2
$\begingroup$

Suppose $X_i\overset{ind}{\sim}\mathcal{E}(\lambda_i)$, where $\lambda_i=(t_i\beta)^{-1}$, where $t_i$'s are positive known values and $\beta$ is positive unknown parameter. Here $i=1,\dots,n$.

It can be calculated that the MLE of $\beta$, $$\hat{\beta}_{MLE}=\frac{1}{n}\sum_{i=1}^n\frac{X_i}{t_i}$$

What is the confidence interval of $\beta$?

I have tried the likelihood ratio test, which gives the test statistic, namely $$\frac{p(\hat{\beta})}{p(\beta_0)}=\left(\frac{\hat{\beta}}{\beta_0}\right)^n\exp\left\{\left[\frac{1}{\beta_0}-\frac{1}{\hat{\beta}}\right]\sum_{i}^n\frac{X_i}{t_i}\right\},$$ and by taking log we get $$n[-\log(\hat{\beta})+\log(\beta_0)-1+\frac{\hat{\beta}}{\beta_0}].$$ I was trying to use the central limit theorem and delta method, but it seems not working. What can we do to prove that we can get the confidence interval for $\beta$?

$\endgroup$
3
  • $\begingroup$ MLE is probably asymptotically normal using $\sqrt n(\hat\beta-\beta)\stackrel{d}\to N(0,1/I(\beta))$. $\endgroup$ Mar 22, 2021 at 14:27
  • $\begingroup$ Thanks. But I am not sure if CLT can be useful here. $\endgroup$
    – Tan
    Mar 23, 2021 at 16:34
  • $\begingroup$ I was referring to the asymptotic normality of MLE under quite general conditions. $\endgroup$ Mar 23, 2021 at 20:22

1 Answer 1

0
$\begingroup$

I am answering my own question. Please correct me if I make any mistakes.

Continuing with $$T=n[-\log(\hat{\beta})+\log(\beta) - 1 +\frac{\hat{\beta}}{\beta}],$$ we can write $$T=n[\log\frac{\hat{\beta}}{\beta} - 1 +\frac{\hat{\beta}}{\beta}]$$

Now, let $$h(x)=\log(x)-1+x,$$ and we see $h(x)$ achieves the minimum when $x=1$. This means $T$ is the smallest when $\frac{\hat{\beta}}{\beta}=1$.

In order words, it is equivalent to accept the null hypothesis ($\beta=\beta_0$) when $T$ is small or when $|\frac{\hat{\beta}}{\beta}-1|$ is small.

Thus, the confidence interval of $\beta$ is simply

$$-c<\frac{\hat{\beta}}{\beta}-1<c,$$ which gives the confidence interval of $\beta$.

$\endgroup$
1
  • $\begingroup$ You will need to define $c$ in terms of the known quantities. Furthermore, there should probably be no $\beta_0$ - there is no null hypothesis behind the concept of a confidence interval. $\endgroup$
    – Michael M
    Mar 23, 2021 at 16:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.