veleumnihryz

2022-04-24

Point and interval estimation of normal variance $\theta$ when the mean $\mu$ is known.

Let$X}_{1},\dots ,{X}_{n$ are independent observations over the random variable ${X}_{i}\sim N(\mu ,\theta )$ where $\mu$ is known and $\theta >0$ . We are given the following estimator for $tehta$ :

$\hat{\theta}=\frac{\sum _{i=1}^{n}\left\{{({X}_{i}-\mu )}^{2}\right\}}{\theta}$

Let

Alice Harmon

Beginner2022-04-25Added 12 answers

Step 1

If$\mu$ is known, then an unbiased estimator of the population variance $\theta ={\sigma}^{2}$ is given by

$\hat{\theta}=\frac{\sum _{i=1}^{n}{({X}_{i}-\mu )}^{2}}{n}.$

With this definition, we have

$\frac{n\hat{\theta}}{\theta}=\sum _{i-1}^{n}{\left(\frac{{X}_{i}-\mu}{\sigma}\right)}^{2}=\sum _{i=1}^{n}{Z}_{i}^{2}\sim \mathsf{C}\mathsf{h}\mathsf{i}\mathsf{s}\mathsf{q}(df=n),$

where${Z}_{i}\stackrel{iid}{\sim}\mathsf{N}{\textstyle \phantom{\rule{1em}{0ex}}}\mathsf{\text{or}}{\textstyle \phantom{\rule{1em}{0ex}}}\mathsf{m}(0,1),$ and the last step is the definition of $\mathsf{C}\mathsf{h}\mathsf{i}\mathsf{s}\mathsf{q}\left(n\right).$

With this distribution for$Q=\frac{n\hat{\theta}}{\theta}$ , one can find constants L and U such that

$P(L\le Q=\frac{n\hat{\theta}}{\theta}\le U)=0.95.$

Then by manipulating the inequalities, we have

$P\left(\frac{n\hat{\theta}}{U}<\theta <\frac{n\hat{\theta}}{L}\right)=0.95,$

so that a$95\mathrm{\%}$ confidence interval for $\theta$ is of the form $(\frac{n\hat{\theta}}{U},\text{}\frac{n\hat{\theta}}{L})$

If

With this definition, we have

where

With this distribution for

Then by manipulating the inequalities, we have

so that a

Leia Wiggins

Beginner2022-04-26Added 18 answers

It is not magic, it is just the sample Variance if you change $\theta $ to n in the question. There are different approaches to finding a desired estimator.

One of them is the maximum likelihood estimator. If you try to estimate the variance of a normally distributed random variable via maximizing the likelihood function, you end up with the sample variance.

As mentioned in the other answer there are two cases, mean is known or unknown. Have a look at this or for the full derivation better this one.

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