Question

Let (X1, ... , Xn) be an iid sample from the exponentially distributed X with pdf given by f(x;0) = -e ô, x > 0, 0 >0. Use the Neyman-Pearson Lemma to find the a-level most powerful test (MPT) of Ho : 0 = 2 vs H : 0 = 3.

Answer 1

Let us consider n random variable X1, X2, ... , Xn which follows pdf given by f(x,0) = exp(-6.0< x < 00.

For tesing the hypothesis

Ho:= 2 us H:8= 3

We have to obtain the likelihood function of the pdf and consequently the critical region is constructed based on the likelihood ration under the alternative and null hypothesis.

$Let f(x_1,x_2, ... ,x_n|\theta ) \ be \ the \ pdf \ of \ the \ distribution.\ Then \ the \ likelihood \ ratio \ of \ the \\ distribution \ is \ given \ by \\\lambda(x_1,x_2, ... ,x_n;\theta_0,\theta_1)= \frac{f(x_1,x_2, ... ,x_n|\theta_1 )}{f(x_1,x_2, ... ,x_n|\theta_0 )} \\\Rightarrow \lambda(x_1,x_2, ... ,x_n;\theta_0,\theta_1)=\frac{\left ( \prod_{i=1}^{n}\frac{1}{\theta_1}exp(-\frac{x}{\theta_1}) \right )}{\left ( \prod_{i=1}^{n}\frac{1}{\theta_0}exp(-\frac{x}{\theta_0}) \right )} \\\Rightarrow \lambda(x_1,x_2, ... ,x_n;\theta_0,\theta_1)=\frac{\left ( \frac{1}{\theta_1} \right )^nexp\left (\frac{ -\sum_{i=1}^{n}x_i }{\theta_1}\right )}{\left ( \frac{1}{\theta_0} \right )^nexp\left (\frac{ -\sum_{i=1}^{n}x_i }{\theta_0}\right )}>k$Therefore the critical region is given by

$C(x_1,x_2, ... ,x_n,\theta_0, \theta_1)=\left \{ {(x_1,x_2,...,x_n): \lambda(x_1,x_2, ... ,x_n;\theta_0,\theta_1)>k} \right \}$

$\\\lambda(x_1,x_2, ... ,x_n;2,3)= \frac{f(x_1,x_2, ... ,x_n|3 )}{f(x_1,x_2, ... ,x_n|2 )} \\\Rightarrow \lambda(x_1,x_2, ... ,x_n;2,3)=\frac{\left ( \prod_{i=1}^{n}\frac{1}{3}exp(-\frac{x}{3}) \right )}{\left ( \prod_{i=1}^{n}\frac{1}{2}exp(-\frac{x}{2}) \right )} \\\Rightarrow \lambda(x_1,x_2, ... ,x_n;2,3)=\frac{\left ( \frac{1}{3} \right )^nexp\left (\frac{ -\sum_{i=1}^{n}x_i }{3}\right )}{\left ( \frac{1}{2} \right )^nexp\left (\frac{ -\sum_{i=1}^{n}x_i }{2}\right )}>k \\\lambda(x_1,x_2, ... ,x_n;2,3)=\left ( \frac{2}{3} \right )^nexp\left (\frac{ -\sum_{i=1}^{n}x_i }{3}+\frac{ \sum_{i=1}^{n}x_i }{2} \right) \\\\\lambda(x_1,x_2, ... ,x_n;2,3)=\left ( \frac{2}{3} \right )^nexp\left (\frac{ \sum_{i=1}^{n}x_i }{6} \right)$

Therefore the critical region is given by

$C(x_1, x_2, ... ,x_n ,2,3)=\left \{ (x_1, x_2, ... ,x_n): \lambda(x_1, x_2, ... ,x_n)>k \right \}\\ C(x_1, x_2, ... ,x_n ,2,3)=\left \{ (x_1, x_2, ... ,x_n: \left ( \frac{2}{3} \right )^nexp\left (\frac{ \sum_{i=1}^{n}x_i }{6} \right)>k \right \} \\\Rightarrow C(x_1, x_2, ... ,x_n ,2,3)=\left \{(x_1, ... ,x_n)= \sum_{i=1}^{n}x_i >k_1\right \}$

where we have

$\left ( \frac{2}{3} \right )^nexp\left (\frac{ \sum_{i=1}^{n}x_i }{6} \right)>k \\ exp\left (\frac{ \sum_{i=1}^{n}x_i }{6} \right)>k\left ( \frac{3}{2} \right )^n=a \\ Taking \ log \ on \ both \ side \\ logexp\left (\frac{ \sum_{i=1}^{n}x_i }{6} \right)>loga \\\frac{ \sum_{i=1}^{n}x_i }{6} >loga \\\sum_{i=1}^{n}x_i >6loga=k_1 \\\sum_{i=1}^{n}x_i >k_1$

where $k_1$ is obtained by size condition with

$P(\sum_{i=1}^{n}x_i>k_1)=\alpha.$