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

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

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

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

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

$\begin{cases} \psi'(t) =e^{3 \theta t} \\<br /> \psi''(t) =3 \theta e^{3 \theta t} \end{cases}$

$\begin{cases} \psi'(0) =1 \\<br /> \psi''(0) =3 \theta \end{cases}$

Hence, M.E. is

$\widehat{\theta} = \frac{\psi''(0)}{3} = \frac{E(X^2)}{3}$

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

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

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

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

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

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

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

$\begin{cases} \psi'(t) =e^{3 \theta t} \\<br /> \psi''(t) =3 \theta e^{3 \theta t} \end{cases}$

$\begin{cases} \psi'(0) =1 \\<br /> \psi''(0) =3 \theta \end{cases}$

Hence, M.E. is

$\widehat{\theta} = \frac{\psi''(0)}{3} = \frac{E(X^2)}{3}$

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

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

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?