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Stochastic processes

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data
    1.PNG

    I know that per definition [tex]E(N)= \sum P(N=k) \cdot k [/tex]. But how can I rewrite the above expectation towards the 'usual definition'?
     
  2. jcsd
  3. Sep 13, 2008 #2

    statdad

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    The expression

    [tex]
    E(\mathbf{N}) = \sum_k \Pr(N=k) \cdot k
    [/tex]

    is the usual definition for the expectation of a discrete random variable. I'm not sure what alternative you refer to. You will see, in books on probability theory, this sum written in the form

    [tex]
    E(\mathbf{N}) = \int x \, dP(x)
    [/tex]

    Edit: I was unable to view your attachment.

    (a Riemann-Stieltjes integral), which reduces to the sum you (and I) have written. I doubt this is what you seek.
     
  4. Sep 14, 2008 #3
    I'm getting seriously tired of getting approval for the attachements. I'll postphone the first exercise for now.

    But can someone help me with the second exercise (X is a stochast that attains non-negative values):

    Proof the following:

    [tex] E(X) = \int_0^{\infty} \overline{F}(x) \mbox{d}x[/tex]

    [tex]
    \overline{F}(x) = 1- F(x) = 1-P(X \leq x) = P(X>x)
    [/tex]


    What I've got so far

    [tex]
    \int_{0}^{\infty} \overline{F}(x)\ \mbox{d}x = \int_0^{\infty} \int_x^{\infty} f(y)\ \mbox{d}y = \int_0^{\infty} \int_0^y f(x)\ \mbox{d}x = \int_0^{\infty} F(y)-F(0)\ \mbox{d}y = y \left( F(y) - F(0) \right)_{0}^{\infty} +E(y)
    [/tex]


    What's going wrong in my solution?
     
  5. Sep 14, 2008 #4

    statdad

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    First, I've never seen the result you are trying to prove - but that doesn't prove it is written incorrectly. I know of the following result:

    If the random positive random variable (so that [tex] F(0-) = 0 [/tex], then

    [tex]
    E(X) = \int_0^\infty \Pr(X > x) \, dx = \int_0^\infty \Pr(X \ge x) \, dx
    [/tex]

    although the integral may be infinite. Is this what you are discussing?

    Second, your integrals, as written, don't make any sense - you need to integrate with respect to two variables, not just one. To point:

    [tex]
    \int_0^\infty \, \int_x^\infty f(y) \, dy
    [/tex]

    is meaningless without a [tex] dx [/tex] as well.
     
  6. Sep 14, 2008 #5
    Well... its the expectation just written in another form which I'll have to proof.




    Yes you're right so we get:

    [tex]\int_{0}^{\infty} \overline{F}(x)\ \mbox{d}x = \int_0^{\infty} \int_x^{\infty} f(y)\ \mbox{d}y\ \mbox{d}x = \int_0^{\infty} \int_0^y f(x)\ \mbox{d}x\ \mbox{d}y = \int_0^{\infty} F(y)-F(0)\ \mbox{d}y = y \left( F(y) - F(0) \right)_{0}^{\infty} +E(y) [/tex]


    Statdad, what am I doing wrong here?
     
  7. Sep 14, 2008 #6

    statdad

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    You are getting caught up in notation.

    When you write (in the middle of your work)

    [tex]
    \int_0^\infty \, \int_x^\infty f(y) \, dy dx = \int_0^\infty \, \int_0^y f(x) \,dx dy
    [/tex]

    why do you change the integrand from [tex] f(y) [/tex] to [tex] f(x) [/tex]? You are correct in saying that the order of integration changes, so the inner integral is w.r.t. [tex] x [/tex], but do you really need to write [tex] f(x) [/tex]? :smile:
     
  8. Sep 15, 2008 #7
    What do you propose then?
     
  9. Sep 15, 2008 #8

    statdad

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    To answer
    what you're doing in the double integral is not a change of variable - if that were the case, your step could be correct.
    Try going through the same steps without changing from [tex] f(y) [/tex] to [tex] f(x) [/tex] at the aforementioned point. (And remember that when you integrate from [tex] 0 [/tex] to [tex] y [/tex] w.r.t. [tex] x, f(y) [/tex] will act like a constant.
     
  10. Sep 15, 2008 #9
    Thanks statdad It worked. Now returning to the first exercise:

    I want to proof that (note that my sum runs to a finite value n)

    [tex] E[N]= \sum_{k=1}^{n} P(N \geq k) = \sum k \cdot P(N=k)[/tex]


    So rewrite it to the usual definition.



    Here's what I got so far (note the interchanging sum):

    [tex] \sum_{k=1}^n P(N \geq k) = \sum_{k=1}^n \sum_{l=1}^n P(N=l) = \sum_{l=1}^n n \cdot P(N=l) [/tex]

    I'm stuck here! What going wrong here?
     
  11. Sep 15, 2008 #10

    statdad

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    When you write

    [tex]
    \sum_{k=1}^n P(N \geq k) = \sum_{k=1}^n \sum_{l=1}^n P(N=l) = \sum_{l=1}^n n \cdot P(N=l)
    [/tex]

    you are essentially writing

    [tex]
    \Pr(N \ge k) = \sum_{l=1}^n \Pr(N=l)
    [/tex]

    This is not correct - the sum on the right here equals 1. In short, the expression for [tex] \Pr(N \ge k)[/tex] needs to be fixed.

    Once that is one, you will have a double sum: reversing the order of summation (watch the indices) will get you where you need to be.
     
  12. Sep 15, 2008 #11
    Yes, you're right it should be:

    [tex]\sum_{k=1}^n P(N \geq k) = \sum_{k=1}^n \sum_{l=k}^n P(N=l) = \sum_{l=k}^n n \cdot P(N=l) [/tex]

    Is this correct?
     
  13. Sep 15, 2008 #12

    statdad

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    This

    [tex]
    \sum_{k=1}^n P(N \geq k) = \sum_{k=1}^n \sum_{l=1}^n P(N=l)
    [/tex]

    portion of your most recent post is still not correct (I haven't even included the final equality)

    When you write out the sum for [tex] \Pr(N \ge k) [/tex], where should the summation begin - at [tex] 1 [/tex] or some other value?
     
  14. Sep 15, 2008 #13
    It should start at k but that's present in my post (#11) right?
     
  15. Sep 15, 2008 #14

    statdad

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    I'm sorry, you are correct - I'm preparing for class and in my haste looked at the wrong post.

    Yes, the first portion of your statement is correct.

    [tex]
    \sum_{k=1}^n P(N \geq k) = \sum_{k=1}^n \sum_{l=k}^n P(N=l)
    [/tex]

    Your error is in your next equality (which I did not include above)

    Chew on this: The final double sum above can be written as

    [tex]
    \sum_{\substack{k=1\dots,n\\l \ge k}} \Pr(N = l)
    [/tex]

    Similar to the other question dealing with double integration, think about how you can reverse the order of summation (inner sum over some values of [tex] k [/tex], outer sum of [tex] l [/tex] while maintaining the same inequality relationships between the two variables of summation.
     
  16. Sep 16, 2008 #15
    Thanks statdad I got it! You'll eventually get a summation that runs to 1 to l giving the l*P(N=l). I appreciate your effort in helping me!
     
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