Question 9v) of specimen paper

Discussion in 'CS2' started by Ross Bennett, Sep 9, 2019.

  1. Ross Bennett

    Ross Bennett Member

    I was just doing question 9 part 5 of the specimen paper, and comparing my method to the solutions. My second integral ranged from 5 to infinity, given the first integral already covered 0 to 5, and the term for the probability of surviving until time 5 is already there in the second term. Why is this not correct?
     
  2. Andrew Martin

    Andrew Martin ActEd Tutor Staff Member

    Hi Ross

    It is possible to write the integral with either the range 0 to infinity or 5 to infinity, depending on how you define the variable of integration. I'll show you what I meant. Let's start with the definition of the complete expectation of life, as the integral of tpx, and split it out into the first 5 years and the subsequent 5 years:

    \[ e^o_0 = \int_0^{\infty} {}_tp_0 \,dt = \int_0^5 {}_tp_0 \,dt + \int_5^{\infty} {}_tp_0 \,dt\]
    The expression inside the first integral is just exp(-mu*t). The expression inside the second integral can be rewritten as follows, using the principle of consistency (for when \( t > 5 \) ).

    \[ {}_tp_0 = {}_5p_0 * {}_{t-5}p_5 \]
    Thus we have:

    \[ e^o_0 = \int_0^{\infty} {}_tp_0 \,dt = \int_0^5 e^{-\mu * t} \,dt + \int_5^{\infty} {}_5p_0 * {}_{t-5}p_5 \,dt\]

    Now, substituting in expressions with exponentials for the probabilities in the second integral we get:

    \[ e^o_0 = \int_0^{\infty} {}_tp_0 \,dt = \int_0^5 e^{-\mu * t} \,dt + \int_5^{\infty} e^{-\mu * 5} * e^{-\lambda * (t-5)} \,dt\]
    Being careful to include the (t-5) in the exponent for the second probability, as this is the duration, not t.

    You could evaluate the integral from there and you should get the same answers as the specimen paper solutions.

    However, we can represent the second integral slightly differently by doing a substitution (in the second integral only):

    \[ s = t - 5 \]
    this gives:

    \[ e^o_0 = \int_0^{\infty} {}_tp_0 \,dt = \int_0^5 e^{-\mu * t} \,dt + \int_0^{\infty} e^{-\mu * 5} * e^{-\lambda * (s)} \,ds\]
    This is the form in the specimen paper and is equally valid.

    Hope this helps

    Andy
     

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