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Math Help - Three proofs.

  1. #1
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    Three proofs.

    Hi! I've got a few problems that I wonder if someone could help me with.

    1) If Y is a discrete random variable that assigns positive probabilites to only the positive integers, show that.



    I don't have a clue about how to solve this one.

    2) Let Y denote a geometric random variable with probability of success p.
    a) Show that for a positive integer a,




    b) Show that for positive integers a and b



    I've been trying a lot with this one, but don't really know how to prove this.

    3) Find E[Y(Y-1)] for a geometric random variable Y by finding d^2/dq^2 for



    I don't know about this one either. My suggestion is that the second derivative is equal to this:



    , but how do I use that to find E[Y(Y-1)]?

    Would be nice if someone could help me with any of these problems.
    Last edited by approx; September 22nd 2008 at 08:22 AM.
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    Quote Originally Posted by approx View Post
    Hi! I've got a few problems that I wonder if someone could help me with.

    1) If Y is a discrete random variable that assigns positive probabilites to only the positive integers, show that.



    I don't have a clue about how to solve this one.
    [snip]
     E(Y)
    = \sum_{j=1}^\infty j \,P(Y=j)
    = \sum_{j=1}^\infty \sum_{k=1}^j P(Y=j)
    = \sum_{k=1}^\infty \sum_{j=k}^\infty P(Y=j)
    = \sum_{k=1}^\infty P(Y \geq k)
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  3. #3
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    Thanks for your help. But could you please motivate the steps? And how come you end up with the index starting with k=1? Is it just a generalised thing?
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    Quote Originally Posted by approx View Post
    Thanks for your help. But could you please motivate the steps? And how come you end up with the index starting with k=1? Is it just a generalised thing?
    The trick is simple, once you see it: express the sum as a double sum, then interchange the order of summation.

    To see how to get from
    \sum_{j=1}^\infty \sum_{k=1}^j
    to
    \sum_{k=1}^\infty \sum_{j=k}^\infty
    take a piece of graph paper, label the axes j and k, and blacken the (j,k) points involved in the first sum. Then think about summing things in the other direction -- first k and then j.
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  5. #5
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    Quote Originally Posted by approx View Post
    Hi! I've got a few problems that I wonder if someone could help me with.

    [snip]

    2) Let Y denote a geometric random variable with probability of success p.
    a) Show that for a positive integer a,



    [snip]
    You should know that \Pr(Y = n) = (1 - p)^{n-1} p = q^{n-1} (1 - q).

    Then:

    \Pr(Y > a) = 1 - \Pr(Y \leq a)  = 1 - \left[ ~ \Pr(Y = 1) + \Pr(Y = 2) + \Pr(Y = 3) + \, ... \, + \Pr(Y = a) ~ \right]

    = 1 - \left[ ~ (1 - q) + q ( 1- q)  + q^2 (1 - q) \, ... \, + q^{a-1} (1 - q) ~ \right]

    = 1 - (1 - q) \left[ 1 + q  + q^2 \, ... \, + q^{a-1} \right]

    Sum the stuff in the square brackets using the formula for the sum of a geometric series. Simplify.

    = 1 - [1 - q^a] = q^a.
    Last edited by mr fantastic; September 24th 2008 at 04:56 AM. Reason: Replaced X's with Y's
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  6. #6
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    Quote Originally Posted by approx View Post
    ]snip]

    2) Let Y denote a geometric random variable with probability of success p.

    [snip]

    b) Show that for positive integers a and b



    I've been trying a lot with this one, but don't really know how to prove this.

    [snip]
    \Pr( Y > a + b ~ | ~ Y > a) = \frac{\Pr(Y > a + b ~ \text{and} ~ Y > a)}{\Pr(Y > a)} = \frac{\Pr(Y > a + b)}{\Pr(Y > a)}

    Use the result from a):

     = \frac{q^{a+b}}{q^a} = q^b.
    Last edited by mr fantastic; September 24th 2008 at 04:57 AM.
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  7. #7
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    Quote Originally Posted by approx View Post
    [snip]
    3) Find E[Y(Y-1)] for a geometric random variable Y by finding d^2/dq^2 for



    I don't know about this one either. My suggestion is that the second derivative is equal to this:



    , but how do I use that to find E[Y(Y-1)]?

    Would be nice if someone could help me with any of these problems.
    By definition:

    E[Y(Y-1)] = \sum_{y=1}^{\infty} y (y-1) q^{y-1} p = q p \sum_{y=1}^{\infty} y(y-1) q^{y-2} = q (1 - q) \sum_{y=1}^{\infty} y(y-1) q^{y-2} .... (1)


    Sum of geometric series: S = \sum_{y=1}^{\infty} q^{y} = \frac{q}{1-q}.

    Therefore: \frac{d^2 S}{d q^2} = \sum_{y=1}^{\infty} y (y - 1) q^{y-2} = \frac{2}{(1 - q)^3} .... (2)

    Substitute equation (2) into equation (1) and simplify.


    You can check this result by noting that:

    E(Y(Y-1)] = E[Y^2] - E[Y]  = E[Y^2] - \left( E[Y] \right)^2 + \left( E[Y] \right)^2 - E[Y] = Var(Y) +  \left( E[Y] \right)^2 - E[Y].

    And for a geometric distribution you should know that E[Y] = \frac{1}{p} = \frac{1}{1 - q} and Var[Y] = \frac{1-p}{p^2} = \frac{q}{(1 - q)^2}.
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  8. #8
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    Thanks both of you, Mr F and awkward, for your help! Much appreciated.
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