Re: [R] Integral of PDF
That does not remedy the situation in any case, take the following function fun - function(x) dnorm(x, -500, 50) + dnorm(x, 500, 50) that has a 'mode' of 0 again. Interestingly, if I transform it by 1/x, to integrate I again have to reduce the error tolerance to at least 1e-10: f - function(x) fun(1/x)/x^2 integrate(f, 0, 1) #- 0, should be 1 while cubature::adaptIntegrate does not require to change the tolerance: require(cubature) adaptIntegrate(f, 0, 1)#- 1 What I find *dangerous* about 'integrate' is not that it returns a wrong result, but that it --- w/o warning --- gives an error estimate that leads completely astray. Hans Werner But that only postpones the misery, Hans Werner! You can always make the mean large enough to get a wrong answer from `integrate'. integrate(function(x) dnorm(x, 700,50), -Inf, Inf, subdivisions=500, rel.tol=1e-11) integrate(function(x) dnorm(x, -700,50), -Inf, Inf, subdivisions=500, rel.tol=1e-11) The problem is that the quadrature algorithm is not smart enough to recognize that the center of mass is at `mean'. The QUADPACK algorithm (Gauss-Kronrod quadratutre) is adaptive, but it does not look to identify regions of high mass. Algorithms which can locate regions of highest density, such as those developed by Alan Genz, will do much better in problems like this. Genz and Kass (1997). J Computational Graphics and Statistics. There is a FORTRAN routine DCHURE that you might want to try for infinite regions. I don't think this has been ported to R (although I may be wrong). There may be other more recent ones. A simple solution is to locate the mode of the integrand, which should be quite easy to do, and then do a coordinate shift to that point and then integrate the mean-shifted integrand using `integrate'. Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu That does not remedy the situation in any case, take the following function fun - function(x) dnorm(x, -500, 50) + dnorm(x, 500, 50) that has a 'mode' of 0 again. Interestingly, if I transform it by 1/x, to integrate I again have to reduce the error tolerance to at least 1e-10: f - function(x) fun(1/x)/x^2 integrate(f, 0, 1) #- 0, should be 1 while cubature::adaptIntegrate does not require to change the tolerance: require(cubature) adaptIntegrate(f, 0, 1)#- 1 What I find *dangerous* about 'integrate' is not that it returns a wrong result, but that it --- w/o warning --- gives an error estimate that leads completely astray. Hans Werner Ravi Varadhan wrote: But that only postpones the misery, Hans Werner! You can always make the mean large enough to get a wrong answer from `integrate'. integrate(function(x) dnorm(x, 700,50), -Inf, Inf, subdivisions=500, rel.tol=1e-11) integrate(function(x) dnorm(x, -700,50), -Inf, Inf, subdivisions=500, rel.tol=1e-11) The problem is that the quadrature algorithm is not smart enough to recognize that the center of mass is at `mean'. The QUADPACK algorithm (Gauss-Kronrod quadratutre) is adaptive, but it does not look to identify regions of high mass. Algorithms which can locate regions of highest density, such as those developed by Alan Genz, will do much better in problems like this. Genz and Kass (1997). J Computational Graphics and Statistics. There is a FORTRAN routine DCHURE that you might want to try for infinite regions. I don't think this has been ported to R (although I may be wrong). There may be other more recent ones. A simple solution is to locate the mode of the integrand, which should be quite easy to do, and then do a coordinate shift to that point and then integrate the mean-shifted integrand using `integrate'. Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu -- View this message in context: http://r.789695.n4.nabble.com/Integral-of-PDF-tp3070243p3070768.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Integral of PDF
Hi,
You need to be careful with -Inf and Inf in R, I suppose they are some large
numbers arbitrarily chosen by R, but not infinite quantities. So the function
integrate can return errors if the function to integrate doesn't have
negligible values beyond those large numbers.
I ran the following to convince myself:
integrate(function(x){return(ifelse(x500,0,1/(x^2)))},-Inf,Inf)
0 with absolute error 0
integrate(function(x){return(ifelse(x500,0,1/(x^2)))},500,Inf)
0.00199 with absolute error 0.11
The second result shouldn't be greater that the first one but...
And to solve your problem:
integrate(function(x) dnorm(x, 500,50), 500-1000, 500+1000)
1 with absolute error 0.74
I centred the interval of integration on 500 instead of on 0 if you integrate
between -Inf and Inf.
