Question

5. Find a method-of-moments estimator (MME) of θ based on a random sample XI, , X, from each of the following distributions (a) f(z; θ)-0( 1-0)1-1 , x-1, 2, . . . . 0 (b) f(z; 0) = (0 + 1)2-0-2, x > 1,0 > 0 (c) fr) re, 0, θ 1

Answer 1

(a)

It is geometric distribution so first population moment is

Suppose X1, X2, X3,...Xn are the independent random variables from the geometric distribution. Then first sample moment is

$M_{1}=\frac{X_{1}+X_{2}+...+X_{n}}{n}=\bar{X}$

Using the method of moments estimates, equating sample moments to distribution moments we have

$E(X)=M_{1}$

$heta=rac{1}{ar{X}}$

hence, required estimate is

(b)

Let us first find the mean of X. So

Now the sample mean is

$\bar{x}=\frac{x_{1}+x_{2}+...+x_{n}}{n}=\frac{1}{n}\sum_{i=1}^{n}x_{i}$

Since according to method of moments estimators, first theortical moment is equal to first sample moment so we have

$\mu=\bar{x}$

hence, required estimate is

(c)

The mean of the pdf is

Let X1, X2, X3....Xn is a random sample from this distribution. So sample mean is

$\bar{X}=\frac{X1+X2+...Xn}{n}$

Since according to method of moments estimators, first theoretical moment is equal to first sample moment so we have

20-1-1

$heta=2/ar{X}$

So required estimate is