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We have $n$ random samples from a population which has normal distribution, and the mean of the population is known.

How do we change the procedure of finding a confidence interval for the population variance, given we know the exact mean of the population?

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If we don't know the mean, the distribution of the point estimator is $$\frac{1}{n-1}\sum_i^n(X_i - \bar{X})^2=\frac{1}{n-1}X'PX \sim \chi^2_{n-1},$$ where $X \sim \mathcal{N}(\mu, I) $ is the vector of observations and $P=[\delta_{ij}-\frac{1}{n}]_{i,j}$ is the centering matrix. The degree of freedom $n-1$ is the rank of this centering matrix.

If we know the mean $\mu$, we can use $(X_i-\mu)$, i.e. the identity matrix $I$ instead of $P$. This matrix has the rank $n$ instead of $n-1$. So the degree of freedom increases and the lower and upper quantiles come closer together. THis makes the CI smaller and thus more informative.

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  • $\begingroup$ I answered a duplicate here before noting this answer, and our answers do not seem to agree. Do you have any thoughts on the discrepancy? $\endgroup$ – Christoph Hanck Oct 8 '15 at 9:51
  • $\begingroup$ @ChristophHanck: The conclusion is the same. I use explicitly the fact $X'PX \sim \chi^2_{\rg(P)}$ with an idempotent matrix $P$. Then I stop at mentioning the quantiles (for me, it's clear then and I'm lazy). You calculate the confidence bounds thoroughly to the end and compare the relative width. This is very good. I would have drawn the conclusion already by comparing $r_n$. If $s^2_0$ and $\hat{\sigma}^2$ differ, the location of the CI will clearly also be different, but the question was about the width. $\endgroup$ – Horst Grünbusch Oct 24 '15 at 23:40

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