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Suppose that $X_1$ ..... $X_n$ are a random sample from a population having an exponential distribution with rate parameter $\lambda$.

Use the Central Limit Theorem to show that, for large n, $\sqrt{n}(\lambda\bar{x}-1) \sim Normal(0,1)$

My attempt: honestly I am really not understanding what this question is asking. I can see that for an exponential distribution, it would have to be shifted by $\lambda$ and scale by a factor of $\sigma$?

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The mean and variance of exponential distribution with parameter $\lambda$ are respectively $\dfrac{1}\lambda$ and $\dfrac{1}{\lambda^2}$,

so central limit theorem says

$$\sqrt{n}(\bar{x} - \frac{1}{\lambda}) \to \mathcal{N}(0,\frac{1}{\lambda^2})$$

Multiply both sides by $\lambda$ to conclude

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  • $\begingroup$ so I would be left with $\sqrt{n}(\lambda\bar{x}-1) \sim Normal(0,\frac{1}{\lambda})$? $\endgroup$ – StatsHelp Nov 12 '14 at 20:26
  • $\begingroup$ @StatsHelp Nope, because $\lambda \mathcal{N}(0, \sigma^2) = \mathcal{N}(0, \lambda^2\sigma^2)$, you would be left with what you want:) $\endgroup$ – Petite Etincelle Nov 12 '14 at 20:26
  • $\begingroup$ @StatsHelp You are welcome! $\endgroup$ – Petite Etincelle Nov 12 '14 at 20:30
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Hint: express $\sqrt{n}(\lambda \overline{x} - 1)$ in terms of the sum of the random variables. What does the Central Limit Theorem say about this sum?

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