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Math Help - Finding the probability

  1. #1
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    Finding the probability

    Let P(X=i)=P(Y=i)=\frac{1}{2^i}
    Where (i=1,2,...).
    X and Y are independent.

    Then how can I find P(min(X,Y)\le i).

    I know that when P(max(X,Y)\le i) means that both X and Y are less than i so I can conclude that P(X\le i,Y\le i)=P(X\le i)P(Y\le i). But how can I do the same for Min?

    Meanwhile, since this is a discrete distribution, I am also facing the problem of finding its mass probability function. e.g. P(X\le i)=\sum_{X_n<i} p(X_n) to solve for some other parts. Any suggestion?

    TIA
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  2. #2
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    Quote Originally Posted by noob mathematician View Post
    Let P(X=i)=P(Y=i)=\frac{1}{2^i}
    Where (i=1,2,...).
    X and Y are independent.

    Then how can I find P(min(X,Y)\le i).

    I know that when P(max(X,Y)\le i) means that both X and Y are less than i so I can conclude that P(X\le i,Y\le i)=P(X\le i)P(Y\le i). But how can I do the same for Min?

    Meanwhile, since this is a discrete distribution, I am also facing the problem of finding its mass probability function. e.g. P(X\le i)=\sum_{X_n<i} p(X_n) to solve for some other parts. Any suggestion?

    TIA
    Read this: http://www.pstat.ucsb.edu/faculty/ca...jointorder.pdf
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  3. #3
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    Thanks for the link.

    The equation I will use is: P(X_{k}\le x)=\sum_{j=0}^{n-k}\dbinom{n}{j}\pi_3^j(1-\pi_3)^{n-j}

    Is it true that for this case, I'm finding:

    \sum_{j=0}^{2-1}\dbinom{2}{j}\pi_3^j(1-\pi_3)^{2-j}

    How do I find \pi_3=P(X>x)?
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  4. #4
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    Quote Originally Posted by noob mathematician View Post
    Thanks for the link.

    The equation I will use is: P(X_{k}\le x)=\sum_{j=0}^{n-k}\dbinom{n}{j}\pi_3^j(1-\pi_3)^{n-j}

    Is it true that for this case, I'm finding:

    \sum_{j=0}^{2-1}\dbinom{2}{j}\pi_3^j(1-\pi_3)^{2-j}

    How do I find \pi_3=P(X>x)?
    I would have thought that n = 2 and k = 2 ...

    Also, \displaystyle \Pr(X > i) = \frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + .... = \frac{1}{2^{i}} \left( \frac{1}{2} + \frac{1}{2^2} + .... \right) and you should know what the sum in the brackets is equal to.
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  5. #5
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    Thanks.

    Would like to extend my question.

    What does it means by P(\text{X divides Y}) in this case?

    Is it possible to generate its probability?
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