For fixed k, the variables \(\displaystyle{\left({2}{k}+{1}\right)}\hat{{{f}}}_{{k}}{\left(\lambda\right)}\) are

Alessandra Carrillo

Alessandra Carrillo

Answered question

2022-04-07

For fixed k, the variables (2k+1)f^k(λ) are asymptotically distributed according the Gamma distribution with shape parameter 2k+1 and mean (2k+1)fX(λ). It turns out that this suggests a confidence interval of the form
((4k+2)f^k(λ)χ{4k+2,1α}2,(4k+2)f^k(λ)χ{4k+2,α}2)
where χ{k,α}2 the α-quantile of the chi-square distribution with k degrees of freedom.
To show that this interval is asymptotically of level 12α

Answer & Explanation

Marin Lowe

Marin Lowe

Beginner2022-04-08Added 18 answers

Step 1
The mean of Γ(α,β) equals to
αβ=2k+1β=(2k+1)fX(λ)  β=1fX(λ).
Rewrite your statement as
(2k+1),f^k(λ)xdΓ(2k+1,1fX(λ)).
Use the PDF of Γ(α,β)
fY(x)=dβαΓ(α)xα1eβx,,x>0
and check that the multiplying YΓ(α,β) by c>0 results in changing β to βc only: cYΓ(α,βc)
So, in order to get asymptotically chi-squared distribution
χ4k+22=Γ(2k+1,12),
divide original estimator
(2k+1)f^k(λ) by fX(λ) and multiply by 2:
2(2k+1),f^k(λ)fX(λ)xdΓ(2k+1,12)=χ4k+22.
((4k+2)f^k(λ)χ4k+2,1α2fX(λ)(4k+2)f^k(λ)χ4k+2,α2)=
=(χ4k+2,α2(4k+2)f^k(λ)fX(λ)dχ4k+22χ4k+2,1α2)12α

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