let Y denote a geometric random variable with probability of success p. (a) Show that for a positive integer a, P(Y > a) = (1 − p)a (b) Showthatforapositiveintegeraandb,P(Y >a+b|Y >a)=(1−p)a =P(Y >b). (c) Why do you think the property in (b) is called the memoryless property of the geometric distribution? (d) In the development of the distribution of the geometric random variable, we assume
23. Let Y denote a geometric random variable with probability of success p. (a) Show that for a positive integer a, P(Y > a) = (1 − p)a (b) Showthatforapositiveintegeraandb,P(Y >a+b|Y >a)=(1−p)a =P(Y >b). (c) Why do you think the property in (b) is called the memoryless property of the geometric distribution? (d) In the development of the distribution of the geometric random variable, we assume
MATLAB: An Introduction with Applications
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Author:Amos Gilat
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23. Let Y denote a geometric random variable with probability of success p.
(a) Show that for a positive integer a, P(Y > a) = (1 − p)a
(b) Showthatforapositiveintegeraandb,P(Y >a+b|Y >a)=(1−p)a =P(Y >b).
(c) Why do you think the property in (b) is called the memoryless property of the geometric distribution?
(d) 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 is the result in part (b) “obvious”?
(e) Show that P (Y = an odd integer) = p 1−q2
24. Given that the random variable W is binomial distribution with n trials and success probability p in each trial and P (W = w) = h(w), show that
(a) h(w) =p(n−w+1),w>0 h(w − 1) (1 − p)w
(b) E [W(W − 1)]= n(n − 1)p2. ? 1 ? 1−(1−p)n+1
(c)E W+1 = (n+1)p
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