- #1

sanctifier

- 58

- 0

## Homework Statement

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].

## Homework Equations

Nothing special.

## 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: