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Let $X_1,...,X_n$, $n\geq5$ be a random sample from a distribution with the density function:

$f(x,\theta)= e^{-(x-\theta)}$ where $x\geq\theta$. Then, which of the following statements are correct:

  1. A $95%$% confidence interval of $\theta$ has to be of finite length.

  2. [$X_{(1)}+\frac{1}{n}ln(0.05),X_{(1)}]$ is a 95% confidence interval for $\theta$.

  3. A $95%$ confidence interval of $\theta$ can be of length 1.

  4. A $95%$ confidence interval of $\theta$ can be length 2.

Here, $X_{(1)}$ is the first ordered Statistic.

My approach

Clearly, the second option is correct. It can be checked by adjusting the term and integrating the density of first ordered Statistic. But I am not sure, what does the finite length mean. If I proceed from the interval, that we obtained from the second option, then the length of $95%$ Confidence interval is turning out to be $\frac{1}{n}ln(0.05)$. Also, the value of $n$ cannot be less than five. So, clearly the length cannot be 1 or two. Thus, we conclude that the only option that seam reasonable is $2$. Please check my steps and correct me if I am wrong. Thanks

userNoOne
  • 1,204
  • (2) looks like the best response. For the others, you might consider whether $[X_{(1)}+\frac{1}{n}\ln(0.05),\infty)$ or $[X_{(1)}+\frac{1}{n}\ln(0.05),X_{(1)}+\frac{1}{n}\ln(0.05)+1]$ or $[X_{(1)}+\frac{1}{n}\ln(0.05),X_{(1)}+\frac{1}{n}\ln(0.05)+2]$ could be called $95%$ confidence intervals by pedants – Henry Feb 20 '18 at 13:32
  • Actually, this question came in my exam. It was in multiple select based. It means, any combination of answer can be correct. I ticked second option only. Do you think it will be correct? Thanks – userNoOne Feb 20 '18 at 13:35

0 Answers0