Stochastic

If we suppose that all the n and n listed above are

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Unformatted text preview: omes smoggy. It stays smoggy for an exponentially distributed number of days with mean 4, then rain comes. The rain lasts for an exponentially distributed number of days with mean 1, then sunshine returns. Remembering that for an exponential the rate is 1 over the mean, the verbal description translates into the following Q-matrix 1 1/ 3 0 1 1 2 3 2 1/3 1/ 4 0 3 0 1/4 1 The relation ⇡ Q = 0 leads to three equations: 1 3 ⇡1 1 3 ⇡1 1 4 ⇡2 1 4 ⇡2 +⇡3 ⇡3 =0 =0 =0 Adding the three equations gives 0=0 so we delete the third equation and add ⇡1 + ⇡2 + ⇡3 = 1 to get an equation that can be written in matrix form as 0 1 1/3 1/3 1 ⇡1 ⇡2 ⇡3 A = 0 0 1 1/4 1A where A = @ 0 1 0 1 This is similar to our recipe in discrete time. To find the stationary distribution of a k state chain, form A by taking the first k 1 columns of Q, add a column of 1’s and then ⇡1 ⇡ 2 ⇡ 3 = 0 0 1 A 1 i.e., the last row of A 1 . In this case we have ⇡ (1) = 3/8, ⇡ (2) = 4/8, ⇡ (3) = 1/8 To check our...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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