[ENH] explicit/analytic form of energy function for log-normal distribution

[fkiraly]

There does not seem to exist a literature reference for the energy functionals of the log-normal distribution, we should try to derive it, or find a reference.

Collecting discussion below, from #214.

Current state:

• explicit formula for the cross-term $\mathbb{E}[|X-c|]$ (almost) derived
• no progress yet on the self-term $\mathbb{E}[|X-X'|]$

@bhavikar04 used Wolfram Alpha to derive the following indefinite integral related to the cross-term $\mathbb{E}[X-c]$:

My reply:
this looks correct. Now you need to add the limits. That should be an easy substitution, no? I recommend, do that manually. Use that

$\lim_{x\rightarrow -\infty} \mbox{erf}(x) = -1$, and $\lim_{x\rightarrow \infty} \mbox{erf}(x) = 1$. You need to be careful with the sign, but that should be it?

The number 0.707 etc should be $\frac{1}{2} \sqrt{2}$, but it doesn't matter for the limits.

[bhavikar04]

For the cross-Energy, when using limits x=0--->inf, it seems to evaluate to 0 except when c=0.

[fkiraly]

For the cross-Energy, when using limits x=0--->inf, it seems to evaluate to 0 except when c=0.

That cannot be, you must be making an error of sign or similar...

There exists no distribution for which the cross-term is almost always 0.

[bhavikar04]

For the cross-Energy, when using limits x=0--->inf, it seems to evaluate to 0 except when c=0.

That cannot be, you must be making an error of sign or similar...

There exists no distribution for which the cross-term is almost always 0.

same thoughts but rechecking gives me the same answer, to confirm: as x-->inf erf functions in the expression become -1 because of the (-logx) term and similarly become 1 when x-->0, right?

[fkiraly]

yes, but the sgn at the start also flips, cancelling these out. So you get a factor of 2, not a factor of 0.

(what bothers me is that cancellation seems to occur for $c<0$ still, which does not seem right, but this should deal with $c>0$)

update, sorry, I think I'm wrong above. The sign swaps thrice:

• once for the bound
• once for 0 vs $\infty$
• once for the sgn

So it cancels to zero?

That cannot be right.

[fkiraly]

There might be sth going on with the wolfram heuristic being buggy. What is that strange "plus zero" in the sign function?

I would suggest taking the Wolfram guess and computing its derivative. Then see what matches up or not.