please help me

[paper pollen]

#1020426321261756536

[zenith heron]

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[modest garden]

Looks like a normal pdf with something a bit added on lemme check

[modest garden]

Pretty sure this is all there is to it you can verify the arithmetic

[paper pollen]

Sorry sir but I didn't understand what you wrote

[modest garden]

As in my handwriting or just the math content

[paper pollen]

As in your handwriting

[paper pollen]

But the answer is 1/2

[modest garden]

Hmm ill recheck my arithmetic then

[manic pulsar]

uv - int vdu?

[grim pebble]

I think I have a much simpler idea.

No integration required at all.

Ah, though...

Nah, maybe not here.

Let me try, though. Maybe something will come from it, hold on.

[craggy willow]

@modest garden it's non zero

looks like the schrodinger wave function

$|\psi_o|^2$

[lean furnace]

Coffeycharas Chichundarcharan

[grim pebble]

To me this looks like a convolution of a logistic RV and a normal RV.

The problem here is that k.

If k was 0, then the integral would be 1/2, since it would be the P(X + Y < 0), which is obviously 1/2 since both X and Y would be symmetric around 0.

Hmm...

[modest garden]

Ohh yes that makes more sense

Ok i just rechecked it the u int vdx term was wrong it evaluates to 1/2 not 1

[grim pebble]

Are you sure the integral doesn't depend on k?

Oh, wait!

You're right!

[modest garden]

mu? no idts I'm trying to retype it rn

[grim pebble]

This is just the convolution of N(0, s^2) and Logistic (0, 1).

[grim pebble]

So, the integral is 1/2 for this reason.

Cool!

I think this is the needed approach.

Calculating this integral directly would be rather difficult.

[modest garden]

Ok actually calculating hte integral seems really stupid\ $\int f(x) dx = \frac{1}{2} \mathrm{erf}\left( \frac{x-\mu}{\sqrt{2} \sigma} \right)$\
and let $g(x)=\frac{1}{1+e^{-x}}$so\
$\int_{-\infty}^\infty f(x)g(x) , dx
= \right. g(x) \int f(x) \left|{-\infty}^{\infty} - \int{-\infty}^\infty \frac{e^{-x}}{(1+e^{-x})^2} \frac{1}{2} \mathrm{erf}\left( \frac{x-\mu}{\sqrt{2} \sigma} \right) dx $

g(x) int f(x) evaluated at lim at inf - lim at -inf is 1/2-0=1/2

The right integral seems kinda stupid to solve theres probably a better way to do it

[lean furnace]

Boris
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[grim pebble]

Ah, actually, no. I spoke too soon. k still makes things bad.

Again, if k was 0, then the whole thing is 1/2.

[modest garden]

Yea you cant solve this like this I've defo made a mistake

[grim pebble]

But this reduces to P(X + Y < k), where X is normal and Y is logistic.

And calculating this probability directly seems like hell.

[modest garden]

Maybe use the taylor series instead but thats kinda lazy and idk if well get a closed form with that

[modest garden]

Ohh then i dont see how it reduces to 1/2

[grim pebble]

If k = 0, then we get P(X + Y < 0). But since X and Y are both symmetric distributions around 0, their sum is also symmetric around 0, hence P(X + Y < 0) = 1/2.

[modest garden]

Ahh I see

Yea I think youre meant to do it with deifnition of the pdfs and cdf I didnt realise that was cdf of logistic initially

[grim pebble]

Well, I can at least more or less confidently say that if the integral is I(k), then I(-∞) = 0, I(0) = 1/2 and I(+∞) = 1 😄

It is still a distribution function, after all.

[modest garden]

maybe the rhs here can be evaluated with law of total probability?

[grim pebble]

Convolution is the law of total probability, really 😄

[modest garden]

Yeaaaa

Ok well any ideas?

[grim pebble]

Not really 😔

I'm not sure this is even possible to integate in elementary functions. Or even with erf.

I mean, I could just search in Gradshtein-Ryzhik, but that will likely take a loooong time.

Though...

Let me try.

Maybe we will get lucky and I find it in there quickly, who knows.

