Question

Question 1: (10 marks) Let Y, Y....,Y, be a random sample from the beta distribution with a = B = 4, and I2 = { u u = 1,2). Write the likelihood ratio test statistic A for testing Ho : H = 1 versus H:u= 2. Note that the pdf of a beta(a,b) distribution is as follows: com_(a+b)/2-1(1 - 0)8-1, 0<I<1. f(x) = f(a)(B)"

Answer 1

Let x1, x2, x3,..., xn be a random sample from a beta distribution with parameter $(\alpha ,\beta )$ .

Suppose that $\alpha$ =$\beta$ = $\mu$ . To decide between two simple hypotheses

H₀: $\mu$ =1,

H₁: $\mu$ =2

One way to decide between H₀ and H₁ is to compare the corresponding likelihood functions:

l₀=L(x1,x2,...,xn ; $\mu$ 0 ) ,

l₁=L(x1,x2,...,xn ; $\mu$ 1 ).

More specifically, if l₀ is much larger than l₁, we should accept H₀. On the other hand if l₁ is much larger, we tend to reject H₀. Therefore, we can look at the ratio of l₀ and l₁ to decide between H₀ and H₁. This is the idea behind likelihood ratio tests

we define ratio

(x1,x2,...,xn)
= l₀/l₁ =   
L(x1,x2,..., xn;O) L(x1,x2,...,xn;/1)
    =
Lx1,x2,...,xN; 10)
   /
Lx1,x2,...,xN;41)

To perform a likelihood ratio test (LRT), we choose a constant c. We reject H₀ if $\lambda$ <c and accept it if $\lambda$ $\geq$ c. The value of c can be chosen based on the desired $\alpha$ .

We have given Beta distribution with $\alpha$ =$\beta$ = $\mu$

$\lambda$ = l₀ / l₁

For H_0: $\mu$ =1 thus   l₀ =
Г(а+ 3)x=-1(1 – x) 3-1 г(а)Г (3)
=
1(1+1)x1-(1 - x)- (1)(1)
= $\Gamma( 2 ) .1$

                                  l₀ = 1

and H_1: $\mu$ =2 thus   l₁ =
Г(а+ 3)x=-1(1 – x) 3-1 г(а)Г (3)
   =
(2+2) x2-'(1 - x)2- (2)(2)

                                    =

r(4)x'(1 – x)' 1* 1

Now $\Gamma(4 )$ = 6

Thus l₁ = 6 * x * (1 - x )

Hence $\lambda$ = l₀ / l₁    = 1 / 6 * x * (1 - x )

Now we accept H₀ if $\lambda$ $\geq$ c

        $\lambda$ = 1 / 6 * x * (1 - x ) $\geq$ c

              $\rightarrow$ 6 * x * (1 - x ) $\leq$ 1 / c

              $\rightarrow$ x - x² $\leq$ 1 / (6c)

where c is the threshold

Thus, we accept H0 if

x * (1 - x ) $\leq$ 1 / (6c)

Let us define c'=1 / (6c), where c' is a new threshold.

Remember that x is the observed value of the random variable x.

Thus, we can summarize the decision rule as follows. We accept H₀ if

x * (1 - x )≤c'.

To choose c' we use the required Prob ( Type I error ) = alp ( given level of significance )

P(type I error) = P( Reject H₀ | H₀ : $\mu$ =1 ) = P( [x * (1 - x )] > c' | H0 ) = alp