last day (17 days later) » 

Toxyd
Toxyd
10:46 AM
@Mike L. Are you around?
 
Toxyd
Toxyd
11:13 AM
@MikeL.
 
 
3 hours later…
Toxyd
Toxyd
2:37 PM
@HDE226868
@HDE226868 You know lots of tricks :O
I looked every button and bolt of the website in order to do that 'link room' thingie.
I might end the day with a nice plot of star masses generated by the use of the IMF based random distribution :)
Should smooth the rest of the sail
 
 
2 hours later…
Mike L.
Mike L.
4:56 PM
@Toxyd hi
what seems to be the problem?
 
Toxyd
Toxyd
5:51 PM
Hello mike
I'll tell you all the details, so you can tell me what did I do wrong
The base function is x^(-a)
a takes the values 0.3, 1.3 and 2.3
the limts of those values are x < 0.08 and x < 0.5
x can take any value from 0.01 to 150
The value does not add up to 1, as desirable for a PDF, so whenever I integrate from one point to another, I divide the result by the integral from 0.01 to 150
the PDF gives me the results: 0.0367 for the first section, 0.05912 for the second and 0.0371 for the third section
So I know, whatever the result is, it should return two thirds of the values under 0.5
@MikeL. Now, here comes my basic confusion
What is the inverse CDF??
I found the CDF(x): it's just the piecewise integral from the minimum value to x
all right
Using this, I find the cut values, 0.0367 and 0.6279
It makes sense, 0.627 of the values are still lower than 0.5
Now arises the confusion
Is the inverse CDF just the inverse of the indefinite integral?
(Should be (10·x^07)/7 for the smallest portion, 10/(3·y^0.3) for the next and 10 / (13·y^1.3) for the last)
* Sorry, a correction, the PDF gives the results: 0.0367 for the first section, 0.5912 for the second and 0.371 for the third section
Or is the inverse CDF the inverse of the indefinite integral AND the correction to make it fit within the range [0,1]?
or is the inverse of the definite integral? :O
If it's the inverse of the definite integral... it's integrated from the cut point up to the value, from 0.0 up to the value, or from the cut point up to the value plus the complete minor sections from 0 to it's respective cut points??
What a mess :O
Thank you for your time. I don't seem to justify it!
Oh, the indefinite integral I'm using is x^(1-a)/(1-a)
But that looks ok to me. just telling so we are both on the same page
I ended up with the value from the range 0..1, knowing to which section it belongs, and ready to be fed to the inverse of the CDF
I've also got the inverse of the integral calculated, gave you the results around up there
But have no clue about what exactly is the inverse of the CDF. Maybe I should make each of the CDFs range from 0 to 1....
Oh. Spent all day on it. The only results I got that made sense... (no negatives, no 389020 or similars...) scaled pretty fast up to 5 and were slightly up from there. Not the expected evolution
It should escalate sloooowly up to 0.5, then rocket up to 150 on the last third of the 0-1 range
 
Mike L.
Mike L.
7:24 PM
@Toxyd well, if the PDF doesn't integrate to 1 then the relative probabilities of the piecewise PDFs might potentially be off
so you might want to take care of that somehow
but if you know the relative probabilities, then you can just assign them and scale all the CDFs appropriately
as for the inverse CDF, it's just that; the inverse of the CDF function
like when you have function f(x)=y and you solve for x
 
Toxyd
Toxyd
7:56 PM
I'll tell you how that turns out. The pdf integrates to something around 5
so let's say it's exactly 5
One of the problems is that, well, it's easy to scale down on the PDF (or the CDF for that matter)
I can either divide the input or the output. But the inverse... well, Should I multiply the input or output by 5 and that's it?
And the confusion about the CDF is that, well, it's not a single function, it's something like pdf(x)-pdf(lowest)
So I should just invert the pdf, or the 'combo' of f(x) - f(lowest)
 
Toxyd
Toxyd
8:12 PM
Uf... Let's see if tomorrow sheds new light onto this
 
Toxyd
Toxyd
8:31 PM
Let's see... CDF for a high value is the integration of PDF1 from the low value to the first limit, the PDF2 from the first limit to the second and the PDF3 from the second limit up to X. The result, divided by the same sum of the three PDFs up to the maximum value instead of up to x, so the result is normalized
That first part is right, isn't it?
Something that could be summarized to look like this ( Abreviating PDFn(x1, x2) as Pn(x1, x2) in order to make it legible) :
@MikeL. CDF(x) = [P1(low limit, limit1) + P2(limit1, limit2) + P3(limit2, x)] / [P1(low limit, limit1) + P2(limit1, limit2) + P3(limit2, limit max)]
If I express the previous as y = [ P1(full) + P2(full) + P3(l2, x) ] / Pn(full)
 
Toxyd
Toxyd
8:54 PM
The inverse should be... CDF-1(y) = PDF3-1( y · [ P1(full) + P2(full) ] - Pn(full) )
Woha... that looks like it's not gonna be easy to read. Well. I'll keep the rest to myself.
You may think that you wasted your time, but I appreciat your uninterested help.
I'll fight with it on my own from here, I think I've already asked too much
Thanx again!
 
Mike L.
Mike L.
9:51 PM
@Toxyd Hey, sorry I'm not here all that often, it's Sunday and I was mostly busy sewing
You only really need to normalize to find the relative probabilities of your piecewise segments. So integrate all those PDFs into CDFs individually, then divide them all by the sum of the whole thing to get your relative probabilities.
Once you have those, you might as well forget about the normalisation factor and treat each of the CDFs as if it were the only one
and here you might just as well forget about the integrals
if the PDF is in the form P(x)=a x^b for some range of x_min x_max, the integral will be in the form CDF(x)=a/(b+1) * x^(b+1) - C where C is just P(x_min)
all you really need to do is rewrite this as y=a/(b+1) * x^(b+1) and solve for x; that right there is your inversion
or to spell it out, x=(y (b+1)/a)^(1/(b+1))
if it's not maxing out at 1 in the CDF step, you just need to manipulate a until it does
sorry, C in CDF(x) is the result of the preceding expression for x_min
 

  last day (17 days later) »