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Weather Forecasting

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|
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. 


               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
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Associated Topics:
College Probability

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