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I have a sample $X_1,X_2,\ldots,X_n$

  1. If the sample is from exponential distribution, I want to use MLE to estimate the parameter $\beta$. I know the result that $$\hat{\beta}=\frac{X_1+X_2+\ldots+X_n}{n}$$But how can I calculate the $95\%$ confidence interval of $\beta$?
  2. If the sample is from normal distribution, I want to use MLE to estimate the parameter $\mu$ and $\sigma$. I also know the result that $$\hat{\mu}=\frac{X_1+X_2+\ldots+X_n}{n}, \quad \hat{\sigma}=\left[\frac{n-1}{n}\cdot S^2(n)\right]^{1/2}$$ But how can I calculate the $95\%$ confidence interval of $\mu$ and $\sigma$?

    1. Step1 to calculate confidence interval, $$\hat{θ}\pm z_{1-\frac\alpha2}\sqrt{\frac{\delta(\hat{θ})}{n}}$$
    2. Step2 to calculate confidence interval $$\delta(\hat{θ})=-n\left(E\left[\frac{d^2}{dθ^2}\ln \mathcal L(θ)\right]\right)^{-1}$$

I know the equation, but I am still confuzed how to calculate $$\left(E\left[\frac{d^2}{dθ^2}\ln \mathcal L(θ)\right]\right)$$

I want to know the equation in the red frame of this picture. enter image description here

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  • $\begingroup$ I get the sigma hat from my text book and i also calculate to get this. i think that my MLE is right. But i don't know exactly how to calculate confidence interval. $\endgroup$ – 周庆特 Dec 26 '15 at 10:07
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The term you cannot calculate is (essentially) the Fischer information:

$$I(θ)=-E\left[\frac{d^2}{dθ^2}\ln \mathcal L(θ)\right]$$

Then $δ(θ)=n(I(θ))^{-1}$. So, to calculate $I(θ)$ for the exponential:

$$\mathcal L(θ)=\prod\limits_{j=1}^nθ e^{-θx_j}=θ^n e^{-θ\sum_{j=1}^{n}x_j}$$ and therefore $$\ln \mathcal L(θ) = \ln{\left(θ^n e^{-θ\sum_{j=1}^{n}x_j}\right)} = n\ln{(θ)}-θ\sum_{j=1}^{n}x_j$$ which you can differentiate twice with respect to $θ$: \begin{align}\frac{d}{dθ} \ln (\mathcal L(θ)) &= \frac{n}{θ}-\sum_{j=1}^n x_j \\ \frac{d^2}{dθ^2} \ln (\mathcal L(θ)) &= -\frac{n}{θ^2}\end{align} So, $$I(θ)=-E\left[\frac{d^2}{dθ^2}\ln \mathcal L(θ)\right]=\frac{n}{θ^2}$$ and therefore $δ(θ)=n(n/θ^2)^{-1}=θ^2$. To make calculations simpler use that

If $X_j$ are i.i.d. for $j=1,2,\ldots,n$ you can take the $I(θ)$ for a single observation $X_j$ and obtain the Fisher information for $X$ with $nI(θ)$.


For the normal distribution $N(\mu,θ^2)$ you should find: $I(θ)=n/(2θ^2)$ (single parameter) and if you estimate both parameters, i.e. $Ν(μ,σ^2)$ with $θ=[μ,σ^2]^Τ$ then $I(θ)$ is a matrix: $$I(θ)=\begin{bmatrix}\frac{1}{2σ^2}&0 \\ 0& \frac{1}{2σ^4}\end{bmatrix}$$

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  • $\begingroup$ I use the parameter beta of exp distribution to do the fisher information. And i differcentiate twice with respect to beta, and i get the equation that still contain X1+X2+,...,+Xn. How should i do with X1+X2+,...,+Xn? $\endgroup$ – 周庆特 Dec 26 '15 at 11:58
  • $\begingroup$ You should do the differentiation again, because the $X_i$'s go away if you do it correctly. It is in my answer above. But even if they did not go away, that would not be a problem. This $X_i$'s are your sample and therefore known. $\endgroup$ – Jimmy R. Dec 26 '15 at 13:08
  • $\begingroup$ I add my additional problem in my original question. That is from my text book. Can you help me explain that red frame in my picture? $\endgroup$ – 周庆特 Dec 26 '15 at 13:16
  • $\begingroup$ Yes, $E[X_i]=\frac{1-p}{p}$ when $X$ geometric, so by linearity $E[\sum X_i]=n(1-p)/p$. The author is missing an $E$ symbol before the sum. But please: do not edit so much an already posted question. A, and stop posting the same question again. $\endgroup$ – Jimmy R. Dec 26 '15 at 13:21
  • $\begingroup$ I'm sorry. I shouldn't change my problem so much and shouldn't post that problem again. But i still wonder that if we meet ∑Xi in the calculation proccess of fisher information, we have to replace E(∑Xi) with the mean of that predetermined distribution? $\endgroup$ – 周庆特 Dec 26 '15 at 13:27

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