Parameter Estimate Example: MSDA 3ED Ch8 Q4

Example) Mathematical Statistics and Data Analysis 3RD Edition Chapter 8, Q4  

Suppose that X is discrete random variable with P(X=0)=$\frac{2}{3}\theta$, P(X=1)=$\frac{1}{3}\theta$, P(X=2)=$\frac{2}{3}(1-\theta)$, P(X=3)=$\frac{1}{3}(1-\theta)$. The following 10 independent observations were taken from such a distribution. (3, 0, 2, 1, 3, 2, 1, 0, 2, 1)

a) Find the method of moment estimate of $\theta$.

b) Find the approximate standard error for your estimate.

▷Think first!

- Based on the sample observations with size 10, what is the sample mean?

- How can you find the expected value in discrete case?

- From the results from two questions above, what is your conclusion?

- We can find the sample variance by using the second moment. So what is variance of the estimate?

▷Solution (a)

Sample mean $=\frac{3+0+2+1+3+2+1+0+2+1}{10}$=$\frac{15}{10}$=$\frac{3}{2}$

$E[X^2]=0\cdot P(X=0) + 1\cdot P(X=1)+2\cdot P(X=2) + 3\cdot P(X=3)$ 

     = $0\cdot \frac{2}{3}\theta + 1\cdot \frac{1}{3}\theta+2\cdot \frac{2}{3}(1-\theta)+3\cdot \frac{1}{3}(1-\theta)$

     = $-2\theta + \frac{7}{3}$

$\therefore$ $\frac{3}{2} = -2\theta + \frac{7}{3} \rightarrow \theta = \frac{5}{12} \rightarrow \hat{\theta}=\frac{5}{12}$

▷Solution (b)

The variance of estimate, $Var(\hat{\theta}) = Var (\frac{7}{6}-\frac{1}{2}\bar{X})$ =$({\frac{1}{2}})^2 + Var(\bar{X}) = \frac{1}{4}\cdot \frac{\sigma^2}{n}$ = hummm... what is the variance of X?

Variance = $\sigma ^2=E[X^2] - [E(X)]^2$, so we need the second moment!

$E[X^2]=0^2\cdot P(X=0) + 1^2\cdot P(X=1)+2^2\cdot P(X=2) + 3^2\cdot P(X=3)$

         $=\theta (\frac{1}{3}-\frac{8}{3}-\frac{9}{3})+ \frac{8}{3}+\frac{9}{3}= -\frac{16}{3}\theta +\frac{17}{3}$

$\sigma ^2=E[X^2] - [E(X)]^2 = -\frac{16}{3}\theta +\frac{17}{3} -(-2\theta+\frac{7}{3})^2 = -2\theta ^2+4\theta+\frac{2}{9}$

Let's go back the variance of estimate, $Var(\hat{\theta})=\frac{1}{4}\cdot \frac{\sigma^2}{n}=\frac{1}{40}(-4\theta ^2+4\theta +\frac{2}{9})$

Now substituting $\theta$ by its moment estimate $\hat{\theta}$, the estimated variance is

$Var(\hat{\theta})=\frac{1}{40}(-4\hat{\theta}^2+4\hat{}\theta +\frac{2}{9})$ =  <$\hat{\theta}=\frac{5}{12}$ from (a)>

           $=\frac{1}{40}(-4\theta ^2+4\theta +\frac{2}{9})=\frac{1}{40}\cdot \frac{172}{144}=0.029861$

$\therefore$ $SE(\hat{\theta})=\sqrt{Var(\hat{\theta})}=\sqrt{0.029861}=0.1728$