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Math Help - Don't know where to start...

  1. #1
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    Don't know where to start...

    This is what I need to do but don't know where to start, please give me some pointers. It has to do with expected value of nonnegative random variable.

    Let X\in {0,1,2,...}
    Show that E(X)=\sum_{k=0}^{\infty}1-F_X(k)

    I know that E(X)=\sum_{k=0}^{\infty}xp(k),
    but what is p(k) and F_X(k)?

    Where do i even start?
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  2. #2
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    Hello,
    Quote Originally Posted by synclastica_86 View Post
    This is what I need to do but don't know where to start, please give me some pointers. It has to do with expected value of nonnegative random variable.

    Let X\in {0,1,2,...}
    Show that E(X)=\sum_{k=0}^{\infty}1-F_X(k)

    I know that E(X)=\sum_{k=0}^{\infty} {\color{red}k} p(k),
    but what is p(k) and F_X(k)?

    Where do i even start?
    I don't understand.. You're given this problem, but you don't even know what p(k) and F_X(k) are ?
    That's not normal !!

    Anyway, in general, p(k)=P(X=k) is the probability distribution, and F_X(k)=P(X \leq k) is the cumulative distribution function.

    Hence 1-F_X(k)=P(X>k)
    Since X \in \{0,1,\dots\}, \forall k \in \{0,1,\dots\}, ~ 1-F_X(k)=P(X>k)=p(k+1)+p(k+2)+\dots=\sum_{j=k+1}^\in  fty p(j)


    <br />
\begin{aligned}<br />
1-F_X(0) &=\sum_{j=1}^\infty p(j)= &p(1)+ &p(2)+&p(3)+\dots+p(n)+\dots \hfill \\<br />
1-F_X(1) &=\sum_{j=2}^\infty p(j)=& &p(2)+&p(3)+\dots+p(n)+\dots \hfill \\<br />
1-F_X(2) &=\sum_{j=3}^\infty p(j)= & & &p(3)+\dots+p(n)+\dots \hfill \\<br />
\vdots \hfill & & & & \hfill \\<br />
\end{aligned}

    Can you see a pattern when you sum these ?

    (1-F_X(0))+(1-F_X(1))=p(1)+2 \sum_{j=2}^\infty p(j)
    (1-F_X(0))+(1-F_X(1))+(1-F_X(2))=p(1)+2p(2)+3 \sum_{j=3}^\infty p(j)


    Now it looks logical that \sum_{k=0}^\infty (1-F_X(k))=\sum_{k=1}^\infty kp(k)=\sum_{k=0}^\infty kp(k)=E(X)

    But I can't manage to find a way to prove it closely. Maybe I'll find tomorrow, or someone else will

    -----------------------------------------------------
    You can prove by induction, for n \in \{0,1,2,\dots\}, that :
    \sum_{k=0}^n (1-F_X(k))=\left[\sum_{k=1}^n kp(k)\right]+(n+1) \sum_{k=n+1}^\infty p(k)
    Then take n to infinity... But I don't know how to deal with it :s
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  3. #3
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    Quote Originally Posted by Moo View Post
    You can prove by induction, for n \in \{0,1,2,\dots\}, that :
    \sum_{k=0}^n (1-F_X(k))=\left[\sum_{k=1}^n kp(k)\right]+(n+1) \sum_{k=n+1}^\infty p(k)
    Then take n to infinity... But I don't know how to deal with it :s
    If E[X]=\infty, then the right-hand side goes to infinity (it is greater than \sum_{k=1}^n kp(k)), so that the equality holds.

    Suppose now E[X]<\infty. In order to take the limit as n\to\infty, we have to justify n \sum_{k=n}^\infty p(k)\to_n 0.

    This can be written n P(X\geq n)\to_n 0. However, we have n P(X\geq n)\leq E[X {\bf 1}_{(X\geq n)}], and the bounded convergence theorem shows that the right-hand sides converges to 0 since E[X]<\infty.

    -------------

    There's another solution without taking a limit. This involves Fubini theorem for double series.

    Indeed, if you write

    \sum_{k=0}^\infty (1-F_X(k))=\sum_{k=0}^\infty \sum_{i=k+1}^\infty p(i) = \sum_{k=0}^\infty \sum_{i=0}^\infty {\bf 1}_{(i>k)} p(i),

    the equality amounts to interchange the sum signs, I let you get convinced of this.
    Last edited by Laurent; February 15th 2009 at 02:59 AM. Reason: A 'k' was mistakenly written instead of an 'i', cf. post#6
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  4. #4
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    Quote Originally Posted by Laurent View Post
    If E[X]=\infty, then the right-hand side goes to infinity (it is greater than \sum_{k=1}^n kp(k)), so that the equality holds.

    Suppose now E[X]<\infty. In order to take the limit as n\to\infty, we have to justify n \sum_{k=n}^\infty p(k)\to_n 0.

