Let $X_1,...,X_n$ random sample of $X$~$exp(\theta)$.

i) Find a exact confidence interval for $\theta$ with coefficient of confidence equal to $\gamma$

ii)Find a asymptotic confidence interval for $\theta$, with coefficient of confidence approximately $\gamma$

What I did

i)Let $Q(X;\theta)=2\theta\sum X_i$~$\chi^2_{2n}$ then $P(q_1\leq Q(X;\theta)\leq q_2)=\gamma$, that $q_1$ and $q_2$ are founded from values of the chi-square distribution. $$P(q_1\leq 2\theta\sum X_i\leq q_2)=P(\frac{q_1}{\sum X_i}\leq\theta\leq\frac{q_2}{\sum X_i})=\gamma$$ then $IC[\gamma;\theta]=[\frac{q_1}{\sum X_i};\frac{q_2}{\sum X_i}]$

ii) Here I am a little lost on how to proceed, I have to try to approach by the normal using delta or something method?

EDIT: If $\overline{X}$~$N(\frac{1}{\theta},\frac{1}{n\theta^2})$ if I take $\theta\overline{X}$ $$E[\theta\overline{X}]=\theta E[\frac{1}{n}\sum X_i]=\theta E[X_1]=\theta\frac{1}{\theta}=1$$ $$Var(\theta\overline{X})=\theta^2 Var(\overline{X})=\theta^2 Var(\frac{1}{n}\sum X_i)=\frac{\theta^2}{n^2}Var(\sum X_i)=\frac{\theta^2}{n}Var(X_1)=\frac{1}{n}$$

It's a $N(1,\frac{1}{n})$?

That's a guess not know if I can do this and not if it's right, waiting for someone more experienced.


The asymptotic confidence interval may be based on the (asymptotic) distribution of the mle. The Fisher information for this problem is given by $\frac{1}{\theta^2}$. Hence an asymptotic CI for $\theta$ is given by

$$\bar{X} \pm 1.96 \sqrt{\frac{\bar{X}^2}{n}}$$

where we have replaced $\theta^2$ by its mle, since we do not know the population parameter.

And here is a very simple R-simulation of the coverage for the case of a sample of size fifty from an exponential distribution with parameter $2$.

for(i in 1:1000){
if(1/2<=mle+qnorm(0.975)*sqrt((mle^2)/50) & 1/2>=mle+qnorm(0.025)*sqrt((mle^2)/50)){r[i]<-1}
 [1] 948
  • $\begingroup$ The approach I've done is wrong? As you built the confidence interval from fisher information? $\endgroup$
    – user72621
    Jun 6 '15 at 21:01
  • $\begingroup$ I'm a little lost on how to build asymptotic intervals, what I did was try to find a pivotal quantity with based on $\overline{X}$ $\endgroup$
    – user72621
    Jun 6 '15 at 21:08
  • 1
    $\begingroup$ @askazy Okay now I see what you have done. No it's not wrong. Pivot the quantity $\theta \bar{X}$ to get the CI in terms of $\theta$. $\endgroup$
    – JohnK
    Jun 6 '15 at 21:11
  • $\begingroup$ But this is also an asymptotic interval? This is my main question. $\endgroup$
    – user72621
    Jun 6 '15 at 21:22
  • 1
    $\begingroup$ @askazy Yes, this is also an asymptotic CI. $\endgroup$
    – JohnK
    Jun 6 '15 at 21:28

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