## Automatize LISREL jobs

LISREL routine can run in DOS or in command line mode of windows (windows-key + R -> CMD) . The command line is just like --

D:\My Documents>"C:\Program Files\lisrel87\lisrel87.exe" "C:\Program Files\lisrel87\LS8EX\EX61.LS8" D:\myOutput.out

1. You only need edit and input the bold part.
2. Quotation marks are used wherever the paths or filenames include blanks.
3. The 2nd argument is the output file. You can still specify more output options in your .ls8 file.
4. A .bat file can automatize batches of such lisrel jobs.

## Developing normal pdf from symmetry & independence

When I was in the 3rd grade of my middle school, I enjoyed my town bookstore as a standing library. There a series of six math-story books by Zhang Yuan-Nan impressed me a lot. I cited a case from one in my PPT when I taught the normal distribution -- the normal pdf can be derived from simple symmetry & independence conditions.

Today I can even google out an illegal pdf of its new edition to verify the case (2005, pp. 89). Actually I have bought the new edition series (now 3 books) and lent them to students. Those conditions are as instinctive as--

1. For white noise errors on 2-D, the independence means pdf at $(x,y)$ is the product of 1-D pdf, that is, $\phi\left(x\right)\phi\left(y\right)$.

2. The symmetry means pdf at $(x,y)$ is just a function of $x^{2}+y^{2}$, nothing to do with direction. That is, $\phi\left(x\right)\phi\left(y\right)=f\left(x^{2}+y^{2}\right)$.

So, $f\left(x^{2}\right)f\left(y^{2}\right)=f\left(x^{2}+0\right)f\left(0+y^{2}\right)=\phi^{2}\left(0\right)f\left(x^{2}+y^{2}\right)$.

For middle school students, the book stated a gap here to arrive at the final result $f\left(x^{2}\right)=ke^{bx^{2}}$, which is $\phi\left(x\right)=\frac{1}{\phi\left(0\right)}f\left(x^{2}+0\right)=\frac{k}{\phi\left(0\right)}e^{bx^{2}}$.

I think non-math graduate students with interests can close the gap by themselves with following small hints.

Denote $\alpha=x^{2},\beta=y^{2}$.
We have
$\log f\left(\alpha\right)+\log f\left(\beta\right)=\log\phi^{2}\left(0\right)+\log f\left(\alpha+\beta\right)$,
or
$\;\;\left[\log f\left(\alpha\right)-\log\phi^{2}\left(0\right)\right]+\left[\log f\left(\beta\right)-\log\phi^{2}\left(0\right)\right]$  $=\left(\log f\left(\alpha+\beta\right)-\log\phi^{2}\left(0\right)\right)$.
Denote $g\left(\alpha\right)=\log f\left(\alpha\right)-\log\phi^{2}\left(0\right)$.
That is, $g\left(\alpha\right)+g\left(\beta\right)=g\left(\alpha+\beta\right)$.

Now to prove $g\left(\frac{m}{n}\right)=\frac{m}{n}g\left(1\right),\forall m,n\in\mathbb{N}$. With continuousness, it gets $g\left(\alpha\right)=\alpha g\left(1\right),\forall\alpha\in\mathbb{R}$.