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Homework Help: Statistics: How to prove the consistent estimator of theta?

  1. Feb 26, 2014 #1
    1. The problem statement, all variables and given/known data

    If the probability density function (p.d.f.) of the random variable X is

    [itex] f(x| \theta ) =\begin{cases} \frac{1}{3\theta} & 0 < x \leq 3\theta
    \\0 & otherwise\end{cases} [/itex]

    Where [itex] \theta > 0 [/itex] is an unknown parameter, and [itex] X_1, X_2 … X_n [/itex] is a sample from X where [itex] n > 2 [/itex]

    Question 1: What is the moment estimator (M.E.) of [itex] \theta [/itex]?

    Question 2: What is the maximum likelihood estimator (M.L.E) of [itex] \theta [/itex]?

    Question 3: Prove [itex] \widehat{\theta} = \frac{1}{3} max\{X_1,X_2...X_n\} [/itex] is the consistent estimator of [itex] \theta [/itex].

    2. Relevant equations

    Nothing special.

    3. The attempt at a solution

    Answer 1:

    Moment generating function (m.g.f.) of X is

    [itex] \psi (t) = E(e^{tx}) = \int_0^{3 \theta } \frac{e^{tx}}{3\theta} dx= \frac{1}{3 \theta t} \int_0^{3 \theta }de^{tx}= \frac{1}{3 \theta t}e^{tx}|_{x=0}^{x=3 \theta}=\frac{1}{3 \theta t}(e^{3\theta t} - 1) [/itex]

    [itex] \begin{cases} \psi'(t) =e^{3 \theta t} \\
    \psi''(t) =3 \theta e^{3 \theta t} \end{cases} [/itex]

    [itex] \begin{cases} \psi'(0) =1 \\
    \psi''(0) =3 \theta \end{cases} [/itex]

    Hence, M.E. is

    [itex] \widehat{\theta} = \frac{\psi''(0)}{3} = \frac{E(X^2)}{3} [/itex]

    Answer 2:

    Let [itex] X [/itex] be a vector whose components are [itex] X_1, X_2 … X_n [/itex], then the joint distribution of [itex] X_1, X_2 … X_n [/itex] is

    [itex] f(X| \theta ) = \frac{1}{(3\theta)^n} \;\;when\;\; 0<X_i \leq 3\theta \;\; for \;\; i=1,2,...,n [/itex]

    Because [itex] X_i \leq 3\theta [/itex], when [itex] \widehat{\theta} = \frac{1}{3} min\{X_1,X_2...X_n\} [/itex], [itex] f(X| \theta ) [/itex] is maximized.

    Hence, M.L.E of [itex] \theta [/itex] is [itex] \frac{1}{3} min\{X_1,X_2...X_n\} [/itex].

    Answer 3:

    I have no idea to even start the proving.
    Last edited: Feb 26, 2014
  2. jcsd
  3. Feb 26, 2014 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your expression for ##EX## is incorrect: it should not be ##\theta/3##. I suggest you avoid moment-generating functions, since you seem to be mis-using them, and they are totally unnecessary in a question of this type. If you want ##EX##, just do the integration, or use familiar elementary results that you should have seen already in a first course but might have forgotten.
    Last edited: Feb 26, 2014
  4. Feb 28, 2014 #3
    Thank you for your replay, Ray.

    Actually, the moment estimator denotes the method of moments estimator, you can find it here: http://en.wikipedia.org/wiki/Method_of_moments_(statistics [Broken])

    What I’m concerning is the estimator found by using method of moments.

    Answer 2 is wrong.

    It should be [itex] \frac{1}{3} max\{X_1,X_2...X_n\} [/itex], since to maximize [itex] f(X| \theta ) = \frac{1}{(3\theta)^n} [/itex], [itex] \theta [/itex] needs to be its minimal value which should be [itex] \frac{1}{3} max\{X_1,X_2...X_n\} [/itex] with respect to the constraint [itex] X_i \leq 3\theta [/itex].

    What about answer 1 and answer 3?
    Last edited by a moderator: May 6, 2017
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