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

    I know that $\displaystyle E(X)=\sum_{k=0}^{\infty}xp(k)$,
    but what is $\displaystyle p(k) $and$\displaystyle 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 $\displaystyle X\in {0,1,2,...}$
    Show that $\displaystyle E(X)=\sum_{k=0}^{\infty}1-F_X(k)$

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

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

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

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


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

    Can you see a pattern when you sum these ?

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


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

    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 $\displaystyle n \in \{0,1,2,\dots\}$, that :
    $\displaystyle \sum_{k=0}^n (1-F_X(k))=\left[\sum_{k=1}^n kp(k)\right]+(n+1) \sum_{k=n+1}^\infty p(k)$
    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 $\displaystyle n \in \{0,1,2,\dots\}$, that :
    $\displaystyle \sum_{k=0}^n (1-F_X(k))=\left[\sum_{k=1}^n kp(k)\right]+(n+1) \sum_{k=n+1}^\infty p(k)$
    Then take n to infinity... But I don't know how to deal with it :s
    If $\displaystyle E[X]=\infty$, then the right-hand side goes to infinity (it is greater than $\displaystyle \sum_{k=1}^n kp(k)$), so that the equality holds.

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

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

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

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

    Indeed, if you write

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

    the equality amounts to interchange the sum signs, I let you get convinced of this.
    Last edited by Laurent; Feb 15th 2009 at 01:59 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 $\displaystyle E[X]=\infty$, then the right-hand side goes to infinity (it is greater than $\displaystyle \sum_{k=1}^n kp(k)$), so that the equality holds.

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

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

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

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

    Indeed, if you write

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

    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; Feb 14th 2009 at 06:11 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:

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

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

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

    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 $\displaystyle a_{ij}\geq 0$ (like in this case), then we can interchange the sum signs (whether the sum is finite or not):


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

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

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

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

    Then you need to take the limit as $\displaystyle n$ tends to $\displaystyle \infty$. I gave a possible justification of the limits in the first part of my previous post.
    Last edited by Laurent; Feb 15th 2009 at 12:33 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... :
    $\displaystyle \sum_{k=0}^\infty (1-F_X(k))=\sum_{k=0}^\infty \sum_{i=k+1}^\infty p(i) = \sum_{k=0}^\infty \sum_{i=0}^\infty {\bf 1}_{(i>k)} p({\color{red}k})$
    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; Feb 15th 2009 at 12:48 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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