Question

5. Let X1, X2, ..., Xn be a random sample from a distribution with pdf of f(x) = (@+1)xº,0<x<1. a. What is the moment estimator for 0 using the method of moments technique? b. What is the MLE for @ ?

Answer 1

(a)

Let us first find the mean of X. So

$\mu=E(X)=\int_{0}^{1}xf(x)dx=\int_{0}^{1}(\theta &plus;1)x^{\theta&plus;1 }dx=\left [ \frac{(\theta&plus;1) x^{\theta &plus;2}}{\theta &plus;2} \right ]_{0}^{1}=\frac{\theta&plus;1 }{\theta &plus;2}$

Now the sample mean is

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

According to method of moments estimators, first theoretical moment is equal to first sample moment so from above we have

$\mu=\bar{x}$

$\frac{\theta&plus;1 }{\theta &plus;2}=\bar{x}$

$\theta &plus;1=\bar{x}\theta &plus;2\bar{x}$

$\theta =\frac{2\bar{x}-1}{1-\bar{x}}$

Hence, required estimate MOM is

$\tilde{\theta }=\frac{2\bar{x}-1}{1-\bar{x}}$

(b)

From pdf we have

$f(x_{1})=(\theta &plus;1)(x_{1})^{\theta}$

$f(x_{2})=(\theta &plus;1)(x_{2})^{\theta}$

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$f(x_{n})=(\theta &plus;1)(x_{n})^{\theta}$

The likelihood function is

$L(\theta)=f(x_{1})f(x_{2})f(x_{3})...f(x_{n})=(\theta&plus;1)^{n} \left ( \prod_{i=1}^{n}x_{i} \right )^{\theta}$

Taking log of both sides of above equation gives

$ln\left [L(\theta) \right ]=nln\left [\theta&plus;1\right ]&plus;(\theta) \sum_{i=}^{n}ln(x_{i})$

Now differentiating both sides gives

$\frac{d}{d\theta}ln\left [L(\theta) \right ]=\frac{n}{\theta&plus;1}&plus;\sum_{i=1}^{n}ln(x_{i})$

Equating it equal to zero gives

$\frac{n}{\theta&plus;1}&plus;\sum_{i=1}^{n}ln(x_{i})=0$

$\theta=-\left [1&plus;\frac{n}{\sum_{i=1}^{n}ln(x_{i})} \right ]$

Hence, required MLE is

$\tilde{\theta}=-\left [1&plus;\frac{n}{\sum_{i=1}^{n}ln(x_{i})} \right ]$