Question

Let X1,..., Xn be a simple random sample from a distribution with density function Svorvo-1 O<331 fx (2:0) = otherwise where e > 0 is an unknown parameter.
(a) Find a MOM estimator for 0.
(b) If the observations are 1 1 1 2'3'2 Determine the point estimate with the estimator you find in part (a).

Answer 1

Answer:

The method of moment (MOM) estimator as:  $\mathbf{\hat {\theta} =(\dfrac{\overline X}{1-\overline X})^2}$

$\overline X = \dfrac{4}{9}$

$\mathbf{\hat {\theta} =\dfrac{16}{25} }$

Step-by-step explanation:

From the question, the correct format for the probability density function is:

$fx(x ; \theta) = \left \{ {{\sqrt{\theta x}^{\sqrt{\theta}-1}}\ \ 0 \leq x \leq 1 \atop {0} \ \ \ \ \ \ \ otherwise } \right.$

where θ > 0 is an unknown parameter.

(a) The MOM estimator can be calculated as follows:

$E(X) = \int ^1_0x. \sqrt{\theta} \ x^{\sqrt{\theta}-1} \ dx$

$E(X) = \int ^1_0 \sqrt{\theta} \ x^{\sqrt{\theta}} \ dx$

$E(X) = \dfrac{\sqrt{\theta} }{\sqrt{\theta} +1 } ( x ^{\sqrt{\theta}+1})^1_0$

$E(X) = \dfrac{\sqrt{\theta} }{\sqrt{\theta} +1 }$

suppose E(X) = $\overline X$

Then;

$\overline X = \dfrac{\sqrt{\theta} }{\sqrt{\theta} +1 }$

$\dfrac{1}{\overline X} = \dfrac{\sqrt{\theta} +1 }{\sqrt{\theta}}$

$\dfrac{1}{\overline X} =1 + \dfrac{1}{\sqrt{\theta}}$

making $\dfrac{1}{\sqrt{\theta}}$ the subject of the formula, we have:

$\dfrac{1}{\sqrt{\theta}} =\dfrac{1}{\overline X} - 1$

$\dfrac{1}{\sqrt{\theta}} =\dfrac{1-\overline X}{\overline X}$

$\sqrt{\theta} =\dfrac{\overline X}{1-\overline X}$

squaring both sides, we have:

The method of moment (MOM) estimator as:  $\mathbf{\hat {\theta} =(\dfrac{\overline X}{1-\overline X})^2}$

b) If the observations are $\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{2}$

Then,

$\overline X = \dfrac{\dfrac{1}{2}+ \dfrac{1}{3}+\dfrac{1}{2}}{3}$

$\overline X = \dfrac{\dfrac{3+2+3}{6}}{3}$

$\overline X = \dfrac{\dfrac{8}{6}}{3}$

$\overline X = \dfrac{8}{6} \times \dfrac{1}{3}$

$\overline X = \dfrac{8}{18}$

$\overline X = \dfrac{4}{9}$

Finally, the point estimate of the estimator $\theta$ is

$\mathbf{\hat {\theta} =\begin {pmatrix} \dfrac{\dfrac{4}{9}}{1-\dfrac{4}{9}} \end {pmatrix}^2}$

$\mathbf{\hat {\theta} =\begin {pmatrix} \dfrac{\dfrac{4}{9}}{\dfrac{5}{9}} \end {pmatrix}^2}$

$\mathbf{\hat {\theta} =\begin {pmatrix} \dfrac{4}{5} \end {pmatrix}^2}$

$\mathbf{\hat {\theta} =\dfrac{16}{25} }$