Question

~ fo(x) On the basis of observing , find the most powerful size versus Hi: X~ fi(x), where fo(x) is the standard normal pdf and 0.05 test of H0 : that is fi(x) is the standard LaPlace pdf. Can you redo this problem when you have observed X1, X2 iid?

Answer 1

$MP~level~0.05~critical~region:~\{x:~f_1(x)>kf_0(x)\}\\ where~f_1(x)>kf_0(x)\\ or,~\frac{1}{2}e^{-|x|}>k\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\\ or,~e^{\frac{1}{2}(|x|^2-2|x|&plus;1)}>k^*\\ or,~(|x|-1)^2>k^{**}\\ or,~|x|>c_1~or~|x|<c_2\\ where~c_1~and~c_2~are~obtained~from~size~condition~i.e.\\ P(|X|>c_1|H_0)&plus;P(|X|<c_2|H_0)=0.05.\\ Therefore~the~level~0.05~MP~critical~region:~\{X:|X|>c_1~or~|X|<c_2\}\\ For~X_1,~X_2\\ MP~level~0.05~critical~region:~\{x:~f_1(x_11,x_2)>kf_0(x_1,x_2)\}\\ where~f(x_1,x_2)~is~the~joint~distribution~of~X_1,X_2.\\ f_1(x_1,x_2)>kf_0(x_1,x_2)\\ or,~\frac{1}{2^2}e^{-|x_1|-|x_2|}>k\frac{1}{2\pi}e^{-\frac{x_1^2&plus;x_2^2}{2}}\\ or,~e^{\frac{1}{2}(|x_1|^2-2|x_1|&plus;1)&plus;(|x_2|^2-2|x_2|&plus;1)}>k^*\\ or,~(|x_1|-1)^2&plus;(|x_2|-1)^2>k^{**}\\ where~k^{**}~is~obtained~from~size~condition.$