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quantile from quantile table calculation without original data

11 messages · PIKAL Petr, Abby Spurdle, David Winsemius +2 more

Original
PIKAL Petr
# 
Dear all

I have table of quantiles, probably from lognormal distribution

 dput(temp)
temp <- structure(list(size = c(1.6, 0.9466, 0.8062, 0.6477, 0.5069,
0.3781, 0.3047, 0.2681, 0.1907), percent = c(0.01, 0.05, 0.1,
0.25, 0.5, 0.75, 0.9, 0.95, 0.99)), .Names = c("size", "percent"
), row.names = c(NA, -9L), class = "data.frame")

and I need to calculate quantile for size 0.1

plot(temp$size, temp$percent, pch=19, xlim=c(0,2))
ss <- approxfun(temp$size, temp$percent)
points((0:100)/50, ss((0:100)/50))
abline(v=.1)

If I had original data it would be quite easy with ecdf/quantile function but without it I am lost what function I could use for such task.

Please, give me some hint where to look.


Best regards

Petr
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Jeff Newmiller
# 
Your example could probably be resolved with approx. If you want a more robust solution, it looks like the fBasics package can do spline interpolation. You may want to spline on the log  of your size variable and use exp on the output if you want to avoid negative results.
On March 5, 2021 1:14:22 AM PST, PIKAL Petr <petr.pikal at precheza.cz> wrote:

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      David Winsemius
    

    

      
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On 3/5/21 1:14 AM, PIKAL Petr wrote:

            

      
      
      
    

        
The quantiles are in reverse order so tryoing to match the data to 
quantiles from candidate parameters requires subtracting them from unity:

 > temp$size
[1] 1.6000 0.9466 0.8062 0.6477 0.5069 0.3781 0.3047 0.2681 0.1907
 > qlnorm(1-temp$percent, -.5)
[1] 6.21116124 3.14198142 2.18485959 1.19063854 0.60653066 0.30897659 
0.16837670 0.11708517 0.05922877
 > qlnorm(1-temp$percent, -.9)
[1] 4.16346589 2.10613313 1.46455518 0.79810888 0.40656966 0.20711321 
0.11286628 0.07848454 0.03970223
 > qlnorm(1-temp$percent, -2)
[1] 1.38589740 0.70107082 0.48750807 0.26566737 0.13533528 0.06894200 
0.03756992 0.02612523 0.01321572
 > qlnorm(1-temp$percent, -1.6)
[1] 2.06751597 1.04587476 0.72727658 0.39632914 0.20189652 0.10284937 
0.05604773 0.03897427 0.01971554
 > qlnorm(1-temp$percent, -1.6, .5)
[1] 0.64608380 0.45951983 0.38319004 0.28287360 0.20189652 0.14410042 
0.10637595 0.08870608 0.06309120
 > qlnorm(1-temp$percent, -1, .5)
[1] 1.1772414 0.8372997 0.6982178 0.5154293 0.3678794 0.2625681 
0.1938296 0.1616330 0.1149597
 > qlnorm(1-temp$percent, -1, .4)
[1] 0.9328967 0.7103066 0.6142340 0.4818106 0.3678794 0.2808889 
0.2203318 0.1905308 0.1450700
 > qlnorm(1-temp$percent, -0.5, .4)
[1] 1.5380866 1.1710976 1.0127006 0.7943715 0.6065307 0.4631076 
0.3632657 0.3141322 0.2391799
 > qlnorm(1-temp$percent, -0.55, .4)
[1] 1.4630732 1.1139825 0.9633106 0.7556295 0.5769498 0.4405216 
0.3455491 0.2988118 0.2275150
 > qlnorm(1-temp$percent, -0.55, .35)
[1] 1.3024170 1.0260318 0.9035201 0.7305712 0.5769498 0.4556313 
0.3684158 0.3244257 0.2555795
 > qlnorm(1-temp$percent, -0.55, .45)
[1] 1.6435467 1.2094723 1.0270578 0.7815473 0.5769498 0.4259129 
0.3241016 0.2752201 0.2025322
 > qlnorm(1-temp$percent, -0.53, .45)
[1] 1.6767486 1.2339052 1.0478057 0.7973356 0.5886050 0.4345169 
0.3306489 0.2807799 0.2066236
 > qlnorm(1-temp$percent, -0.57, .45)
[1] 1.6110023 1.1855231 1.0067207 0.7660716 0.5655254 0.4174793 
0.3176840 0.2697704 0.1985218

Seems like it might be an acceptable fit. modulo the underlying data 
gathering situation which really should be considered.

