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Thread: Distribution of MLE

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    Distribution of MLE




    Hey guys how are you? I have the following question:

    Let X1,X2,...,Xn be a random sample from a Pareto distribution having pdf
    f(x|b)= (a*b^a)/x^(a+1) where x>=b (1)

    Determine the maximum likelihood estimator for b, say b' on (0,infinity) and by considering P(b'>x) or otherwise show that b' has the Pareto distribution with pdf given by (1) but with a replaced by an.


    My attempt: I found the MLE as b'=min Xi where 1<=i<=n, since our pdf is monotonically increasing w.r.t b.

    After that I know how to find the asymptotic distribution of the MLE using the formula including the expected information but then we say that MLE follows a normal distribution for large n.

    How do I show that the MLE follows a Pareto distribution in this case? I am so struggled, any help would be much appreciated!



    P.S The hint tells us to consider P(b'>x) but how can I find P(min Xi >x) and why should it help me?

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    Re: Distribution of MLE

    Quote Originally Posted by Darktranquillity View Post
    P.S The hint tells us to consider P(b'>x) but how can I find P(min Xi >x) and why should it help me?
    It's a hint on deriving the distribution of the sample minimum (which in this case is the distribution of your mle).

    Think about this - if X is your sample minimum... what can you say about each and every one of the values in your sample? How can you relate this to the hint?

    Edit: If you're looking for the interesting off-topic discussion that was contained in this thread before then you might want to turn your attention... here.

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    Re: Distribution of MLE

    The sample minimum is every time the smallest of our observations. Where does this lead me? Consider also that x>=b for the previous pdf, so the probability of b' to be >x wouldn't be 0, using that x>=b??

    Really I am so struggled. Can you please help me more? I can't think anything else at all!
    Last edited by Darktranquillity; 10-27-2011 at 07:06 AM.

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    Re: Distribution of MLE

    If you wanted to know P(X(min) >= b) this is the same as P(X1 >= b, X2 >= b, X3 >= b, ..., Xn >= b) right? So use independence and a little bit of math to see if you can get the distribution from that.

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    Re: Distribution of MLE

    Quote Originally Posted by Dason View Post
    If you wanted to know P(X(min) >= b) this is the same as P(X1 >= b, X2 >= b, X3 >= b, ..., Xn >= b) right? So use independence and a little bit of math to see if you can get the distribution from that.
    P(min Xi >x) is the same as P(min Xi >= b), since x>=b? But then why P(minXi >= b) is the same as P(X1>=b, X2>=b,.....)? I mean onle one is the sample minimun, why do we use every Xn?
    Is it possible to provide me the solution since there have been to many days I am trying to find a solution? Thank you very much, I hope I am not asking for too much!
    Last edited by Darktranquillity; 10-27-2011 at 12:18 PM.

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    Re: Distribution of MLE

    If b < Min of Xi then b has to be less than EVERY one of the Xi. So if we're looking at the cdf of X(min) we want to know P(X(min) <= b) = 1 - P(X(min) > b) = 1 - P(X1 > b, X2 > b, X3 > b, ..., Xn > b)

    Can you see where to go from here?

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    Re: Distribution of MLE

    Quote Originally Posted by Dason View Post
    If b < Min of Xi then b has to be less than EVERY one of the Xi. So if we're looking at the cdf of X(min) we want to know P(X(min) <= b) = 1 - P(X(min) > b) = 1 - P(X1 > b, X2 > b, X3 > b, ..., Xn > b)

    Can you see where to go from here?
    So if I understand well: We want to find the pdf of x(min) = b'. But we know that b<=x so we need b<=min(x). The hint of the question tells me to consider P(b'>x) i.e
    P(xmin>x). But you tell me to look at P(xmin<=b) not P(xmin>x) that's what I am confused about...

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    Re: Distribution of MLE

    It doesn't matter. P(xmin <= b) = 1 - P(xmin > b)... so it's the same thing as the hint really...

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    Re: Distribution of MLE

    But we know that for a Paretto distribution any x is >= b. So P(x>=b) = 1, since is always true if we are talking for a Paretto distribution. So P(xmin>=b) = 1, since xmin is any of the xn. What am I thinking incorrectly?!

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    Re: Distribution of MLE

    But we know that for a Paretto distribution any x is >= b. So P(x>=b) = 1, since is always true if we are talking for a Paretto distribution. So P(xmin>=b) = 1
    Yes you are correct, but back to the hints in the first post:

    P.S The hint tells us to consider P(b'>x)
    So now you know how to calculate this? for a general x > b.

    (you already know when x \leq b the probability is 1)

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    Re: Distribution of MLE

    Quote Originally Posted by BGM View Post
    Yes you are correct, but back to the hints in the first post:



    So now you know how to calculate this? for a general x > b.

    (you already know when x \leq b the probability is 1)
    But a Pareto distribution needs x>=b so P(x<b)=0 [for any x] so P(xmin<b)=0 since we are talking for a Pareto distribution! But P(x>b)=1 and P(xmin>b) or P(b'>b)=1. How do I find P(xmin>x) or P(b'>x) where b' = xmin?

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    Re: Distribution of MLE

    First of all try to think about the following relationship :

    1. If \min_i x_i > x, then x_1 > x, x_2 > x, ..., x_n > x

    2. If x_1 > x, x_2 > x, ..., x_n > x, then \min_i x_i > x

    i.e. The two events are equivalent (if and only if).

    And thus the probability of the corresponding events are equal.

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    Re: Distribution of MLE

    Yeah I got sloppy with notation and was just trying to get you to think about P(Xmin > b) for any constant b (not necessarily the b that is in the pareto distribution).

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    Re: Distribution of MLE

    Quote Originally Posted by BGM View Post
    First of all try to think about the following relationship :

    1. If \min_i x_i > x, then x_1 > x, x_2 > x, ..., x_n > x

    2. If x_1 > x, x_2 > x, ..., x_n > x, then \min_i x_i > x

    i.e. The two events are equivalent (if and only if).

    And thus the probability of the corresponding events are equal.
    So I have to find P(x1>x, x2>x,....,xn>x) to find the CDF of the sample minimum and then integrate it to find its pdf?? If this is the way how do I find P(x1>x, x2>x,....,xn>x) ?

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    Re: Distribution of MLE


    Ok assuming you agree on that fact. Several things to point out:

    1. \Pr\left\{\min_{i=1,2,...,n}X_i > x\right\} is not the CDF; it is the survival function. \Pr\left\{\min_{i=1,2,...,n}X_i \leq x\right\} is the CDF.

    2. Usually in these problems, the random sample X_1, X_2, ..., X_n are independent unless otherwise specified.

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