Question

At time t = 0, there is one individual alive in a certain population. A pure birth process then unfolds as follows. The time until the first birth is exponentially distributed with parameter 1. After the first birth, there are two individuals alive. The time until the first gives birth again is exponential with parameter 1, and similarly for the second individual. Therefore, the time until the next birth is the minimum of two exponential (a) variables, which is exponential with parameter 21. Similarly, once the second birth has occurred, there are three individuals alive, so the time until the next birth is an exponential rv with parameter 31, and so on (the memoryless property of the exponential distribution is being used here). Suppose the process is observed until the sixth birth has occurred and the successive birth times are 25.6, 42.9, 50.7, 56.5, 59.4, 62.9 (from which you should calculate the times between successive births). Derive the mle of 1. [Hint: the likelihood is a product of exponential terms.] (Round your answer to four decimal places.)

Answer 1

from the observed of f(x XiN PMF data -dæ de x, F LÀ : x - 3)... X, - ai La TT . + 6x6) of los 6! #(x1,1) 6! db el x, + 2 x2 + ... 6x6) Logl 6 log! 1(x, +242 +. - a Logh an da (x, t2ly tou+686) maximum dikelihood estimate of 't is co 6 17 6 Thus (X, 42X2 +--- +686)

The Successive birth times 25.6, 42.9, 50-7,56.5, 59.4,62.9 Total = 298 25.6 42-9 25.6 17-3 Xy = 50.7 7.8 42.9 < 3 5.8 56-5 -50-7 . Xu X5 X6 59. 4 -56.5 = 2.9 62-9 –59.4 = 3.5 6 6 role of a (x, +2*2 + ... + 6xb) 25-6+2(17-3) +3(78)+4 (58) + 5(2.9) +613.5) 6 142-3 0.042164441 -0.0421 * this I hope please answer will helpful for you give up - vole for my work