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I'd like to derive a $100\%(1-\alpha)$ confindence interval for $\sigma$ in a linear model $Y=X\beta+\epsilon$, $X$ - $n\times p$. I thought that I could make use of the fact that:

$\frac{RSS}{\sigma}=\frac{\sum_1^n(Y_i-X_i\beta)^2}{\sigma}\sim \mathcal X_{n-p}^2 $

and take it as a pivot, then:

$\mathbb P(a<\frac{\sum_1^n(Y_i-X_i\beta)^2}{\sigma}<b)=1-\alpha$

and get constants $a,b$ somehow. Am I right or not really?

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    $\begingroup$ What is $\sigma$ here? Is it the variance of the error term? Also you are using $\beta$ istead of $\hat{\beta}$. Do you know the true parameter vector $\beta$ of the model? $\endgroup$
    – Dayne
    Nov 11 '20 at 7:51
  • $\begingroup$ @Dayne correct, variance of the error term, $\epsilon \sim N(0, \sigma^2I_{nxn})$. Well, I was just given this exact instruction as in the question. $\endgroup$
    – thesecond
    Nov 11 '20 at 10:19
  • $\begingroup$ So constructing a confidence internal makes sense when you have a sample to estimate a population parameter. Given the question it i safe to assume that $\sigma$ is unknown. How about $\beta$? $\endgroup$
    – Dayne
    Nov 11 '20 at 17:05
  • $\begingroup$ @Dayne there was nothing mentioned about it. $\endgroup$
    – thesecond
    Nov 11 '20 at 21:52
  • $\begingroup$ You can do so, but note that this is very unrobust! $\endgroup$ Nov 5 '21 at 12:36
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Consider the model

$$Y=X\beta+\varepsilon\,,$$

where $Y$ is an $n\times 1$ response vector, $X$ is an $n\times p$ matrix of covariates (fixed) with full column rank, $\beta$ is a $p\times 1$ vector of parameters and $\varepsilon$ is an $n\times 1$ error vector. Also assume $n>p$.

The residual sum of squares is then

$$\text{RSS}=(Y-X\hat\beta)^T(Y-X\hat\beta)\,,$$

where $\hat\beta$ is the OLS estimator of $\beta$.

If $\varepsilon \sim N_n(0,\sigma^2 I_n)$, then an appropriate pivot for $\sigma^2$ is

$$\frac{\text{RSS}}{\sigma^2} \sim \chi^2_{n-p}$$

This implies

$$P\left(\chi^2_{1-\alpha/2,n-p}<\frac{\text{RSS}}{\sigma^2}<\chi^2_{\alpha/2,n-p}\right)=1-\alpha\quad,\forall\,\sigma^2$$

where $\chi^2_{\alpha,n-p}$ is an upper quantile of a $\chi^2_{n-p}$ distribution, i.e. $P(\chi^2_{n-p}>\chi^2_{\alpha,n-p})=\alpha$

In this setup, a $100(1-\alpha)\%$ confidence interval for $\sigma^2$ is

$$\left(\frac{\text{RSS}}{\chi^2_{\alpha/2,n-p}},\frac{\text{RSS}}{\chi^2_{1-\alpha/2,n-p}}\right)$$

The corresponding interval for $\sigma$ is

$$\left(\sqrt\frac{\text{RSS}}{\chi^2_{\alpha/2,n-p}},\sqrt\frac{\text{RSS}}{\chi^2_{1-\alpha/2,n-p}}\right)$$

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Yes you are on the good path. a is $\chi^2_{1-\frac{\alpha}2, n-p}$ and bis $\chi^2_{\frac{\alpha}2, n-p}$ enter image description here

You can find an example here

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  • $\begingroup$ Thanks! But what about $\beta$? Shall I use least squares estimate? $\endgroup$
    – thesecond
    Nov 11 '20 at 10:44

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