Question

This statistics is insanely hard. >:(\[z=f\left(a_1,a_2,\ldots,a_n\right),\sigma_{z}=\sum_{i=1}^n\frac{\partial f\left(a_n\right)}{\partial a_n}\sigma_n\]Solve when:\[\sigma=\sqrt{\int_\textbf{text{X}}{\left(x-\int_\textbf{text{X}}xp\left(x\right)\,dx\right)}^2p\left(x\right)\,dx}\]And \(p\left(x\right)\) is the probability distribution of the Gaussian:\[f\left(x\right)=ae^{-frac{\left(x-b\right)^2}{2x^2}}\]

Answer 1

Retyping the question... made a few TeX errors.\[z=f\left(a_1,a_2,\ldots,a_n\right),\sigma_{z}=\sum_{i=1}^n\frac{\partial f\left(a_n\right)}{\partial a_n}\sigma_n\]Solve for \(\sigma_{z^2}\) when:\[\sigma=\sqrt{\int_\textbf{X}{\left(x-\int_\textbf{X}xp\left(x\right)\,dx\right)}^2p\left(x\right)\,dx}\]And \(p\left(x\right)\) is the probability distribution of the Gaussian:\[f\left(x\right)=ae^{-frac{\left(x-b\right)^2}{2x^2}}\]

Answer 2

Ugh, I'm so mad, I have no idea what's going on. I can simplify it only a bit, but it doesn't help whatsoever.

Answer 3

There's some dumb trick here, because there's no way I'm doing all of this out.

Answer 4

ok all i can say is.... good luck with that....

Answer 5

@TuringTest HELP

Answer 6

yeah right!
\(maybe\) @Zarkon or @JamesJ want to look at this, but I think badref is just playing with the pretty LaTeX symbols ;)

Answer 7

They're so pretty. :D

But, no, srsly, what am I supposed to do. Now, since I'm looking for error, and error is independent of the mean, \(\int_\textbf{X}xp\left(x\right)\,dx=0\), and, on average, the P-density should evaluate to \(0.64\). I'm not sure what this tells me.