In a Poisson distribution the mean equals the variance. I would like to find a confidence interval of the variance. Is my reasoning below correct?
Using the central limit theorem I construct a 95% confidence interval for the mean $\mu$
$L \leq \mu \leq U$
$L \leq \sigma^2 \leq U$
It would seem to me that the inequality should work like any other inequality in mathematics but statistics can throw a curve ball sometimes so I'm not certain about it. I can't find any papers discussing if this approach is valid.
Another good example of this is a confidence interval for the mean and median of a normal distribution. The mean confidence interval is smaller but the median confidence interval is more robust so either might be preferred as an estimate of the other.

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    $\begingroup$ You're forming a confidence interval for the parameter $\mu$. That confidence interval applies equally whether you're looking at the variance, the mean or any other population quantity which is equal to $\mu$. Your last paragraph suggests you're confusing the parameter and a sample statistic. A confidence interval for $\mu$ in the normal (which is the population median) is just a confidence interval for $\mu$, however you construct it. When you say 'median confidence interval' do you mean 'a confidence interval for that median, $\mu$ based on sample quantiles'? $\endgroup$
    – Glen_b
    Commented Oct 1, 2013 at 22:56

1 Answer 1


Your approach is basically correct but heavily depends on the strong distributional assumption you are making. If it is violated, even for very large samples, the confidence regions wont have the stated coverage probabilities. That's why statisticians try to avoid such reasoning if there are more robust methods available.

There is actually an example (not related to confidence intervals but point estimation) where your approach is frequently used by applied statisticians: Assume you want to estimate the true 97.5% quantile e.g. to detect outliers. Often, instead of calculating the sample 97.5% quantile, researchers assume normality and estimate the true quantile by sample mean plus two standard deviations. If the underlying distribution is normal (which it usually has no reason to be), this estimate is more efficient than the one based on sample quantiles.


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