# Confidence interval for exponential distribution

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 $2$.

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

• The approach I've done is wrong? As you built the confidence interval from fisher information?
– user72621
Commented Jun 6, 2015 at 21:01
• I'm a little lost on how to build asymptotic intervals, what I did was try to find a pivotal quantity with based on $\overline{X}$
– user72621
Commented Jun 6, 2015 at 21:08
• @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 $\theta$. Commented Jun 6, 2015 at 21:11
• But this is also an asymptotic interval? This is my main question.
– user72621
Commented Jun 6, 2015 at 21:22
• @askazy Yes, this is also an asymptotic CI. Commented Jun 6, 2015 at 21:28

For ii) you have several options but two of them are particularly appealing. One option is to go for a Wald-type confidence interval the other is to go with an interval based on the log-likelihood ratio statistic.

For the Wald-type you first get the limiting distribution of the MLE of $$\theta$$, thus something along the line $$\hat \theta \sim N(\theta, \text{se}(\hat\theta)^2)$$, where $$\text{se}(\hat\theta)^2$$ is the estimated variance of the distribution of $$\hat\theta$$, the MLE of $$\theta$$. The Wald-type interval of approximate confidence level $$1-\alpha$$ is (the usual)

$$\hat\theta \pm z_{\alpha/2}\text{se}(\hat\theta).$$ where $$z_{\alpha/2}$$ is s.t. $$P(Z\leq z_{\alpha/2}) = \alpha/2$$ and $$Z\sim N(0,1)$$. This is what JohnK is suggesting in his answer.

For the confidence interval based on the log-likelihood ratio statistic, you find the full explanation here with the associated R code. You'll have to adapt it to your situation; i.e. just replace the log-likelihood function used there with yours. Note that with this method you'll have to use a computer to compute the interval since the method entails the inversion of a non-linear function.

These two confidence intervals are approximate in that the coverage probability for a given fixed sample will not typically be exactly $$1-\alpha$$; nor will it be always $$\geq 1-\alpha$$. But, the guarantee is that for increasing sample size, we expect the coverage probability to converge at $$1-\alpha$$.

I will explain a bit more deeply the process @JohnK gone through:

$$\hat\lambda_\text{MLE} = \frac{1}{\bar{X}}$$

$$l''=\log(L)'' = \frac{-n}{\lambda^2}$$

Therefore, $$-E[l''] =$$ Fisher's Information $$= \frac{1}{\lambda^2}$$. ($$n=1$$ for a single sample variance)

Now: $$I(\hat{\lambda}_\text{MLE}) = \bar{X}^2$$

The variance of the estimator: $$\frac{1}{I({\lambda})n } = \frac{1}{\frac{n}{\lambda^2}}=\frac{1}{\frac{n}{\hat{\lambda}_\text{MLE}^2}} = \frac{\bar{X}^2}{n}$$