4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $N(0,\sigma^2)$,

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

Therefore,

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

Therefore,

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

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

Therefore,

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

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.