I have exponential distributed data $Exp(\lambda)$ with sample n = 50. Also, The sample mean = 2.17. I need to find the estimator of parameter $\lambda$ by the method of moments and to build 95% confidence interval.

We know, that by the method of moments we got the estimation the parameter $\lambda$ = $1/\bar X$.

Now for confidence interval of my estimator i must calculate the variance of estimator: $Var(\hat \lambda)$ = $Var(1/\bar X)$

But now i am not sure what to do. My attempt: $1/Var(\bar X) = 1/Var(\sum (1/n) *X_i)$ and by the propery o variance i got the result $\lambda^2 * n$.

But i doubt, that is correct result. Can you please help me? many thanks

  • $\begingroup$ Please provide some context: how were your data obtained and what distributional assumptions are you making? $\endgroup$ – whuber Apr 21 '16 at 16:27
  • $\begingroup$ If this is a homework problem, please add the [self-study] tag. $\endgroup$ – Greenparker Apr 21 '16 at 17:18
  • $\begingroup$ Added to the main post $\endgroup$ – Daniel Yefimov Apr 21 '16 at 17:24

$$Var\left(\dfrac{1}{\bar{X}} \right) \ne \dfrac{1}{Var(\bar{X})}.$$

So what you have is indeed incorrect. To solve the problem, go through the following steps:

  1. What is the distribution of $\sum_{i=1}^{n} X_i$?
  2. What is the distribution of $\bar{X}$?
  3. What is the distribution of $Y = 1/\bar{X}$?
  4. What is $Var(Y)$?
  • $\begingroup$ 1. This is surely gamma distribution 2.Expected value of gamma distribution 3. It will be 1/E(gamma distribution) Am i on the right way of thinking? $\endgroup$ – Daniel Yefimov Apr 21 '16 at 18:47
  • $\begingroup$ @DanielYefimov No, E(1/Gamma) is not 1/E(Gamma). You need to find the distribution of the "inverse" of a Gamma. And then, find the expectation of that inverse. $\endgroup$ – Greenparker Apr 21 '16 at 18:49
  • $\begingroup$ It should be the inverse-gamma distribution. en.wikipedia.org/wiki/Inverse-gamma_distribution And then expecation of this distribution. Finally, the variance of the inverse-gamma distribution. I think, that i am right in this case. Sorry for stupid questions:) And thanks for a help $\endgroup$ – Daniel Yefimov Apr 21 '16 at 19:25
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    $\begingroup$ @DanielYefimov That sounds correct. $\endgroup$ – Greenparker Apr 21 '16 at 23:21
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    $\begingroup$ @Daniel I presume you realize by now, but just in case you didn't catch on to this issue yet, your original answer to Greenparker's q2 is wrong; the distribution of $\bar{X}$ is not the expected value of the variable you're averaging. $\endgroup$ – Glen_b Apr 22 '16 at 0:09

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