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Math Help - Probability Mass Function that goes to infinity

  1. #1
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    Exclamation Probability Mass Function that goes to infinity

    In a game the bank will pay nothing with probability 0.8 = f(0), and will pay $n with p.m.f. (probability mass function) f(n) = af(n-1)/n for n=1,2,3... (infinity)


    1- find the constant a.
    2- Find the probability, that the bank pays $1.
    3- Find the probability the bank pays $1 or more.
    4- Find the probability the expected payment of the bank.
    5- Find the variance of the bank.
    6- If it costs a quarter to play, should you play?

    I'm so lost and not sure what kind of infinite sum this is. Any help will be greatly appreciated!!!
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  2. #2
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    Quote Originally Posted by Intsecxtanx View Post
    In a game the bank will pay nothing with probability 0.8 = f(0), and will pay $n with p.m.f. (probability mass function) f(n) = af(n-1)/n for n=1,2,3... (infinity)


    1- find the constant a.
    2- Find the probability, that the bank pays $1.
    3- Find the probability the bank pays $1 or more.
    4- Find the probability the expected payment of the bank.
    5- Find the variance of the bank.
    6- If it costs a quarter to play, should you play?

    I'm so lost and not sure what kind of infinite sum this is. Any help will be greatly appreciated!!!
    I will do 1. The rest are left for you to do (or at least to show some working and then give a clear statement of where you're still stuck).

    f(0) = 0.8

    f(1) = \frac{a f(0)}{1} = \frac{0.8 a}{1}

    f(2) = \frac{a f(1)}{2} = \frac{0.8 a^2}{1\times 2} = \frac{0.8 a^2}{2!}

    f(3) = \frac{a f(2)}{2} = \frac{0.8 a^3}{3 \times 2!} = \frac{0.8 a^3}{3!}.

    The pattern is clear: f(n) = \frac{0.8 a^n}{n!}.

    Therefore you require 1 = f(0) + f(1) + f(2) + .... + f(n) + .... = 0.8 \sum_{i=0}^{\infty}\frac{a^n}{n!} = 0.8 e^a.

    Solve for a.
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  3. #3
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    Thumbs up Thank you for your response, you beautiful mathematician!

    Haha, this was fantastic. It helped so much. I worked out the whole problem now and found E[X] to be the same as the constant a= ln(5/4). However, I can't seem to figure out how to find Var[X] because I'm not sure how to square X? Thank you for helping me out by the way. I really appreciate it.
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  4. #4
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    Quote Originally Posted by Intsecxtanx View Post
    Haha, this was fantastic. It helped so much. I worked out the whole problem now and found E[X] to be the same as the constant a= ln(5/4). However, I can't seem to figure out how to find Var[X] because I'm not sure how to square X? Thank you for helping me out by the way. I really appreciate it.
    n = 0 \Rightarrow n^2 = 0.

    n = 1 \Rightarrow n^2 = 1.

    n = 2 \Rightarrow n^2 = 4.

    n = 3 \Rightarrow n^2 = 9.

    etc.
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