Linear model – confidence interval for $\sigma$

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}

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

• 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? Nov 11 '20 at 7:51
• @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. Nov 11 '20 at 10:19
• 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 $\beta$? Nov 11 '20 at 17:05
• @Dayne there was nothing mentioned about it. Nov 11 '20 at 21:52
• You can do so, but note that this is very unrobust! Nov 5 '21 at 12:36

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

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

Yes you are on the good path. a is $$\chi^2_{1-\frac{\alpha}2, n-p}$$ and bis $$\chi^2_{\frac{\alpha}2, n-p}$$ You can find an example here

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