Linear regression with a given (non zero) intercept

Question

If I have a simple linear regression model with the intercept $\beta_0$ known, would the least squares estimator of $\beta_1$ be

$\frac{\sum(y_ix_i)}{\sum({x_i}^2)} - \frac{\beta_0*\sum(x_i)}{\sum(x_i^2)}?$

Because I am treating $\beta_0$ as a constant?

I have also got that the variance of $\hat\beta_1$ is variance of $\left(\frac{\sum y_ix_i}{\sum x_i^2}\right)$? But how do I compute this?

Also how do I find a 100(1-$\alpha$)% confidence interval for $\beta_1$? How do I find MSE?

Thanks

Answer 1

I assume that $y_i=\beta_0+\beta_1 x_i +\varepsilon_i$

Estimation of beta 1

why would

$$\hat \beta_1=\frac{\sum(y_ix_i)}{\sum({x_i}^2)} - \frac{\beta_0*\sum(x_i)}{\sum(x_i^2)}?$$

the intercept $\beta_0$ known, the least squares estimator of $\beta_1$ is still \begin{align*} \hat\beta_1 &=\frac{cov(x,y)}{var(x)}\\ &=\frac{\sum (x_i-\bar x)(y_i-\bar y)}{\sum (x_i-\bar x)}\\ &=\frac{\sum (x_i y_i-\bar xy_i-x_i\bar y+\bar x\bar y)}{\sum (x_i-\bar x)}\\ &=\frac{\sum (x_i y_i)-\bar x\bar y}{\sum (x_i-\bar x)}\\ \end{align*}

even if: $\bar y=\beta_0+\beta_1\bar x+\bar\varepsilon$ I can't see where you can place $\beta_0$

variance of beta 1

I don't remember that

$var(\hat \beta_1)=\left(\frac{\sum y_ix_i}{\sum x_i^2}\right)$

To my mind it is rather:

$$var(\hat\beta_1)=\frac{\sigma^2}{(\sum x_i-\bar x)^2}$$

If you want to compute it, please make you confortable and use the following set of data. $$\begin{array}{lll} \hline t & y_i & x_i \\ \hline 1 & 100 & 100 \\ 2 & 106 & 104 \\ 3 & 107 & 106 \\ ... & ... & ... \\ 9 & 137 & 126 \\ \hline \end{array}$$

I have not yet studied confidence tests.