What is the standard error of the coefficient in a linear regression model performed by a $\chi^2$ regression, without an intercept present?
I've determined $b$ and $\sigma_b$ as follows:
We want to fit a set of data points $(x_i, y_i)$ to the model $y = b x$. To find $b$ set $\frac{d \chi^2}{d b} = 0$
$$ 0 = \frac{d \chi^2}{d b} = \frac{d}{d b} \sum _i \frac{(y_i - b x_i)^2}{\sigma_i^2} = -\sum_i 2 x_i \frac{y_i - b x_i}{\sigma_i^2} $$
$$ b = \frac{\sum _i \frac{x_i y_i}{\sigma_i^2}}{\sum_i \frac{x_i^2}{\sigma_i^2}} := \frac{S_{xy}}{S_{xx}} $$
To find $\sigma_b$ we must perform error propagation and set $\sigma_b^2 = \sum_i \sigma_i^2 \left( \frac{\partial b}{\partial x_i} \right)^2$
Using the quotient rule
$$ \frac{\partial b}{\partial x_i} = \frac{y_i S_{xx} - 2 x_i S_{xy}}{S_{xx}^2 \sigma_i^2} $$
$$ \sigma_i^2 \left(\frac{\partial b}{\partial x_i} \right)^2 = \frac{y_i^2 S_{xx}^2 - 4 x_i y_i S_{xx} S_{xy} + 4 x_i^2 S_{xy}^2}{S_{xx}^4 \sigma_i^2} $$
$$ \sum _i \sigma_i^2 \left(\frac{\partial b}{\partial x_i} \right)^2 = \frac{S_{yy}}{S_{xx}^2} - 4 \frac{S_{xy}^2}{S_{xx}^3} + 4 \frac{S_{xy}^2}{S_{xx}^3} = \frac{S_{yy}}{S_{xx}^2} $$
From googling I'm quite sure that my formula $b = \frac{S_{xy}}{S_{xx}}$ is correct.
Is $\sigma_b^2 = \frac{S_{yy}}{S_{xx}^2}$ correct?
Thank you ! :):):)
edit: The reason I ask is because I used that formula on my data, and it basically fails the sanity check of being similar to the uncertainty of the quantity $b$ as derived differently, using a simple $\chi^2$ weighted average and standard deviation on the formula $b = y/x$.
Asked my professor, and he said I was actually supposed to take $\frac{\partial b}{\partial y_i}$ as my partial derivative, which gives
$$\sigma_b^2 = \frac{1}{S_{xx}}$$