"We have the Random Variable X, which is Gamma (p,lambda) distributed with the density: fp,lambda (x)=lambda p Gamma(p)*xp−1*e−lambda x with p=10 and H_0:lambda=2 or H_1:lambda=4 and alpha=0.001

Winston Todd 2022-10-18 Answered
We have the Random Variable X, which is Γ ( p , λ ) distributed with the density:
f p , λ ( x ) = λ p Γ ( p ) x p 1 e λ x
with p=10 and H 0 : λ = 2 or H 1 : λ = 4 and α = 0.001
I want to apply the Lemma of Neyman Pearson which states:
Be c>0 fixed and chosen in the way that A ( c ) = { x B : f 0 ( x ) f 1 ( x ) c } such that P H 0 ( X A ( c ) ) = α
Then the test with the region A(c) among all tests with significance level α is the most powerful.
I am now trying to calculate A(c), but got stuck. I have:
A ( c ) f 0 ( x ) d x = A ( c ) λ p Γ ( p ) x p 1 e λ x d x = α .
But I don't know how to get A(c) from this integral...
f 0 ( x ) f 1 ( x ) = 1 1024 e 2 x
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Answers (1)

Phillip Fletcher
Answered 2022-10-19 Author has 21 answers
For the hypotheses H 0 : λ = λ 0 and H 1 : λ = λ 1 you have:
ln f 0 ( x ) ln f 1 ( x ) = p ( ln λ 0 ln λ 1 ) x ( λ 0 λ 1 ) .
Hence, we have:
d d x ( ln f 0 ( x ) ln f 1 ( x ) ) = λ 1 λ 0 .
If λ 1 < λ 0 then the likelihood ratio is an increasing function of x so the region A(c) is of the form [ x c , ). (Alternatively, if λ 1 < λ 0 then the log-likelihood ratio is a decreasing function of x so the region A(c) is of the form [ 0 , x c ].) In the former case we have:
α = P ( X A ( c ) | H 0 ) = λ 0 p Γ ( p ) x c x p 1 e λ 0 x   d x = λ 0 Γ ( p ) λ 0 x c t p 1 e t   d t = λ 0 Γ ( p , λ 0 x c ) Γ ( p ) ,
where the last line in this equation uses the incomplete gamma function. Hence, the boundary value x c is the value that solves:
Γ ( p , λ 0 x c ) = α λ 0 Γ ( p ) .
This value can be obtained numerically from the incomplete gamma function.
Application to your example: In your question you have p=10, λ 0 = 2 and λ 1 = 4. Since λ 1 > λ 0 the likelihood ratio is an increasing function of x and so you have A ( c ) = [ x c , ) with the boundary value x c solving:
Γ ( 10 , 2 x c ) = α 2 Γ ( 10 ) .
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