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Convergence in Probability

  1. Jan 26, 2013 #1
    I was a bit confused with the pages that I attached...

    1) "An intuitive estimate of [itex]\theta[/itex] is the maximum of the sample". But we are only taking random samples, so even the maximum might be far from [itex]\theta[/itex], right?

    2) I don't understand how [itex]E(Y_n) = (n/(n+1))\theta[/itex]. I thought that [itex]E(Y_n) = (Y_n)*pdf = (Y_n)(\frac{nt^n-1}{\theta^n})[/itex].

    3) "Further, based on the cdf of Y_n, it is easily seen that [itex]Y_n \rightarrow \theta[/itex]". Does that mean that E(Y_n) converges to theta, so Y_n must also converge to theta?


    Thank you in advance
     

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  3. Jan 26, 2013 #2

    Stephen Tashi

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  4. Jan 26, 2013 #3
    Thank you for the link. It was very helpful...but I'm still a bit confused about my first and last questions...
     
  5. Jan 26, 2013 #4

    Stephen Tashi

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    Correct. The story of life in probability theory is that there is no deterministic connection between probability and actuality. The important theorems that mention random variables and actual oucomes only speak of the probability of certan actualities (which has a circular ring to to it). The best you can do is find an actuality that has a probability of 1 as some sort of limit is approached.
     
  6. Jan 26, 2013 #5

    mathman

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    For your first question - you are right. The maximum is a good guess, but it could easily be wrong.

    For you third question, define Yn.
     
  7. Jan 26, 2013 #6
    [itex]Y_n[/itex] is the maximum of [itex]X_1, ... , X_n[/itex]. Do we need to look at what [itex]E(Y_n)[/itex] approaches in order to see what [itex]Y_n[/itex] approaches to?
     
  8. Jan 28, 2013 #7

    mathman

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    Since E(Yn) -> θ and θ is the maximum of the distribution, the probability that lim Yn is < θ must be 0.

    The proof is straightforward. Let A be the event that the limit is < θ, then:

    E(Yn) = E(Yn|A)P(A) + E(Yn|A')P(A') -> θ only if P(A) = 0.
     
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