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Homework Help: Statistics- unbiased estimator #2

  1. Mar 1, 2010 #1
    1. The problem statement, all variables and given/known data
    There is some question, I solved it but am not sure I got the right answer.

    Let Y1, Y2..... Yn be a random sample of size n from the pdf fY(y;[tex]\Theta[/tex]= [tex]\frac{1}{\Theta}[/tex]*e-y/[tex]\Theta[/tex] , y>0
    Let [tex]\Theta[/tex]_hat=n*Ymin is tex]\Theta[/tex]_hat for [tex]\Theta[/tex] ?

    2. Relevant equations

    Ymin=n*(1-FY(y))n-1fY(y)

    Also, it looks like an exponential distribution.


    3. The attempt at a solution

    What I need to find is E[n*Ymin]

    I get [(-n)/(n+1)]/(1/ [tex]\Theta[/tex] )

    Is this the correct answer ?
     
  2. jcsd
  3. Mar 1, 2010 #2
    where did you get the eqn for Ymin it makes no sense.
     
  4. Mar 1, 2010 #3
    i take that back give me a sec
     
  5. Mar 1, 2010 #4
    okay here it is. the PDF of Ymin is f(y)=(-n/theta)*exp(-yn/theta). To get this we need to use the following (questionable?) sequence of logic: the probability that ymin is less than y is 1 minus the probability that ymin >= x. for this inequality to hold we need Y(i)>=x for all i. i.e. Y(1)>=x AND Y(2)>=x ... Y(n)>=x. this will give us the CDF for Ymin from which we derive the PDF of Ymin given above. Now check that E(nYmin)= theta. (I actually get
    -theta but there may be some minor issue with my calculation see what you get, if
    E(nYmin) =theta (not negative theta) then it is unbiased.
     
  6. Mar 1, 2010 #5
    Hello,

    I don't really understand what you are trying to do here.
    you found the pdf for Ymin ? we are given the pdf
    Also, I got the CDF for Ymin, but am not sure I did it the right way.
    And , I don't really understand what you got for the CDF..

    Thanks.
     
  7. Mar 1, 2010 #6

    statdad

    User Avatar
    Homework Helper

    You are given the distribution for the individual values - it is the distribution from which you take the sample. The minimum is a statistic, and you need its distribution.

    The derivation outlined above is correct. If [tex] Y_{\min} [/tex] is the minimum of a sample, then to get its distribution (assuming the sample comes from a continuous distribution, as yours does). I'll use [tex] G, g [/tex] for the CDF and PDF of [tex] Y_{\min}[/tex], and [tex] F, f [/tex] for the CDF and PDF of the population.

    [tex]
    P(Y_{\min} \ge t) = P(X_1 \ge t \text{ and } X_2 \ge t \dots \text{ and } X_n \ge t) = (1-F(t))^n
    [/tex]

    because of independence, so the CDF of [tex] Y_{\min} [/tex] is

    [tex]
    G(t) \equiv P(Y_{\min} \le t) = 1 - (1 - F(t))^n
    [/tex]

    and the density [tex] Y_{\min}[/tex]is

    [tex]
    g(t) = G'(t) = n(1-F(t))^{n-1} f(t)
    [/tex]

    In your problem the original data come from an exponential distribution. Use the CDF for that in place of F, the PDF in place of f, to get the density of [tex] Y_{min} [/tex] in this particular case. The expected value of the minimum is

    [tex]
    \int t g(t) \, dt
    [/tex]

    and this will be a function of [tex] \theta [/tex]. You should be able to finish the problem from there.
     
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