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I am trying to solve this problem: Given $X_1, \ldots, X_n$ a random sample of a population of random variables with p.d.f. $f(x, \theta) = e^{-(x-\theta)} I(x)_{x\geq\theta}$, find a confidence interval of level $1-\alpha$ for $\theta$.

Following a hint, I first looked into the distribution of $Y = \text{min}(X_i) - \theta$, which I found to be Exponential($n$).

Taking into account that the indicator function in the pdf limits to values of $x$ greater than $\theta$, and using the exponential's cdf. and some bounds, I got that the confidence interval should be $I_{1-\alpha}(X_1, \ldots, X_n) = \left[0, \frac{\ln(\alpha)}{n}\right]$.

Would that be correct? I am not sure I am using the hint correctly. If it is wrong, what else could I try?

Thanks a lot.

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  • $\begingroup$ I think it is not correct, because for $0<\alpha<1$, $ln(\alpha)$ is negative. $\endgroup$ – ALPHA Jul 3 '16 at 20:16
  • $\begingroup$ Y has Exponential distribution with parameter $n$, that is Y has Gamma(1,n). You can obtain Chi square distribution with a change variable. Therefore you can easily find confidence interval for $\theta$. $\endgroup$ – ALPHA Jul 3 '16 at 20:28
  • $\begingroup$ You're right! What I did was take $\mathbb{P}(0 \leq \text{min}(X_i) \leq \theta) = 1-e^{n\theta} = 1-\alpha$. Am I in the right direction using the cdf. of $Y$? $\endgroup$ – johnny Jul 3 '16 at 20:31
  • $\begingroup$ Would $Z = 2n^2(\text{min}(X_i) - \theta) \sim \gamma(n, \frac{1}{2})$ do the trick, or is there a simpler way? $\endgroup$ – johnny Jul 3 '16 at 20:35
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I think you were on the right track initially, but may have confused the arithmetic of the exponential distribution

If $Y \sim \text{Exp}(n)$ then $\mathbb{P}(Y \le y)= 1-e^{-ny}$ so

  • there is a probability $\alpha$ that $0 \le Y \le -\frac1n \log_e(1-\alpha)$,
  • and also a probability $\alpha$ that $-\frac1n \log_e(\frac{1+\alpha}{2}) \le Y \le -\frac1n \log_e(\frac{1-\alpha}{2})$

This gives confidence intervals for $\theta$

  • $[\min(X_i) + \frac1n \log_e(1-\alpha), \min(X_i)]$
  • or $[\min(X_i) + \frac1n \log_e(\frac{1-\alpha}{2}), \min(X_i)+\frac1n \log_e(\frac{1+\alpha}{2})]$

The first of these alternatives might be seen as "one-sided" but is narrower than the second which might be seen as "two-sided". As an example, suppose $\alpha=95\%$, $n=2$, $X_1=5$ and $X_2=6.2$ then this would give confidence intervals for $\theta$ of about $[3.502,5]$ or $[3.156,4.987]$

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