Confidence interval of the parameter of exp and normal distribution from MLE?I have a sample X1,X2,…,Xn1. If the sample is from exponential distribution, I want to use MLE to estimate the parameter β. I know the result thatβ^=X1+X2+…+XnnBut 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^=[n−1n⋅S2(n)]12But how can I calculate the 95% confidence interval of μ and σ?Step1 to calculate confidence interval, θ^±z1−α2δ(θ^)nStep2 to calculate confidence intervalδ(θ^)=−n(E[d2dθ2ln⁡L(θ)])−1I know the equation, but I am still confuzed how to calculate(E[d2dθ2ln⁡L(θ)])

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Confidence interval of the parameter of exp and normal distribution from MLE?I have a sample X1,X2,…,Xn1. If the sample is from exponential distribution, I want to use MLE to estimate the parameter β. I know the result thatβ^=X1+X2+…+XnnBut 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^=[n−1n⋅S2(n)]12But how can I calculate the 95% confidence interval of μ and σ?Step1 to calculate confidence interval, θ^±z1−α2δ(θ^)nStep2 to calculate confidence intervalδ(θ^)=−n(E[d2dθ2ln⁡L(θ)])−1I know the equation, but I am still confuzed how to calculate(E[d2dθ2ln⁡L(θ)])

Rachettolyf8 · Accepted Answer

The term you cannot calculate is (essentially) the Fischer information:I(θ)=−E[d2dθ2ln⁡L(θ)]Then&amp;nbsp;δ(θ)=n(I(θ))−1. So, to calculate&amp;nbsp;I(θ)&amp;nbsp;for the exponential:KL(θ)=∏j=1nθe-θxj=θne-θ∑j=1nxjand therefore&amp;nbsp;lnL(θ)=ln(θne-θ∑j=1nxj)=nln(θ)-θ∑j=1nxjwhich you can differentiate twice with respect to&amp;nbsp;θ:u1&#039;y1+u2&#039;y2=0u1&#039;y1&#039;+u2&#039;y2&#039;=f(x).So,&amp;nbsp;I(θ)=−E[d2dθ2ln⁡L(θ)]=nθ2and therefore&amp;nbsp;δ(θ)=n(n/θ2)-1=θ2. To make calculations simpler use thatIf&amp;nbsp;Xj&amp;nbsp;are i.i.d. for&amp;nbsp;j=1,2,…,n&amp;nbsp;you can take the&amp;nbsp;I(θ)&amp;nbsp;for a single observation&amp;nbsp;Xj&amp;nbsp;and obtain the Fisher information for X with&amp;nbsp;nI(θ).For the normal distribution&amp;nbsp;N(μ,θ2)&amp;nbsp;you should find:&amp;nbsp;I(θ)=n2θ2&amp;nbsp;(single parameter) and if you estimate both parameters, i.e.&amp;nbsp;Ν(μ,σ2)&amp;nbsp;with&amp;nbsp;θ=[μ,σ2]Τ&amp;nbsp;then&amp;nbsp;I(θ)&amp;nbsp;is a matrix:I(θ)=12σ20012σ4

Confidence interval of the parameter of exp and

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