Answer:
The method of moment (MOM) estimator as: [tex]\mathbf{\hat {\theta} =(\dfrac{\overline X}{1-\overline X})^2}[/tex]
[tex]\overline X = \dfrac{4}{9}[/tex]
[tex]\mathbf{\hat {\theta} =\dfrac{16}{25} }[/tex]
Step-by-step explanation:
From the question, the correct format for the probability density function is:
[tex]fx(x ; \theta) = \left \{ {{\sqrt{\theta x}^{\sqrt{\theta}-1}}\ \ 0 \leq x \leq 1 \atop {0} \ \ \ \ \ \ \ otherwise } \right.[/tex]
where θ > 0 is an unknown parameter.
(a) The MOM estimator can be calculated as follows:
[tex]E(X) = \int ^1_0x. \sqrt{\theta} \ x^{\sqrt{\theta}-1} \ dx[/tex]
[tex]E(X) = \int ^1_0 \sqrt{\theta} \ x^{\sqrt{\theta}} \ dx[/tex]
[tex]E(X) = \dfrac{\sqrt{\theta} }{\sqrt{\theta} +1 } ( x ^{\sqrt{\theta}+1})^1_0[/tex]
[tex]E(X) = \dfrac{\sqrt{\theta} }{\sqrt{\theta} +1 }[/tex]
suppose E(X) = [tex]\overline X[/tex]
Then;
[tex]\overline X = \dfrac{\sqrt{\theta} }{\sqrt{\theta} +1 }[/tex]
[tex]\dfrac{1}{\overline X} = \dfrac{\sqrt{\theta} +1 }{\sqrt{\theta}}[/tex]
[tex]\dfrac{1}{\overline X} =1 + \dfrac{1}{\sqrt{\theta}}[/tex]
making [tex]\dfrac{1}{\sqrt{\theta}}[/tex] the subject of the formula, we have:
[tex]\dfrac{1}{\sqrt{\theta}} =\dfrac{1}{\overline X} - 1[/tex]
[tex]\dfrac{1}{\sqrt{\theta}} =\dfrac{1-\overline X}{\overline X}[/tex]
[tex]\sqrt{\theta} =\dfrac{\overline X}{1-\overline X}[/tex]
squaring both sides, we have:
The method of moment (MOM) estimator as: [tex]\mathbf{\hat {\theta} =(\dfrac{\overline X}{1-\overline X})^2}[/tex]
b) If the observations are [tex]\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{2}[/tex]
Then,
[tex]\overline X = \dfrac{\dfrac{1}{2}+ \dfrac{1}{3}+\dfrac{1}{2}}{3}[/tex]
[tex]\overline X = \dfrac{\dfrac{3+2+3}{6}}{3}[/tex]
[tex]\overline X = \dfrac{\dfrac{8}{6}}{3}[/tex]
[tex]\overline X = \dfrac{8}{6} \times \dfrac{1}{3}[/tex]
[tex]\overline X = \dfrac{8}{18}[/tex]
[tex]\overline X = \dfrac{4}{9}[/tex]
Finally, the point estimate of the estimator [tex]\theta[/tex] is
[tex]\mathbf{\hat {\theta} =\begin {pmatrix} \dfrac{\dfrac{4}{9}}{1-\dfrac{4}{9}} \end {pmatrix}^2}[/tex]
[tex]\mathbf{\hat {\theta} =\begin {pmatrix} \dfrac{\dfrac{4}{9}}{\dfrac{5}{9}} \end {pmatrix}^2}[/tex]
[tex]\mathbf{\hat {\theta} =\begin {pmatrix} \dfrac{4}{5} \end {pmatrix}^2}[/tex]
[tex]\mathbf{\hat {\theta} =\dfrac{16}{25} }[/tex]
