Point and interval estimation of normal variance θ when the mean μ is known.Let X1,…,Xn are independent observations over the random variable Xi∼N(μ,θ) where μ is known and θ&gt;0. We are given the following estimator for tehta:θ^=∑i=1n{(Xi−μ)2}θ

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Point and interval estimation of normal variance θ when the mean μ is known.Let X1,…,Xn are independent observations over the random variable Xi∼N(μ,θ) where μ is known and θ&amp;gt;0. We are given the following estimator for tehta:θ^=∑i=1n{(Xi−μ)2}θ

Alice Harmon · Accepted Answer

Step 1If μ is known, then an unbiased estimator of the population variance θ=σ2 is given byθ^=∑i=1n(Xi−μ)2n.With this definition, we havenθ^θ=∑i−1n(Xi−μσ)2=∑i=1nZi2∼Chisq(df=n),where Zi∼iidNorm(0,1), and the last step is the definition of Chisq(n).With this distribution for Q=nθ^θ, one can find constants L and U such thatP(L≤Q=nθ^θ≤U)=0.95.Then by manipulating the inequalities, we haveP(nθ^U&amp;lt;θ&amp;lt;nθ^L)=0.95,so that a 95% confidence interval for θ is of the form (nθ^U, nθ^L)

Leia Wiggins · Accepted Answer

It is not magic, it is just the sample Variance if you change&amp;nbsp;θ&amp;nbsp;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.

Point and interval estimation of normal variance \theta when the

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