You can fiddle with that result. My statistical hat (not of PhD level 
certification) says that the middle quantiles in this sequence probably 
have the lowest sampling error for a lognormal, but I'm rather unsure 
about that. A counter-argument might be that since there is a hard lower 
bound of 0 for the 0-th quantile that you should be more worried about 
matching the 0.1907 value to the 0.01 order statistic, since 99% of the 
data is know to be above it. Seems like efforts at matching the 0.50 
quantile to 0.5069 for the logmean parameter and matching the 0.01 
quantile 0.1907 for estimation of? the variance estimate might be 
preferred to worrying too much about the 1.6 value which would be in the 
right tail (and far away from your region of extrapolation.)


Further trial and error:


 > qlnorm(1-temp$percent, -0.58, .47)
[1] 1.6709353 1.2129813 1.0225804 0.7687497 0.5598984 0.4077870 
0.3065638 0.2584427 0.1876112
 > qlnorm(1-temp$percent, -0.65, .47)
[1] 1.5579697 1.1309763 0.9534476 0.7167775 0.5220458 0.3802181 
0.2858382 0.2409704 0.1749275
 > qlnorm(1-temp$percent, -0.65, .5)
[1] 1.6705851 1.1881849 0.9908182 0.7314290 0.5220458 0.3726018 
0.2750573 0.2293682 0.1631355
 > qlnorm(1-temp$percent, -0.65, .4)
[1] 1.3238434 1.0079731 0.8716395 0.6837218 0.5220458 0.3986004 
0.3126657 0.2703761 0.2058641
 > qlnorm(1-temp$percent, -0.68, .4)
[1] 1.2847179 0.9781830 0.8458786 0.6635148 0.5066170 0.3868200 
0.3034251 0.2623852 0.1997799
 > qlnorm(1-temp$percent, -0.65, .39)
[1] 1.2934016 0.9915290 0.8605402 0.6791257 0.5220458 0.4012980 
0.3166985 0.2748601 0.2107093
 >
 > qlnorm(1-temp$percent, -0.65, .42)
[1] 1.3868932 1.0416839 0.8942693 0.6930076 0.5220458 0.3932595 
0.3047536 0.2616262 0.1965053
 > qlnorm(1-temp$percent, -0.68, .42)
[1] 1.3459043 1.0108975 0.8678396 0.6725261 0.5066170 0.3816369 
0.2957468 0.2538940 0.1906976


(I did make an effort at searching for quantile matching as a method for 
distribution fitting, but came up empty.)

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      Abby Spurdle
    

    

      
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I note three problems with your data:
(1) The name "percent" is misleading, perhaps you want "probability"?
(2) There are straight (or near-straight) regions, each of which, is
equally (or near-equally) spaced, which is not what I would expect in
problems involving "quantiles".
(3) Your plot (approximating the distribution function) is
back-the-front (as per what is customary).

        
On Fri, Mar 5, 2021 at 10:14 PM PIKAL Petr <petr.pikal at precheza.cz> wrote:

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Abby Spurdle
    

    

      
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I'm sorry.
I misread your example, this morning.
(I didn't read the code after the line that calls plot).

After looking at this problem again, interpolation doesn't apply, and
extrapolation would be a last resort.
If you can assume your data comes from a particular type of
distribution, such as a lognormal distribution, then a better approach
would be to find the most likely parameters.

i.e.
This falls within the broader scope of maximum likelihood.
(Except that you're dealing with a table of quantile-probability
pairs, rather than raw observational data).

I suspect that there's a relatively easy way of finding the parameters.

I'll think about it...
But someone else may come back with an answer first...

        
On Sat, Mar 6, 2021 at 8:17 AM Abby Spurdle <spurdle.a at gmail.com> wrote:

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      Abby Spurdle
    

    

      
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I came up with a solution.
But not necessarily the best solution.

I used a spline to approximate the quantile function.
Then use that to generate a large sample.
(I don't see any need for the sample to be random, as such).
Then compute the sample mean and sd, on a log scale.
Finally, plug everything into the plnorm function:

p <- seq (0.01, 0.99,, 1e6)
Fht <- splinefun (temp$percent, temp$size)
x <- log (Fht (p) )
psolution <- plnorm (0.1, mean (x), sd (x), FALSE)
psolution

The value of the solution is very close to one.
Which is not a surprise.

