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Math Help - Conditional probability

  1. #1
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    Conditional probability

    Let X1,..be independent random variables with the common distribution function F, and suppose they are independent of N, a geometric random variable with parameter p. Let M=max(X1,...Xn).

    Find P(M\leq x) by conditioning on N

    At the first glance I didn't quite understand the information given in this problem. But after a while I wrote down what I believe is asked to solve in the problem

     P(M\leq x)=\sum_{k=0}^{\infty} P(M\leq x|N=k)P(k)

    But this is as far as I can go, I know P(k) = p(1-p)^(k-1), and I'm clueless on what's  P(M\leq x|N=k)
    I want to say that  M\leq x is independ of  N=k , but I'm not sure whether it's true.

    Helps would be appreciated, thanks!
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  2. #2
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    Quote Originally Posted by ChaosticMoon View Post
    Let X1,..be independent random variables with the common distribution function F, and suppose they are independent of N, a geometric random variable with parameter p. Let M=max(X1,...Xn).
    It seems like it should be M=\max(X_1,\ldots,X_N) where N is the given random variable.

    Find P(M\leq x) by conditioning on N

    At the first glance I didn't quite understand the information given in this problem. But after a while I wrote down what I believe is asked to solve in the problem

     P(M\leq x)=\sum_{k=0}^{\infty} P(M\leq x|N=k)P(k)

    But this is as far as I can go, I know P(k) = p(1-p)^(k-1), and I'm clueless on what's  P(M\leq x|N=k)
    I want to say that  M\leq x is independ of  N=k , but I'm not sure whether it's true.
    Since X_1,X_2,\ldots are independent of N, conditioning by N=k just gives P(\max(X_1,\ldots,X_N)\leq x|N=k)=P(\max(X_1,\ldots,X_k)\leq x) (the conditioning doesn't affect the X_i's by independence; it just fixes the value of N).
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