Statistics: How to prove the consistent estimator of theta?

[sanctifier]

Homework Statement

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

f(x| \theta ) =\begin{cases} \frac{1}{3\theta} &amp; 0 &lt; x \leq 3\theta<br /> \\0 &amp; otherwise\end{cases}

Where \theta &gt; 0 is an unknown parameter, and X_1, X_2 … X_n is a sample from X where n &gt; 2

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

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

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

Homework Equations

Nothing special.

The Attempt at a Solution

Answer 1:

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

\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)

\begin{cases} \psi&#039;(t) =e^{3 \theta t} \\<br /> \psi&#039;&#039;(t) =3 \theta e^{3 \theta t} \end{cases}

\begin{cases} \psi&#039;(0) =1 \\<br /> \psi&#039;&#039;(0) =3 \theta \end{cases}

Hence, M.E. is

\widehat{\theta} = \frac{\psi&#039;&#039;(0)}{3} = \frac{E(X^2)}{3}

Answer 2:

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

f(X| \theta ) = \frac{1}{(3\theta)^n} \;\;when\;\; 0&lt;X_i \leq 3\theta \;\; for \;\; i=1,2,...,n

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

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

Answer 3:

I have no idea to even start the proving.

 

[Ray Vickson]

Homework Statement

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

f(x| \theta ) =\begin{cases} \frac{1}{3\theta} &amp; 0 &lt; x \leq 3\theta<br /> \\0 &amp; otherwise\end{cases}

Where \theta &gt; 0 is an unknown parameter, and X_1, X_2 … X_n is a sample from X where n &gt; 2

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

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

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

Homework Equations

Nothing special.

The Attempt at a Solution

Answer 1:

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

\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)

\begin{cases} \psi&#039;(t) =e^{3 \theta t} \\<br /> \psi&#039;&#039;(t) =3 \theta e^{3 \theta t} \end{cases}

\begin{cases} \psi&#039;(0) =1 \\<br /> \psi&#039;&#039;(0) =3 \theta \end{cases}

Hence, M.E. is

\widehat{\theta} = \frac{\psi&#039;&#039;(0)}{3} = \frac{E(X^2)}{3}

Answer 2:

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

f(X| \theta ) = \frac{1}{(3\theta)^n} \;\;when\;\; 0&lt;X_i \leq 3\theta \;\; for \;\; i=1,2,...,n

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

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

Answer 3:

I have no idea to even start the proving.

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.

 

[sanctifier]

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 )

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

Answer 2 is wrong.

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

What about answer 1 and answer 3?