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Math Help - Anyone Understand Random Variables?

  1. #1
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    Anyone Understand Random Variables?

    For example, I am trying to solve the following problem:

    If X1...Xn are iid random variables, then what values for "a" and "b" solve the following equation:

    E[(X1+...+Xn)] = a*E[X1^2] + b*(E[X1])^2

    While we're at it; my car insurance starts at 1000 dollars and reduces by 10% every year that I don't get into an accident. If there is only a 0.05 probability of me getting into an accident any year , does anyone know what the PMF for my total payments up to the year that I get into an accident will be ?

    If anyone can help me, then I owe you one.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by anyt View Post
    For example, I am trying to solve the following problem:

    If X1...Xn are iid random variables, then what values for "a" and "b" solve the following equation:

    E[(X1+...+Xn)] = a*E[X1^2] + b*(E[X1])^2
    Well, gigen that the Xi's are iid RV's:

    E[(X1+...+Xn)] = n E[X1]

    There is no unique solution for a and b such that:

    n E[X1] = a*E[X1^2] + b*(E[X1])^2

    but one that works is a=0, b=n/E[X1].

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by anyt View Post

    While we're at it; my car insurance starts at 1000 dollars and reduces by 10% every year that I don't get into an accident. If there is only a 0.05 probability of me getting into an accident any year , does anyone know what the PMF for my total payments up to the year that I get into an accident will be ?
    If the accident occurs in the n-th year n \ge 1, your payments will have been

    c(n)=1000 + 1000*0.9 + .. + 1000*0.9^{n-1}=1000~\frac{1-0.9^n}{1-0.9}

    and this occurs with probability:

     <br />
p(c(n))=(1-0.05)^{n-1}\times 0.05<br />

    (that is the probability of no accidents in years 1 to n-1, times the probability of an accident in year n) and all other payments have zero probability.

    RonL
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    Thanks

    Thanks for the help
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