Date: 05/24/97 at 11:44:56 From: Brett Subject: Probability (Conditional Prob. and Independence) I have a hard time with theoretical questions and am stuck on a particular problem. As a simplified model for weather forecasting, suppose that the weather (either wet or dry) tomorrow will be the same as the weather today with probability p. If the weather is dry on Jan. 1, show that Pn, the probability that it will be dry n days later, satisfies: Pn = (2p - 1)Pn-1 + (1 - p) n >= 1 P0 = 1 Prove that Pn = 1/2 + 1/2(2p - 1)^n n >= 0
Date: 05/25/97 at 14:43:20 From: Doctor Anthony Subject: Re: Probability (Conditional Prob. and Independence) These problems are best done with probability matrices. The columns must add to 1. Also a probability vector: |a| |b| means a = probability that a day is dry, b = probability that it is wet. When you multiply this vector by the matrix shown below, you get the probabilities for the following day. FROM DRY WET -------- DRY | p 1-p ||Pn | |p*Pn + (1-p)(1-Pn)| TO | || | = | | WET | 1-p p ||1-Pn| |(1-p)Pn + p(1-Pn) | The top entry in the resulting vector is the probability that the (n+1)th day is dry, i.e. it equals P(n+1) So P(n+1) = p*Pn + 1 - p - Pn + p*Pn = (2p-1)Pn + (1-p) This is the expression we were asked to prove, though I have given the expression for P(n+1) from Pn, rather than Pn from P(n-1). If P0 = 1, then P1 = (2p-1)*1 + 1-p = 2p-1 + 1-p = p = 1/2 + (1/2)(2p-1)^1 So this fits the pattern for the nth term. We now assume the expression is true for n = k and see what happens when we apply the iterative formula to find P(k+1): P(k) = 1/2 + 1/2(2p-1)^k and P(k+1) = (2p-1)Pk + (1-p) P(k+1) = (2p-1)[1/2 + 1/2(2p-1)^k] + (1-p) = p-1/2 + (1/2)(2p-1)^(k+1) + 1-p = 1/2 + (1/2)(2p-1)^(k+1) This is the same expression as before, except that k+1 has replaced k. Thus if the expression is true for n = k it is also true for n = k+1. But it is true for n = 1, so it will be true for n = 2, then for n = 3 and so on to all positive integral values of n. By induction the expression is true generally. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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