Here's a plot of everything:

u <- seq (0.000001, 1.65,, 200)
v <- plnorm (u, mean (x), sd (x), FALSE)
plot (u, v, type="l", ylim = c (0, 1) )
points (temp$size, temp$percent, pch=16)
points (0.1, psolution, pch=16, col="blue")

        
On Sat, Mar 6, 2021 at 8:09 PM Abby Spurdle <spurdle.a at gmail.com> wrote:

            

      
      
      
    

        

      
      
    

  
  


  


      

    

    

      
      

        


  

        

      David Winsemius
    

    

      
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On 3/6/21 1:02 AM, Abby Spurdle wrote:

            

      
      
      
    

        
Here's another approach, which uses minimization of the squared error to 
get the parameters for a lognormal distribution.

temp <- structure(list(size = c(1.6, 0.9466, 0.8062, 0.6477, 0.5069,
0.3781, 0.3047, 0.2681, 0.1907), percent = c(0.01, 0.05, 0.1,
0.25, 0.5, 0.75, 0.9, 0.95, 0.99)), .Names = c("size", "percent"
), row.names = c(NA, -9L), class = "data.frame")

obj <- function(x) {sum( qlnorm(1-temp$percent, x[[1]], 
x[[2]])-temp$size )^2}

# Note the inversion of the poorly named and flipped "percent" column,

optim( list(a=-0.65, b=0.42), obj)

#--------------------

$par
 ???????? a????????? b
-0.7020649? 0.4678656

$value
[1] 3.110316e-12

$counts
function gradient
 ????? 51?????? NA

$convergence
[1] 0

$message
NULL


I'm not sure how principled this might be. There's no consideration in 
this approach for expected sampling error at the right tail where the 
magnitudes of the observed values will create much larger contributions 
to the sum of squares.

  
    

      
      
    

  


  


      

    

    
1 day later

    

      
      

        


  

        

      Martin Maechler
    

    

      
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> Your example could probably be resolved with approx. If
    > you want a more robust solution, it looks like the fBasics
    > package can do spline interpolation. 

base R's  spline package does spline interpolation !!

Martin

  
  


  


      

    

    

      
      

        


  

        

      Jeff Newmiller
    

    

      
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I am aware of that... I have my own functions for this purpose that use splinefun. But if you are trying to also do other aspects of probability distribution calculations, it looked like using fBasics would be easier than re-inventing the wheel. I could be wrong, though, since I haven't used fBasics myself.

        
On March 8, 2021 12:41:40 AM PST, Martin Maechler <maechler at stat.math.ethz.ch> wrote:

            

      
      
      
    

        

  
    

      
      
    

  


  


      

    

    

      
      

        


  

        

      PIKAL Petr
    

    

      
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Hallo David, Abby and Bert

Thank you for your solutions. In the meantime I found package rriskDistributions, which was able to calculate values for lognormal distribution from quantiles.

Abby

            

      
      
      
    

        
[1] 9.980823e-06

David

            

      
      
      
    

        
[1] 0.0003120744

rriskDistributions

            

      
      
      
    

        
[1] 1.697379e-05

Bert suggested to ask for original data before quantile calculation what is probably the best but also the most problematic solution. Actually, maybe original data are unavailable as it is the result from particle size measurement, where the software always twist the original data and spits only descriptive results.

All your results are quite consistent with the available values as they are close to 1, so for me, each approach works.

Thank you again.

Best regards.
Petr

            

      
      
      
    

        

      
      
    

        

      
      
    

  
  


  


      

    

    
3 days later

    

      
      

        


  

        

      Abby Spurdle
    

    

      
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Hi Petr,

In principle, I like David's approach the best.
However, I note that there's a bug in the squared step.
Furthemore, the variance of the sample quantiles should increase as
they move away from the modal region.

I've built on David's approach, but changed it to a two stage
optimization algorithm.
The parameter estimates from the first stage are used to compute density values.
Then the second stage is weighted, using the scaled density values.

I tried to create an iteratively reweighted algorithm.
However, it didn't converge.
(But that doesn't necessarily mean it can't be done).

The following code returns the value: 1.648416e-05

qfit.lnorm <- function (p, q, lower.tail=TRUE, ...,
    par0 = c (-0.5, 0.5) )
{   n <- length (p)
    qsample <- q

    objf <- function (par)
    {   qmodel <- qlnorm (p, par [1], par [2], lower.tail)
        sum ( (qmodel - qsample)^2) / n
    }
    objf.w <- function (wpar, w)
    {   qmodel <- qlnorm (p, wpar [1], wpar [2], lower.tail)
        sum (w * (qmodel - qsample)^2)
    }

    wpar0 <- optim (par0, objf)$par
    w <- dlnorm (p, wpar0 [1], wpar0 [2], lower.tail)
    optim (wpar0, objf.w,, w=w)
}

par <- qfit.lnorm (temp$percent, temp$size, FALSE)$par
plnorm (0.1, par [1], par [2])

        
On Tue, Mar 9, 2021 at 2:52 AM PIKAL Petr <petr.pikal at precheza.cz> wrote: