Question

- Let X1, X2, ..., Xn be iid from the pdf fe(x) = 0e-82, > 0. Note that T = 2 , X, is a sufficient statistic. Consider testing the hypothesis H.: 8 = 1 vs H: 8 = 2 using Bayes method. Suppose the prior distribution is P(0 = 1) = ? and P(0 = 2) = 1 - . (a) Show that the Bayes test rejects H, if T < In log(2) + log((1 - ))) (b) Take n = 50. Calculate numerically the power of the Bayes test at 0 = 1 and 0 = 2 for values of = 0.2 0.4 0.6,0.8. How does the power change with the prior probability ?

Answer 1

a) Here in this case we first need to calculate the posterior probability using the formula

$P(H_0|X=x)=\frac{f_X(x|H_0)P(H_0)}{f_X(x)}\\P(H_1|X=x)=\frac{f_X(x|H_1)P(H_1)}{f_X(x)}$

One way to decide about between H0 and H1 is by comparing $P(H_0|X=x)~and~P(H_1|X=x)$ and accept the hypothesis with higher posterior probability. This is basically the Maximum Aposteriori Test (MAP). So according to MAP we reject H0 iff

$P(H_0|X=x)<P(H_1|X=x)$

In this case the

$P(H_0|X=x)=\frac{1^ne^{-1T}\times\pi}{f_X(x)}\\P(H_0|X=x)=\frac{2^ne^{-2T}\times(1-\pi)}{f_X(x)}$

Taking the above inequality we get

e-17 x 2"-27 x (1 – 7) Taking log on both sides to get T + log a <n log 2 – 2T + log(1 – 7) 1-TT T< [n log 2] + log

b) In order to calculate the power of the test, we use this rejection condition as proved above where T follows exponential distribution with $n\theta$ as parameter. So the power will be

$P(T<t|H_1)=\int_{0}^{t}n\theta e^{-n\theta x}dx=1-e^{-n\theta t},where~t=[n\log 2]+\log(\frac{1-\pi}{\pi})\\$

Hence for $\pi$ =0.2,0.4,0.6,0.8 and n=50 and $\theta=2$

The t values corresponding to the above $\pi$ values is calculated as

36.04365,35.06282,34.25189,33.27106 and hence by putting the values of t we get power as 1 for all the values. It is mostly due to large sample size.

For $\theta=1$ also the power is coming 1 owing to the large sample size