1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Statistics: How to prove the consistent estimator of theta?
  1. Consistent Estimator (Replies: 0)

  2. Estimator / Statistics (Replies: 3)

Loading...