Solving for e y 2 using the fact e y e y 2 1p p2 1 p

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Unformatted text preview: e of each trial is modeled by a Bernoulli random variable with parameter p. Here p is assumed to satisfy 0 < p ≤ 1. Thus, the outcome of each trial is one with probability p and zero with probability 1 − p. Let Y denote the number of trials conducted until the outcome of a trial is one. Let us find the pmf of Y. Clearly the range of possible values of Y is the set of positive integers. Also, Y = 1 if and only if the outcome of the first trial is one. Thus, pY (1) = p. The event {Y = 2} happens if the outcome of the first trial is zero and the outcome of the second trial is one. By independence of the trials, it follows that pY (2) = (1 − p)p. Similarly, the event {Y = 3} happens if the outcomes of the first two trials are zero and the outcome of the third trial is one, so pY (3) = (1 − p)(1 − p)p = (1 − p)2 p. By the same reasoning, for any k ≥ 1, the event {Y = k } happens if the outcomes of the first k − 1 trials are zero and the outcome of the k th trial is one. Therefore, pY (k ) = (1 − p)k−1 p for k ≥ 1. 2.5. GEOMETRIC DISTRIBUTION 37 The random variable Y is said to have the g...
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