Consider the linear regression $Y_i=\alpha_0+\beta_0 X_i+\epsilon_0$, where $i=1,2,...,n$, and $\epsilon \sim N(0, \sigma^2)$.

$(\hat{\alpha},\hat{\beta})$ estimate $(\alpha_0,\beta_0)$ and are found by minimizing $\sum_{i=1}^n (Y_i-\alpha-\beta X_i)^2.$

We get that $\hat{\alpha}=\bar{Y}-\hat{\beta} \bar{X}$ and $\hat{\beta}=\frac{\sum_{i=1}^n(X_i-\bar{X})Y_i}{\sum_{i=1}^n(X_i-\bar{X})^2}$.

Let $X_1,X_2,...,X_n$ be i.i.d $N(0,\sigma^2)$.

I have to show that $\sum_{i=1}^n X_i^2$ can be used to construct a pivot to find a confidence interval for $\sigma^2$, not necessarily the shortest confidence integral.

I know that a pivot has to have $N(0,1)$ distribution. Do I just have to show that a form of $\sum_{i-1}^n X_i^2$ has $N(0,1)$ distribution using the fact that $X_i \sim N(0,\sigma^2)$?

I'm not sure how to continue, any help is appreciated.


1 Answer 1


I think there is a typo in your formula, it should be
$$\sum_{i=1}^n X_i^2$$ not $\sum_{i-1}^n X_i^2$

Since $X_i$s are i.i.d $N(0,\sigma^2)$,

$$\frac{X_i}{\sigma}\sim N(0,1)$$


$$\frac{X_i^2}{\sigma^2}\sim \chi^2_{(1)}$$


$$\frac{X_1^2}{\sigma^2}+\frac{X_2^2}{\sigma^2}+...+\frac{X_n^2}{\sigma^2}\sim \chi^2_{(n)}$$

i.e $$\frac{\sum_{i=1}^n X_i^2}{\sigma^2}\sim \chi^2_{(n)}$$


$$P(\chi^2_{(n)}(\frac{\alpha}{2})<\frac{\sum_{i=1}^n X_i^2}{\sigma^2}<\chi^2_{(n)}(1-\frac{\alpha}{2}))=1-\alpha$$

$$\Leftrightarrow P(\frac{1}{\chi^2_{(n)}(1-\frac{\alpha}{2})}<\frac{\sigma^2}{\sum_{i=1}^n X_i^2}<\frac{1}{\chi^2_{(n)}(\frac{\alpha}{2})})=1-\alpha$$ $$\Leftrightarrow P(\frac{\sum_{i=1}^n X_i^2}{\chi^2_{(n)}(1-\frac{\alpha}{2})}<\sigma^2<\frac{\sum_{i=1}^n X_i^2}{\chi^2_{(n)}(\frac{\alpha}{2})})=1-\alpha$$

Therefore, the $(1-\alpha)\%$ confidence intervale is

$$(\frac{\sum_{i=1}^n X_i^2}{\chi^2_{(n)}(1-\frac{\alpha}{2})},\frac{\sum_{i=1}^n X_i^2}{\chi^2_{(n)}(\frac{\alpha}{2})})$$

Note $\chi^2_{(n)}(\frac{\alpha}{2})$ is just quantile function of the $\chi^2_{(n)}$ distribution.


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