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Distribution until the First Increase?

  1. Apr 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Let [tex]X_1, X_2,...[/tex] be independent identically distributed continous random variables. We call [tex]n[/tex] the first time of increase if

    [tex]X_1>X_2>X_3>...>X_{n-1}<X_n[/tex]

    Let [tex]N[/tex] be the time until the first increase. Show that [tex]E[N]=e[/tex].

    2. Relevant equations

    Given for a discrete random variable with values in the positive integers:

    [tex]E(X)=\sum_{k=1}^\infty P(X>=k)[/tex]

    3. The attempt at a solution

    Well I know that [tex]X_i[/tex] are exchangeable so the desired outcome has only 1 in [tex]n![/tex] chance of happening.

    [tex]P(N=n)=P(X_1>\cdots>X_{n-1}\mbox{ and }X_{n-1}<X_n)[/tex]

    [tex]=P(X_1>X_2)P(X_2>X_3)P(X_3>X_4)...P(X_{n-2}>X_{n-1})P(X_{N-1}<Xn)[/tex]

    [tex]=\frac{1}{2} \times \frac{1}{2^2} \times \frac{1}{2^3}...\frac{1}{2^{n-1}}[/tex]

    Ok clearly I don't know what I'm doing. Maybe if someone could tell me what this evaluates to I could find my way to it? I just don't see how an [tex]e[/tex] can pop out whereas the only two ways to calculate [tex]e[/tex] I know are

    [tex]\lim_{k\to\infty}(1+\frac{1}{k})^k[/tex] and [tex]e^z=\sum_{i=1}^\infty \frac{z^i}{i!}[/tex]

    Also, I have no idea how the tail sum formula is suppose to help. How can I make use of it when the random variables in question are continuous?

    Thanks guys!
     
  2. jcsd
  3. Apr 17, 2009 #2
    These sort of questions never stop amazing me...

    First, you gotta keep in mind that although X is determined by a continous random variable, the distribution of the number [tex]k[/tex] for which [tex]X_{k-1}<X_{k}[/tex] (+ the other condition) is, in fact, a discrete random variable. Let's call this discrete random variable [tex]Y[/tex] to avoid confusion with the other random variable [tex]X[/tex]. Second, the false assumption you have made so far is that you can factorize [tex]P(X_1>X_2>X_3)[/tex] into [tex]P(X_1>X_2)P(X_2>X_3)[/tex]. This is not true, since these are not independent outcomes. Namely, there are plenty of possibilites where [tex]X_2[/tex] can be larger than [tex]X_3[/tex], while at the same time be larger than [tex]X_1[/tex] - these are not independent.

    You want to approach the situation a bit different.

    First, note that [tex]Y[/tex] is in fact a discrete random variable, taking on the values [tex]2,3,4,\ldots[/tex]. The first thing you want to determine is: what is the distribution of [tex]Y[/tex]? In other words, what is

    [tex]P(Y=k)[/tex]

    For that we take the first [tex]k[/tex] outcomes [tex]X_i[/tex]. ([tex]i=1,\ldots,k[/tex]). We can always order these from highest to lowest:
    [tex]X_{i_1} > X_{i_2} > \ldots > X_{i_k}[/tex]
    for instance, we can have:
    [tex]X_{4} > X_{8} > \ldots > X_2}[/tex]

    There are in total [tex]k![/tex] ways of possible orderings. The important remark is now that all these orderings are equally likely to happen. And also, what we want is a particular subset of orderings, namely the one where we have
    [tex]X_1 > X_2 > \ldots X_{k-1} < X_k[/tex]
    You should convince yourself (again) that there are [tex]k-1[/tex] orderings corresponding to this statement. In general, the outcome [tex]X_k[/tex] is "plugged in" on the other side:

    [tex]X_1> X_2 >\ldots X_i > X_k > X_{i+1} > \ldots X_{k-1}[/tex]

    Again, there are [tex](k-1)[/tex] ways for this to occure. Since all possible orderings are equally likely to happen, the chance for the above to happen is:

    [tex]P(Y=k) = \frac{k-1}{k!}[/tex]

    Beautiful argument hey?

    The rest of the question is now pretty straightforward, if you use the formula for the expectation value:

    [tex]E[Y] = \sum{k=2}^\infty k P(Y=k)[/tex]
     
  4. Apr 17, 2009 #3
    Ah ok it all makes very good sense. The only thing I cannot grasp is how there are k-1 orderings. The way I see it there is only 1. Consider for k=4

    [tex]X_1>X_2>X_3<X_4[/tex]

    There is only 1 way to order it such that the condition still holds, or am I missing something?

    Edit: Ok, I see for the random draws there are (k-1)! orderings so for

    [tex]X_{(1)}>X_{(2)}>...>X_{(k-1)}[/tex]

    but there is still only 1 ordering of the random variables. Maybe I am misinterpreting the question.

    If you are considering the orderings and not the act
     
    Last edited: Apr 17, 2009
  5. Apr 18, 2009 #4
    I presume that you suggest one ends up with;

    [tex] \sum_{k=0}^{\infty} \frac{1}{k!} =e [/tex]

    right?

    But not E[Y] is asked but E[N] so I immediatley think of Walds equation (though I'm sure if it's applicable here but I'm pretty certain it is).
     
  6. Apr 18, 2009 #5

    matt grime

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    Y is N.
     
  7. Apr 18, 2009 #6
    No, that is because this ordering:

    [tex]X_1>X_2>X_3<X_4[/tex]

    is not a unique one. Namely, you haven't specified the relation between [tex]X_4[/tex] and [tex]X_1[/tex] and [tex]X_2[/tex]. You can have:

    [tex]X_1>X_2>X_4>X_3[/tex]
    [tex]X_1>X_4>X_2>X_3[/tex]
    [tex]X_4>X_1>X_2>X_3[/tex]

    So there are 3 definite orderings corresponding to the one you want. The total number of orderings is [tex]4![/tex], so that

    [tex]P(Y=4) = \frac{3}{4!}[/tex]

    So again, in general you would have [tex]k![/tex] total orderings and [tex](k-1)[/tex] of them correspond to the case where
    [tex]X_1>X_2>\ldots>X_{k-1} <X_k[/tex]
     
  8. Apr 18, 2009 #7
    Yea, the forums were acting fishy last night, and it wouldn't let me edit the post.
     
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