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Math Help - Distribution until the First Increase?

  1. #1
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    Question Distribution until the First Increase?

    Hey,

    On a homework I've been given this problem. I don't even know where to start. Is there a name for this distribution or a way I can look up more information about it? Thanks!

    ----

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

    X_1>X_2>X_3>...>X_{n-1}<X_n

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

    ----

    The hint we were given was to use the tail sum formula for a random variable taking values in the positive integers. But how do you go from a discrete formula to a continuous one? I am completely lost. What is this distribution called?

    Thanks guys!
    Last edited by mbaboy; April 16th 2009 at 06:31 PM. Reason: fixed latex
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  2. #2
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    Quote Originally Posted by mbaboy View Post
    Hey,

    On a homework I've been given this problem. I don't even know where to start. Is there a name for this distribution or a way I can look up more information about it? Thanks!

    ----

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

    X_1>X_2>X_3>...>X_(n-1)<X_n

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

    ----

    The hint we were given was to use the tail sum formula for a random variable taking values in the positive integers. But how do you go from a discrete formula to a continuous one? I am completely lost. What is this distribution called?

    Thanks guys!
    The nice fact is that the distribution of N does not depend on the distribution of the X_i's (provided it is diffuse: P(X_i=x)=0 for any x).

    You should notice that all the orderings of X_1,\ldots,X_n are equally likely. This is just because the r.v. X_1,\ldots,X_n play symmetric roles (you should try to write a full proof). Therefore, there is only 1 chance out of n! to have X_1>\cdots>X_{n-1}>X_n, for instance.

    Use this to compute P(N=n)=P(X_1>\cdots>X_{n-1}\mbox{ and }X_{n-1}<X_n)...

    After this (short) computation, the expected value will be straightforward.
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  3. #3
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    Ok tell me if I am on the right track. I'm not exactly sure how to compute this.

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

    =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)

    =\frac{1}{2} \times \frac{1}{2^2} \times \frac{1}{2^3}...\frac{1}{2^{n-1}}

    Ok clear I make no sense. Maybe if someone could tell me what this evaluates to I could find my way to it? I just don't see how an e can pop out whereas the only two ways to calculate e I know are

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

    I'm guessing the first comes up when evaluating an integral to get the expected value. Any help? I have take only a little probability.
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  4. #4
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    Quote Originally Posted by mbaboy View Post
    Ok tell me if I am on the right track. I'm not exactly sure how to compute this.

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

    =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)

    =\frac{1}{2} \times \frac{1}{2^2} \times \frac{1}{2^3}...\frac{1}{2^{n-1}}

    Ok clear I make no sense.
    Indeed, the events are not independent, so the probability isn't just the product of the probabilities.

    Read again my previous post. It tells that the only property you need is that, for any permutation \sigma of \{1,\ldots,n\}, P(X_{\sigma(1)}>\cdots >X_{\sigma(n)})=\frac{1}{n!}. This is just because these events have same probability and they partition the probability space (one and only one of these events happens).

    Now, you can write the event \{N=n\} using these events: to which orders does it correspond? Each order has probability \frac{1}{n!}, so... It is a one-liner but I would like you to find it.
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  5. #5
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    Ok so tell me if this makes sense.

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

    so E(X)=\sum_{n=1}^\infty \frac{n}{n!}=e

    Is that right? Can I do that summation since n! only takes values on the positive integers? I'm just confused because X_i can be defined continuously on all real numbers, yet somehow I am suppose to end up with a summation that evaluates to e.

    Edit: Ok ok I see what you are saying! What I have above is wrong. The number of orderings is (n-1)!. Give me a few minutes with this I think I got it! Many thanks. I guess what I couldn't fathom is that although X_i are continuous, the distribution for N obviously is not.
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