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 = 0.001I want to apply the Lemma of Neyman Pearson which states:Be c&gt;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 ) ) = αThen 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

Question

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  =  0.001I want to apply the Lemma of Neyman Pearson which states:Be c&amp;gt;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  )  )  =  αThen 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&#039;t know how to get A(c) from this integral...                    f        0            (      x      )                      f        1            (      x      )        =      1    1024    ⋅      e          2      x

Phillip Fletcher · Accepted Answer

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    &amp;lt;      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    &amp;lt;      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    &amp;gt;      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  )  .

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

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