I am trying to prove that for a linear fit relating variables $x$ and $y$, if we have more samples of data pairs ($x_i$, $y_i$) between the range we wish to do a fit, the uncertainty of the slope $B$ in the linear fit $y = A + Bx$ will decrease. I intuitively think this must be true, in a similar way the standard error of the mean will decrease for a larger sample (larger $N$). However, when I try to prove it rigorously, I get stuck in my proof. Could you please help me out?
I suppose that both $x$ and $y$ have a constant random error of $\sigma _x$ and $\sigma _y$.
Taylor (1997) states that the uncertainty in $B$ is:
$$\sigma _B = \sqrt{(\sigma _y)^2 + (B \sigma _x)^2 } \sqrt{\frac{N}{N\sum_{i=1}^Nx_i^2 -(\sum_{i=1}^Nx_i)^2}} $$
My try:
I was approaching the problem the following way. I tried to prove that the second square root monotonically decreases with the number of samples $N$ for the same range. It should decrease independent of the range $x_f - x_o$ and the initial point $x_o$. Each point $x_i$ is then:
$$x_i = x_o + \frac{x_f-x_o}{N-1} (i-1)$$
From $i=1$ to $i=N$. So if we increase $N$ we'll simply have more intermediate points $x_i$ from $x_o$ to $x_f$.
I tried to develop the summations but I reached a point where I couldn't advance more.
Thank you very much for your help!
I've progressed in the proof! The square root term is:
$$ \sqrt{\frac{1/N}{1/N\sum_{i=1}^Nx_i^2 -(1/N\sum_{i=1}^Nx_i)^2}} $$
The denominator is the definition of variance for $x_i$.
$$ \sqrt{\frac{1}{N \sigma_x}} $$
If $R = x_f - x_i$, the variance of $x_i$ goes from $R^2/4$ when $N=2$ to $R^2/12$ when $N=\infty$ (uniform distribution). However I don't know how the variance will change with $N$, so this isn't proof that the uncertainty decreases monotonically with $N$, but it does proof that for large $N$ the uncertainty will eventually decrease, as the variance is bounded.