Solved – Pivot for a confidence interval

confidence intervaldefinitionmathematical-statisticsnormal distributionpivot

I'm looking at an example in my lecture notes where $X_1, X_2,…,X_n$ are iid $N(\mu,\sigma^2)$, where $\sigma^2$ is known. $\bar{X}$ is an unbiased estimator of $\mu$.

The pivot for the confidence interval is given as

$\frac{\bar{X} -\mu}{\frac{\sigma}{\sqrt{n}}}$

and the notes state that this pivot does not depend on $\mu$. I don't really understand how it doesn't. What does it mean for something to 'depend on' something else?

Best Answer

Well, as a function it certainly do depend on $\mu$, simply since the expression for the pivot $$ \frac{\bar{X} -\mu}{\frac{\sigma}{\sqrt{n}}} $$ do contain $\mu$. So in what sense do it not depend on $\mu$? To understand you need to check the definition of a pivot: It is a function of data and model parameters which have a distribution which do not depend on the unknown model parameters. That means that the pivot has a known probability distribution, and that is what makes the probability calculations in constructing the confidence interval work.

So the answer to your question is: The pivot has a probability distribution which do not depend on $\mu$.

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