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 $\beta$?

  • $\begingroup$ 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


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 $\beta$.

  • $\begingroup$ 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

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