Chapter 24 question

Discussion in 'CM1' started by Molly, Sep 3, 2023.

  1. Molly

    Molly Ton up Member

    Hi all,

    Was doing april 2022 question 10 and i think i might be a bit confused on chapter 24 logic.

    why does (aq)^surr_62 not equal ap_62*q^surr_x, and similarly i would have thought (aq)^d_x=(ap)_x*q^d_x but this also isnt the case - why is this?

    im slightly confused on the purpose of the independent probabilities, when would we use these other than to derive \mu and \sigma?

    Thank you
     
  2. Molly

    Molly Ton up Member

    also, when calculating profit signature, is there a difference between using p_x and (ap)_x?
     
  3. Molly

    Molly Ton up Member

    Sorry, am really confused by this question, and have another point to it so am going to restructure my answer here:

    1) in order to calculate (aq)^d_x, i dont understand why we couldnt find the force of decrement \mu by rearranging
    q^d_x=1-e^-\mu_x
    so that \mu_x=-ln(1-q_x^d)
    and then finding (aq)^d_x using:
    (aq)^d_x=\mu/(\sigma+\mu) *(1-e^-1(\mu+\sigma)).
    this gives the wrong values for (aq)^d_x but im unsure why this formula doesnt hold for this situation

    2) also why does (aq)^surr_62 not equal ap_62*q^surr_x, and similarly i would have thought (aq)^d_x=(ap)_x*q^d_x but this also isnt the case - why is this?

    3) when calculating profit signature, is there a difference between using p_x and (ap)_x

    sorry for all the parts to this question, i am confused with this chapter as it seems like the formulas sometimes hold and sometimes do not - i think i must be missing something?

    Thanks so much
     
  4. Michael Clarkson

    Michael Clarkson ActEd Tutor Staff Member

    Hi Molly,

    For 1)

    your approach would be correct if you had constant forces of transition applying throughout the year.
    However, you are told that surrenders only happen at the end of each year.

    Therefore, the dependent probability of death = independent probability of death (because surrenders only happen after the deaths have all occurred). This is reflected in the 2nd and 4th columns of the table.

    why does (aq)^surr_62 not equal ap_62*q^surr_x

    (aq)^surr_62 is the dependent probability of leaving due to surrender over the year.
    Due to the timing mentioned above this equals p_62 (probability of not dying throughout year) * q^surr_x (end of year surrender probability).

    ap_62 is the probability of staying in force until the end of the year = probability of not dying throughout the year or not surrendering at the end of the year.

    (aq)^d_x=(ap)_x*q^d_x

    (aq)^d_x is the dependent probability of leaving due to death through the year. Your formula starts with (ap)_x (1 minus dependent probability of dying - dependent probability of surrendering) then multiplies this by the independent probability of death. So you've applied a death probability twice.

    when calculating profit signature, is there a difference between using p_x and (ap)_x

    yes - p_x is the probability of survival (not dying) and (ap)_x is the probability of staying active (not dying / not surrendering).
    If it's possible to die or surrender then you will need to use (ap)_x when calculating the profit signature because you need the probability of staying active up until the start of each year.

    Thanks,
    Michael
     
  5. Molly

    Molly Ton up Member

    Hi Michael,

    Thank you very much for this, your reply has really cleared this up and i really really appreciate the detail!!

    Please could i confirm one more thing, that q^d_x=1-e^-\mu) is true only when constant transition intensities apply also?

    Thanks so much,
    Molly
     
  6. Michael Clarkson

    Michael Clarkson ActEd Tutor Staff Member

    correct
    it is based on the fact that q^d_x = 1-exp(-INT(mu))... the fact the transition intensities are constant allows you to integrate the constant and simplify the formula
     
    Molly likes this.
  7. Molly

    Molly Ton up Member

    Thats great, thank you :)
     

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