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hpp.lik(x, T1, rate)
hpp.mle
nhpp.lik
X = c(0.17, 0.39, 0.63, 0.78, 0.99) hpp.lik(X, T1 = 1, rate = 4) hpp.lik(X, T1 = 1, rate = 5) hpp.lik(X, T1 = 1, rate = 6) # 5 is the most likely of these three rates
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