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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.

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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 $2$.

r<-rep(0,1000)
for(i in 1:1000){
  x<-rexp(50,2)
mle<-mean(x)
if(1/2<=mle+qnorm(0.975)*sqrt((mle^2)/50) & 1/2>=mle+qnorm(0.025)*sqrt((mle^2)/50)){r[i]<-1}
}
sum(r==1)
 [1] 948
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  • $\begingroup$ 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
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    $\begingroup$ @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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