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Math Help - Geometric distribution

  1. #1
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    Geometric distribution

    Hi, I am studying discrete random variables and have been given the question:

    Suppose that X is a geometrically distributed random variable with parameter p (0,1) i.e for all n=1,2..... P(X=n)=(1-p)^(n-1)p

    Show that for k=0,1,......
    P(X>k)=(1-p)^k

    In the previous part of the question I proved that E(X)=1/p, will the formula
    E(X)=sum(P(X>n)) where n 0,1,2..... help me?

    Thanks for your help
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  2. #2
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    Re: Geometric distribution

    E(X)=sum(P(X>n)) where n 0,1,2..... help me?
    I haven't seen that before and its not true in general. if its true for this distribution (and i haven't checked) then it might help but its probably just as easy to note the following:


    P(X>k) = P(X=k+1) + P(X=k+2) ....
    if you write out the terms you will see a geometric progression. There is a standard formula for summing those.
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  3. #3
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    Re: Geometric distribution

    Quote Originally Posted by SpringFan25 View Post
    I haven't seen that before and its not true in general. if its true for this distribution (and i haven't checked) then it might help but its probably just as easy to note the following:


    P(X>k) = P(X=k+1) + P(X=k+2) ....
    if you write out the terms you will see a geometric progression. There is a standard formula for summing those.
    The "..." makes it a bit more difficult than if we look at 1 - P(X \leq k) = P(X > k). This way we have a finite sum of probabilites up to k for each k.
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  4. #4
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    Re: Geometric distribution

    i guess its a matter of style/backround whether you prefer to sum a finite or infinite geometric series.

    Personally i would consider the infinite case simpler, as there is no risk of counting the wrong number of terms (which i have an irrational fear of doing). In addition I would point out that the formula for the finite geometric sum is just the difference between 2 infinite sums anyway.
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  5. #5
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    Re: Geometric distribution

    Quote Originally Posted by SpringFan25 View Post
    i guess its a matter of style/backround whether you prefer to sum a finite or infinite geometric series.

    Personally i would consider the infinite case simpler, as there is no risk of counting the wrong number of terms (which i have an irrational fear of doing). In addition I would point out that the formula for the finite geometric sum is just the difference between 2 infinite sums anyway.

    thanks for your help guys, I used the infinite case
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