Answer to Question #253056 in Statistics and Probability for adeq

Question

a) Suppose that 𝑋1,𝑋2,...,𝑋𝑛 is a random sample from a distribution with

probability density function:

𝑓(𝑥;𝜆)={

1

6𝜆4𝑥3𝑒−𝑥

𝜆, 𝑖𝑓 𝑥 ≥0 𝑎𝑛𝑑 𝜆>0

0, 𝑒𝑙𝑠𝑒 𝑤ℎ𝑒𝑟𝑒

i) Show whether or not 𝑓(𝑥;𝜆) belongs to the 1-parameter exponential family.

(5)

ii) Derive the Maximum likelihood estimator of 𝜆. (8)

iii) Suppose that 𝑛=200, ∑ 𝑥𝑖200𝑖=1 =20,∑ 𝑥𝑖2200𝑖=1 =100, and ∑ 𝑥𝑖3200𝑖=1 =250.

Determine the Maximum likelihood estimate of 𝜆. (2)

Answer 1

i)

$\int_0^\infty x^3e^{-x\lambda}dx=-\frac{1}{\lambda}x^3e^{-x\lambda}|_0^\infty+\frac{3}{\lambda}\int_0^\infty x^2e^{-x\lambda}dx=0-\frac{6}{\lambda^2}x^2e^{-x\lambda}|_0^\infty+\frac{6}{\lambda^2}\int_0^\infty xe^{-x\lambda}dx=0-\frac{6}{\lambda ^3}xe^{-x\lambda}|_0^\infty+\frac{6}{\lambda^3}\int_0^\infty e^{-x\lambda}dx=-\frac{6}{\lambda ^4}e^{-x\lambda}|_0^\infty=0+\frac{6}{\lambda ^4}=\frac{6}{\lambda ^4}$

Therefore $\int_0^\infty(\frac{1}{6}\lambda^4 x^3 e^{-x\lambda})dx=1$

I.e. $p(x.\lambda)=\frac{1}{6}\lambda^4 x^3 e^{-x\lambda},x>0$ is exponential 1 parametric distribution.

2) Maximum likehood estimator of $\lambda$ :

$L(\lambda)=p(x_1,...,x_n|\lambda)=\prod_{i=1}^np(x_i\lambda)=\frac{\lambda^{4n}}{6^n}(\prod_{i=1}^{n}x_i)^3e^{-\lambda\sum_1^nx_i}$

$lnL(\lambda)=4n\cdot ln(\lambda)-n\cdot ln(6)+3ln$ $(\prod_{i=1}^{n}x_i)$- $\lambda\sum_1^nx_i$ ;

$(lnL(\lambda))'=\frac{4n}{\lambda}-$ $\sum_{i=1}^nx_i=0;$

$\lambda=\frac{4n}{\sum_{i=1}^nx_i}=\frac{4}{\overline x}$ - the likehood estimator.

3)

$\hat\lambda=\frac{4\cdot 200}{\sum_{i=1}^{200}x_i}=\frac{800}{20}=40$