Question

2. Let X=(X,, X2, ,X") be a random sample with size n taken from population has (oo-ox0xc. Find the point estiator o by usin the method of moment

Answer 1

There is only one parameter $\theta$ , and by method of moments we know the sample mean is equal to the expectation of the random variable X.

Here sample mean is $\bar X={X_1+X_2+\cdots +X_n\over n} ={1 \over n} \sum_{i=1}^{n}X_i$

Also

$E(X) =\int_{0}^{1 }x\theta x^{\theta - 1}dx\\~~~\hspace {1cm} ~~~= \theta \left ( x^{\theta +1}\over \theta +1\right )_{0} ^{1} \\~~~\hspace {1 cm} ~~~ = 1-{1 \over \theta+1}$

So equating the two gives

${1 \over n} \sum_{i=1}^{n}X_i= 1-{1 \over \theta+1}\\ \Rightarrow {1 \over \theta +1}=1- \bar X\\ \Rightarrow \theta={1 \over 1-\bar X} - 1$

Thus the point estimator is

$\hat \theta={\bar X\over 1-\bar X}$