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Assume that $X_1,X_2, ..., X_n$ is random sample with exponential distribution and with probability density function:

$$f(x)=\left\{ \begin{array}{c} \lambda e^{-\lambda x} \; x>0, \\ 0 \qquad x <0, \end{array} \right. $$

where $\lambda >0$ is unknown parametr. Find interval estimation of parameter $\lambda$ with central limit theorem

May someone give me an idea, how to find estimation of the parameter?

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closed as off-topic by Did, heropup, Behrouz Maleki, Shailesh, C. Falcon Jan 23 '17 at 0:48

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  • $\begingroup$ Are you looking for a $ 100(1-\alpha) \% $ CI or an asymptotic $ 100(1-\alpha) \% $ CI ? $\endgroup$ – Greg Jan 22 '17 at 17:01
  • $\begingroup$ asymptotic 100(1−α)%100(1−α)% CI $\endgroup$ – Muffy Jan 22 '17 at 17:26
  • $\begingroup$ Please incorporate your clarification into the body of the Question as an Edit. While you are at it, given some indication of what you tried and what difficulty was found, so that Readers can better address this as your level of understanding. $\endgroup$ – hardmath Jan 22 '17 at 21:17
  • $\begingroup$ Okay, I understand. I will do changes in this question. $\endgroup$ – Muffy Jan 22 '17 at 21:40
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First, it is easy to see that $ E(X_1) = \frac{1}{\lambda} $ and $ Var(X_1) = \frac{1}{\lambda^2}$. From CLT it follows that

$$ \frac{\sqrt{n}\left (\overline{X} -\frac{1}{\lambda}\right)}{\frac{1}{\lambda}} \xrightarrow {d} N(0,1)$$

and equivalently,

$$ \sqrt{n} \left (\lambda \overline{X} - 1\right)\xrightarrow {d} N(0,1).$$

Therefore, the asymptotic $ 100(1-\alpha)\% $ CI of $ \lambda $ is

$$ P \left ( - z_{\frac{\alpha}{2}} \leq \sqrt{n} \left (\lambda \overline{X} - 1\right) \leq z_{\frac{\alpha}{2}} \right) = 1 - \alpha $$

and equivalently,

$$ P \left (\frac{ - z_{\frac{\alpha}{2}} + \sqrt{n}}{\overline{X} \sqrt{n}} \leq \lambda \leq \frac{z_{\frac{\alpha}{2}} + \sqrt{n}}{\overline{X}\sqrt{n}}\right) = 1 - \alpha. $$

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