On 3/6/21 1:02 AM, Abby Spurdle wrote:
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")
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
$message
NULL
I'm not sure how principled this might be. There's no consideration in
this approach for expected sampling error at the right tail where the
magnitudes of the observed values will create much larger contributions
to the sum of squares.
--
David.
On Sat, Mar 6, 2021 at 8:09 PM Abby Spurdle <spurdl...@gmail.com> wrote:
I'm sorry.
I misread your example, this morning.
(I didn't read the code after the line that calls plot).
After looking at this problem again, interpolation doesn't apply, and
extrapolation would be a last resort.
If you can assume your data comes from a particular type of
distribution, such as a lognormal distribution, then a better approach
would be to find the most likely parameters.
i.e.
This falls within the broader scope of maximum likelihood.
(Except that you're dealing with a table of quantile-probability
pairs, rather than raw observational data).
I suspect that there's a relatively easy way of finding the parameters.
I'll think about it...
But someone else may come back with an answer first...
On Sat, Mar 6, 2021 at 8:17 AM Abby Spurdle <spurdl...@gmail.com> wrote:
I note three problems with your data:
(1) The name "percent" is misleading, perhaps you want "probability"?
(2) There are straight (or near-straight) regions, each of which, is
equally (or near-equally) spaced, which is not what I would expect in
problems involving "quantiles".
(3) Your plot (approximating the distribution function) is
back-the-front (as per what is customary).
On Fri, Mar 5, 2021 at 10:14 PM PIKAL Petr <petr.pi...@precheza.cz> wrote:
Dear all
I have table of quantiles, probably from lognormal distribution
dput(temp)
temp <- structure(list(size = c(1.6, 0.9466, 0.8062, 0.6477, 0.5069,
0.3781, 0.3047, 0.2681, 0.1907), percent = c(0.01, 0.05, 0.1,
0.25, 0.5, 0.75, 0.9, 0.95, 0.99)), .Names = c("size", "percent"
), row.names = c(NA, -9L), class = "data.frame")
and I need to calculate quantile for size 0.1
plot(temp$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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