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(Survival Model) How did it arrive at the expression

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  • (Survival Model) How did it arrive at the expression

    Hello. I really need some help. I am reading "Survival Models and Their Estimations" 2nd edition by Dick London and my question concerns with the chapter 4 of this book. It was staightforward that I cannot understand how did the author arrive at the formula. Please help me...

    let t* = t + 0.5 represent the midpoint of the interval (t, t + 1]. It is assumed that T is exponentially distributed over (t, t+1].

    It was mentioned that the estimator L(t*) can be written as

    L(t*) = - ln (N_(t+1) / n_t), which is a biased estimator (why? I don't get it)

    It says: the random variable L(t*) is a natural log function of the binomial random variable N_(t+1), so the variance of L(t*), conditional on n_t, can be approximated by the method of statistical differentials using this formula:

    Var{g(X)} = [g'(m)]^2 Var(X), where E[X] = m.

    and so, Var[L(t*) / n_t] = q_t / (p_t n_t). (how did that arrive as that?)

  • #2
    Unfortunately, Mr. London is currently out of the country. We will be happy to contact him on your behalf upon his return during the last week of August.
    Sandi Lynn Scherer
    ACTEX Publications

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    • #3
      The student seems to have two questions. The first question asks why the estimator given by Equation (4.26) is a biased estimator. Although the estimator for pt, being a binomial proportion, is unbiased, the estimator for l(t*) is biased. Recall the definition of unbiased is that the expected value of the estimator equals the quantity being estimated. (E.g., E[p^t] = pt.) But

      E[l^(t*)] = E[- ln p^t] ¹ - ln E[p^t] = - ln pt = l(t*).

      In words, the expected value of the log of an estimator does not equal the log of the expected value of the estimator, so the estimator l^(t*) is biased.

      The second question regards the use of Equation (2.75) to approximate the variance of the estimator. In this application, the random variable X is represented by Nt+1; its mean is nt×pt and its variance is nt×pt×qt since it is binomial. The function g(X) is

      g(X) = g(Nt+1) = - ln (Nt+1/nt),

      so

      g¢(X) = - 1/Nt+1

      and

      g¢(m) = g¢(X)½X=m = - 1/nt×pt .

      Then

      Var[g(X)] = [g¢(m)]2×Var(X) = (- 1/nt×pt)2(nt×pt×qt) = qt/nt×pt .
      Sandi Lynn Scherer
      ACTEX Publications

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      • #4
        Thank you for the explanation.

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