Question

QUESTION 3 17) Let Xi. X. X be a random sample from a distribution with probability density function f(x, ?) | ße_ß, for x >0 elsewhere (a) What is the likelihood (LU) = L (x1.X2. xalß)) of the sample? Simplify it. (b) Use the factorization criterion/theorem to show that ? x, is a sufficient statistic for . 4

Answer 1

According to the random sample of $X_{1},X_{2},.....X_{n}$ froma distribution with probability function ias

$f(x)=\beta e^{-\beta x}$ for $x\geq 0$

$=0$ elsewhere .

(a) Now the likelihood estimation is calculated as

$L(\beta\mid x )=\prod_{1}^{n}f(x,\beta )=\prod_{1}^{n}\beta e^{-\beta x}$

  $=\beta e^{-\beta x_{1}}\times \beta e^{-\beta x_{2}}\times \beta e^{-\beta x_{3}}\times ......\times \beta e^{-\beta x_{n}}$

  $=\beta^{n} e^{-\beta \sum_{1}^{n}x}$

Taking log both side we get

$logL(\beta\mid x )=log(\beta^{n} e^{-\beta \sum_{1}^{n}x})$

  $=nlog\beta -\beta \sum xloge......................(1)$

Taking derivative both side with respect to $\beta$ we get

$\frac{dlogL(\beta\mid x )}{d\beta }=\frac{n}{\beta }-\sum x=0$ first order condition of optimization.

$or,\frac{n}{\beta }=\sum x$

$or,\beta =\frac{n}{\sum x}=\frac{1}{\bar{x}}$ as $\bar{x}=\frac{\sum x}{n}$

Therefore likelihood of the sample is $\beta =\frac{1}{\bar{x}}$ (maximum likelihood estimator)

(b) According to the random sample of $X_{1},X_{2},.....X_{n}$ froma distribution with probability function , joint probability density function is

$f(x_{1},x_{2},.....x_{3},\beta )=f(x_{1},\beta )\times f(x_{2},\beta )\times f(x_{3},\beta )\times .....\times f(x_{n},\beta )$

Therefore by replacing $f(x)=\beta e^{-\beta x}$ we get

$f(x_{1},x_{2},.....x_{3},\beta )=\beta e^{-\beta x_{1}}\times \beta e^{-\beta x_{2}}\times \beta e^{-\beta x_{3}}\times .....\times \beta e^{-\beta x_{n}}$

$f(x_{1},x_{2},.....x_{3},\beta )=\beta^{n} e^{-\beta\sum_{i=1}^{n} x_{i}$

Now according to the factorization theorem we can factored the joint proabbility function as one which is depends on the parameter $\beta$ with the statistics $\sum_{i=1}^{n}X_{i}$ and another is independent on the parameter $\beta$ as

$f(x_{1},x_{2},.....x_{3},\beta )=[\beta^{n} e^{-\beta\sum_{i=1}^{n} x_{i}]\times 1$

Where $[\beta^{n} e^{-\beta\sum_{i=1}^{n} x_{i}]$ depends on parameter $\beta$ and is independent of parameter $\beta$ .

Therefore the factorization theorem shows that  $\sum_{i=1}^{n}X_{i}$is a sufficient statistics for $\beta$ and as $\bar{x}$ is one to one relationship with $\sum_{i=1}^{n}X_{i}$ , then $\bar{x}$ also a sufficient statistics for $\beta$ .