This is a review question I'm trying to solve. I didn't receive a direct answer, only a few tips from my professor, and I want to see if I'm moving in the right direction. It's a very general question.
The question is: Assume X1, ... , Xn independent and identically distributed random variables from the exponential distribution:
f(x) = λe^(-λx), x > 0.
Find an estimate of λ.
I'm thinking we have to use the point estimate formula:
sample mean +/- z*(std dev / sqrt(n) )
Then, we need to obtain E(x^2) and (E(x))^2 in order to find Var(X).
Var(X) = E(x^2) - (E(x))^2
E(x) can be obtained from taking the integral from infinity to 0 on the formula we were given multiplied by X by integrating for X. To get E(x^2) we do the same thing except multiply the formula by X^2 instead of X.
However, my professor said that the sample mean (x-bar) would be 1/λ, but I'm not sure where's she's getting this from.
Are my steps correct? Or am I getting something mixed up somewhere? I always appreciate explanations, especially since this is a more general type of review question.
The population mean $\mu = 1/\lambda$, where $\lambda$ is the rate used to parameterize your exponential distribution. The standard deviation of this distribution is the same as its mean.
See the Wikipedia article on 'exponential distribution.' This article discusses two possible confidence intervals (CIs). The one you seem to be considering is for relatively large sample sizes and takes the form $1/\bar X \pm M,$ where $M$ is a suitable margin of error. This CI is based on the normal distribution. This is probably what your instructor intends.
Note: I think a better kind of CI is based on the exact distribution of the sample mean $\bar X.$ It is the first one discussed in the Wikipedia article. With modern statistical software there is no difficulty calculating the confidence limits. (Some printed tables of the chi-squared distribution may be adequate; this distribution is not symmetrical, so you have to find upper and lower cut-off points of the distribution.)
Addendum: Consider distribution with rate $\lambda = .01$ and $\mu = 1/\lambda = 100.$ A sample of size $n = 100$ might have $\bar X = 93.57$ and thus $\hat \lambda = 0.0107$ (which is close to $\lambda = .01).$ The 95% normal-based CI is $(.0086, .0128),$ as we would hope--95% of the time.
Incidentally, but not relevant to the CI: The sample SD of these 100 observations is 102.59, which is also 'close' to $\mu = 100,$ but it is not generally as good an estimator of $\mu$ as is $\bar X$, and so $\bar X$ is used to find both the center and the margin of error the CI.