svend_tveskaeg wrote:First: This is
not a MWE. If you had tried to compiled it yourself before posting it, you would see that.
Here is a MWE (assuming that I have understood you correct):
Code: Select all
\documentclass{article}
\usepackage{amsmath}
\DeclareMathOperator{\var}{Var}
\begin{document}
\begin{align*}
I(\theta)
&= \frac{2n(1-\theta)^2+2n(\theta(1-\theta)}{(1-\theta)^2}+\frac{2n\theta(1-\theta)+2n\theta^2}{\theta^2}
= \frac{2n}{\theta(1-\theta)}\\
\var(\hat{\theta})
&\approx \frac{1}{I(\theta)}\\
I(\hat{\theta})
&= \frac{2n}{\hat{\theta}(1-\hat{\theta})}
\end{align*}
\end{document}
How was what I posted not a minimum working example? I did compile it before I posted it and it didn't give any errors, and I haven't used any obscure packages.
But anyway, I want the to break the first line at the second equal sign and align it with the first equal sign, as I had in the example I gave:
Code: Select all
\begin{align*}
I(\theta)&=\frac{2n(1-\theta)^2+2n(\theta(1-\theta)}{(1-\theta)^2}+\frac{2n\theta(1-\theta)+2n\theta^2}{\theta^2}
\\&=\frac{2n}{\theta(1-\theta)}
\end{align*}
I then want to align
Code: Select all
Var(\hat{\theta})\approx\frac{1}{I(\theta})\indent I(\hat{\theta}=\frac{2n}{\hat{\theta}(1-\hat{\theta})}
with I(\theta) of the first equation.
Does that make sense? And also, please tell me how that's not a proper MWE...A lot of people have said that in my posts, but I don't understand how...