    This can be written n P(X\geq n)\to_n 0. However, we have n P(X\geq n)\leq E[X {\bf 1}_{(X\geq n)}], and the bounded convergence theorem shows that the right-hand sides converges to 0 since E[X]<\infty.

    -------------

    There's another solution without taking a limit. This involves Fubini theorem for double series.

    Indeed, if you write

    \sum_{k=0}^\infty (1-F_X(k))=\sum_{k=0}^\infty \sum_{i=k+1}^\infty p(i) = \sum_{k=0}^\infty \sum_{i=0}^\infty {\bf 1}_{(i>k)} p(k)" alt="\sum_{k=0}^\infty (1-F_X(k))=\sum_{k=0}^\infty \sum_{i=k+1}^\infty p(i) = \sum_{k=0}^\infty \sum_{i=0}^\infty {\bf 1}_{(i>k)} p(k)" />,

    the equality amounts to interchange the sum signs, I let you get convinced of this.
    Thanks for all the help all. I know this might sound stupid, it's the first time that I did any kind of stats, but I thought Fubini's Theorem is used on integrals in vector calculus. How did you turn that into the 2 sums? I understand the reasoning behind what moo did the first time the most. Am I wrong to think so?
    Last edited by synclastica_86; February 14th 2009 at 07:11 PM.
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  5. #5
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    Quote Originally Posted by synclastica_86 View Post
    Thanks for all the help all. I know this might sound stupid, it's the first time that I did any kind of stats, but I thought Fubini's Theorem is used on integrals in vector calculus. How did you turn that into the 2 sums? I understand the reasoning behind what moo did the first time the most. Am I wrong to think so?
    What Moo did is also "Fubini" (interchanging sums) but for a finite sum; in this case we don't say Fubini because it is always true by simple properties of the sum.

    Let me repeat what Moo wrote in a pretty way:

    \begin{aligned}<br />
1-F_X(0) &=\sum_{j=1}^\infty p(j)= &p(1)+ &p(2)+&p(3)+\dots+p(n)+\dots \hfill \\<br />
1-F_X(1) &=\sum_{j=2}^\infty p(j)=& &p(2)+&p(3)+\dots+p(n)+\dots \hfill \\<br />
1-F_X(2) &=\sum_{j=3}^\infty p(j)= & & &p(3)+\dots+p(n)+\dots \hfill \\<br />
\vdots \hfill & & & & \hfill \\<br />
\end{aligned}

    If you sum line after line (there are infinitely many), you get \sum_{k=0}^\infty (1-F_X(k)).

    If you sum column after column, you get 1p(1)+2p(2)+3p(3)+\cdots  = \sum_{k=1}^\infty k p(k).

    In order to justify why these sums are the same, you need a theorem. Fubini theorem holds for double integrals indeed, but there is a very similar theorem for double sums, and I think it is customary to call it Fubini as well.

    Anyway, this theorem says that if we have nonnegative numbers a_{ij}\geq 0 (like in this case), then we can interchange the sum signs (whether the sum is finite or not):


    \sum_i\sum_j a_{ij}=\sum_j\sum_i a_{ij}.

    This amounts to adding the columns or the lines in the previous sketch.

    NB: The theorem also gives a condition for doing the same in case there are negative numbers:

    if \sum_i \sum_j |a_{ij}|<\infty, then \sum_i \sum_j a_{ij} = \sum_j \sum_i a_{ij}.

    --
    You may prefer not to use this theorem about double series and sum only n+1 lines of the previous sketch to get the equality Moo gives:

    \sum_{k=0}^n (1-F_X(k))=\left[\sum_{k=1}^n kp(k)\right]+(n+1) \sum_{k=n+1}^\infty p(k).

    Then you need to take the limit as n tends to \infty. I gave a possible justification of the limits in the first part of my previous post.
    Last edited by Laurent; February 15th 2009 at 01:33 AM.
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  6. #6
    Moo
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    Hmmm ... I was vainly trying to deal with Fubini, but there was something wrong... :
    \sum_{k=0}^\infty (1-F_X(k))=\sum_{k=0}^\infty \sum_{i=k+1}^\infty p(i) = \sum_{k=0}^\infty \sum_{i=0}^\infty {\bf 1}_{(i>k)} p({\color{red}k})
    Isn't the red k an i ?

    Thereafter, it gives... very easily the solution... That's great ! Thinking of the indicator function was brilliant lol.

    Thanks for that Fubini stuff, Laurent.
    Last edited by Moo; February 15th 2009 at 01:48 AM.
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  7. #7
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    Quote Originally Posted by Moo View Post
    Hmmm ... I was vainly trying to deal with Fubini, but there was something wrong... :

    Isn't the red k an i ?
    Yes, indeed... I'm about to correct the mistake.
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