Confidence interval of the parameter of exp and

svrstanojpkqx

svrstanojpkqx

Answered question

2022-03-25

Confidence interval of the parameter of exp and normal distribution from MLE?
I have a sample X1,X2,,Xn
1. If the sample is from exponential distribution, I want to use MLE to estimate the parameter β. I know the result that
β^=X1+X2++Xnn
But how can I calculate the 95% confidence interval of β?
2. If the sample is from normal distribution, I want to use MLE to estimate the parameter μ and σ. I also know the result that μ^=X1+X2++Xnn,σ^=[n1nS2(n)]12
But how can I calculate the 95% confidence interval of μ and σ?
Step1 to calculate confidence interval,
θ^±z1α2δ(θ^)n
Step2 to calculate confidence interval
δ(θ^)=n(E[d2dθ2lnL(θ)])1
I know the equation, but I am still confuzed how to calculate
(E[d2dθ2lnL(θ)])

Answer & Explanation

Rachettolyf8

Rachettolyf8

Beginner2022-03-26Added 9 answers

The term you cannot calculate is (essentially) the Fischer information:
I(θ)=E[d2dθ2lnL(θ)]
Then δ(θ)=n(I(θ))1. So, to calculate I(θ) for the exponential:
KL(θ)=j=1nθe-θxj=θne-θj=1nxj
and therefore lnL(θ)=ln(θne-θj=1nxj)=nln(θ)-θj=1nxj
which you can differentiate twice with respect to θ:
u1'y1+u2'y2=0u1'y1'+u2'y2'=f(x).
So, I(θ)=E[d2dθ2lnL(θ)]=nθ2
and therefore δ(θ)=n(n/θ2)-1=θ2. To make calculations simpler use that
If Xj are i.i.d. for j=1,2,,n you can take the I(θ) for a single observation Xj and obtain the Fisher information for X with nI(θ).
For the normal distribution N(μ,θ2) you should find: I(θ)=n2θ2 (single parameter) and if you estimate both parameters, i.e. Ν(μ,σ2) with θ=[μ,σ2]Τ then I(θ) is a matrix:
I(θ)=12σ20012σ4

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