[modest garden]

If only that lhs evaluates to 1 it would be a lot easier to solve edit: it is one oops

[lean furnace]

Boris

[modest garden]

@grim pebble is this fine? Idk how we can do anything beyond this tho

[grim pebble]

Not sure. Besides, I couldn't find relevant integrals in Gradstein-Ryzhik or Shlychkov-Dunaev, unfortunately...

[modest garden]

Hmm i think it's correct tho with law of total probability but idk who can check plus not of the form that op wanted

[modest garden]

Hmm ya I've been trying to think about it tho can't progress now tho do you have any ideas

[craggy willow]

graph it, see if you can spot something like the sqrt(2pi)s

or try it with different k and s

as i said it awfully looks like some sort of density function

[modest garden]

Well for standard normal its 1/2 as darpy said

[modest garden]

It essentially is yea just a pdf times cdf

[craggy willow]

I'm highly suspicious as the Aryan person used chat gpt to Anser the riddle.

it might be a troll like the last one

[modest garden]

I doubt it this and the other question are both legit questions it seems

[craggy willow]

I have other things to cry about I'm saving tears

[paper pollen]

Bro why are you saying I used chatgpt to solve that riddle

Chatgpt’s response on this question

My answer on that riddle

[modest garden]

@bleak tusk see this one if you want

Its an integral of a normal pdf times the cdf of standard logistic

[bleak tusk]

yes I've observed it

I know

[modest garden]

the other question is basically the same

[modest garden]

So darpy used convolution I did something similar but simplified it to P(X>Y)

And case for standard normal is piss easy cuz its symmetric its 1/2

The other question shows that by computing the integral manually but you can see it easily with what me and darpy did

Also just got this comment from some dude with 300k rep. I think that validates the proof I did and also implies we cant solve it further

@grim pebble not a satisfying answer yet but its some development

[grim pebble]

Ah, that's a shame...

Can someone maybe look through Gradshteyn-Ryzhik and Schlychkov-Dunaev again? Maybe I just missed an appropriate integral.

Or any other integrals handbooks that you have. In which case, please share them 😁

[paper pollen]

but it can be more difficult to implement. The simplification to P(X>Y) is a clever way to solve the problem for the standard normal distribution.

[modest garden]

Only you're gonna own such hardcore soviet integral lookup books darpy 😭

[modest garden]

Yea but if it was standard normal u needed that was very easy to solve me and darpy both assumed it was the general case

[grim pebble]

I believe those have English translations. At least the first one, it's very well-known.

Well, as I asked you to share yours, I can share these, too! I only have the Russian versions, though.

[modest garden]

Aww I dont use any integral books tho 😦

But I think wolfram mathematica will have a big bunch of these integral handbook stuff coded in no? I tried putting the problem in mathematica once it didnt help

[grim pebble]

Weeeell, that's not the same.

Besides, Wolframalpha isn't too good with dealing with integrals containing parameters.

[modest garden]

Yes but mathematica is the main big app that usally does more stuff

Anywho I think we have to wait hopefully theres someone who has researched this topic and found an answer on stack but it seems unlikely we can solve it with a nice closed form P(X>Y) is the nicest looking thing ig

[grim pebble]

Hm, I see...
Well, if you do find some textbooks or resources, please let me know! It will be a fine addition to my collection.

[modest garden]

I will keep an eye out ! But i think those soviet ones are the best for this

[grim pebble]

Probably.
Anyway, here they are if you want to take a look.

As I said, though, they are in Russian.

[modest garden]

Oh wow what a treasure trove only the index needs to be translated anyway everything else is readable

[grim pebble]

Hm.

I can do that.

Though, it would be better if the books along with my files with translated indices were posted somewhere on the server, so that I'm not doing it just for one thread and a couple of people 😅

[modest garden]

Hahaha don't do it now I'll put a reminder to check this tomorrow there might be a translated version online too

[grim pebble]

Alright.

I think we need to discuss it with the staff members.

If there are any handbooks like that in Russian, I can just translate their index. That doesn't take long, I just want many people to see it, otherwise it isn't worth it.

[modest garden]

I found one darpy !

They acknowledged half of Russia in the start

You can put it in any other channel if you want

I can't find the other one tho

[grim pebble]

Oh, cool!

[grim pebble]

@real forge , is it possible to add this somewhere that will be easy to access? I believe this is an important resource.
Sorry for the ping if you're busy.

[real forge]

We can put this in archived contents

Not me, not right now, though