HTH,
Samuel
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Doran, Harold
Sent: 02 December 2010 21:22
To: r-help@r-project.org
Subject: [R] Integral of PDF
The integral of any probability density from -Inf to Inf should equal 1,
correct? I don't understand last result below.
integrate(function(x) dnorm(x, 0,1), -Inf, Inf)
1 with absolute error 9.4e-05
integrate(function(x) dnorm(x, 100,10), -Inf, Inf)
1 with absolute error 0.00012
integrate(function(x) dnorm(x, 500,50), -Inf, Inf)
8.410947e-11 with absolute error 1.6e-10
all.equal(integrate(function(x) dnorm(x, 500,50), -Inf, Inf)$value, 0)
[1] TRUE
sessionInfo()
R version 2.10.1 (2009-12-14)
i386-pc-mingw32
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] statmod_1.4.6 mlmRev_0.99875-1 lme4_0.999375-35 Matrix_0.999375-33
lattice_0.17-26
loaded via a namespace (and not attached):
[1] grid_2.10.1 nlme_3.1-96 stats4_2.10.1 tools_2.10.1
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__ Information from ESET NOD32 Antivirus, version of virus signature
database 5668 (20101202) __
The message was checked by ESET NOD32 Antivirus.
http://www.eset.com
__ Information from ESET NOD32 Antivirus, version of virus signature
database 5669 (20101203) __
The message was checked by ESET NOD32 Antivirus.
http://www.eset.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Integral of PDF
Your function does NOT have a mode at zero. It is bimodal with 2 modes: -500 and 500. So, my approach still works. Zero is a stationary point where the gradient is zero, but it is not a mode (the second derivative at zero is not negative but it is zero). Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Hans W Borchers Sent: Friday, December 03, 2010 3:30 AM To: r-help@r-project.org Subject: Re: [R] Integral of PDF That does not remedy the situation in any case, take the following function fun - function(x) dnorm(x, -500, 50) + dnorm(x, 500, 50) that has a 'mode' of 0 again. Interestingly, if I transform it by 1/x, to integrate I again have to reduce the error tolerance to at least 1e-10: f - function(x) fun(1/x)/x^2 integrate(f, 0, 1) #- 0, should be 1 while cubature::adaptIntegrate does not require to change the tolerance: require(cubature) adaptIntegrate(f, 0, 1)#- 1 What I find *dangerous* about 'integrate' is not that it returns a wrong result, but that it --- w/o warning --- gives an error estimate that leads completely astray. Hans Werner But that only postpones the misery, Hans Werner! You can always make the mean large enough to get a wrong answer from `integrate'. integrate(function(x) dnorm(x, 700,50), -Inf, Inf, subdivisions=500, rel.tol=1e-11) integrate(function(x) dnorm(x, -700,50), -Inf, Inf, subdivisions=500, rel.tol=1e-11) The problem is that the quadrature algorithm is not smart enough to recognize that the center of mass is at `mean'. The QUADPACK algorithm (Gauss-Kronrod quadratutre) is adaptive, but it does not look to identify regions of high mass. Algorithms which can locate regions of highest density, such as those developed by Alan Genz, will do much better in problems like this. Genz and Kass (1997). J Computational Graphics and Statistics. There is a FORTRAN routine DCHURE that you might want to try for infinite regions. I don't think this has been ported to R (although I may be wrong). There may be other more recent ones. A simple solution is to locate the mode of the integrand, which should be quite easy to do, and then do a coordinate shift to that point and then integrate the mean-shifted integrand using `integrate'. Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu That does not remedy the situation in any case, take the following function fun - function(x) dnorm(x, -500, 50) + dnorm(x, 500, 50) that has a 'mode' of 0 again. Interestingly, if I transform it by 1/x, to integrate I again have to reduce the error tolerance to at least 1e-10: f - function(x) fun(1/x)/x^2 integrate(f, 0, 1) #- 0, should be 1 while cubature::adaptIntegrate does not require to change the tolerance: require(cubature) adaptIntegrate(f, 0, 1)#- 1 What I find *dangerous* about 'integrate' is not that it returns a wrong result, but that it --- w/o warning --- gives an error estimate that leads completely astray. Hans Werner Ravi Varadhan wrote: But that only postpones the misery, Hans Werner! You can always make the mean large enough to get a wrong answer from `integrate'. integrate(function(x) dnorm(x, 700,50), -Inf, Inf, subdivisions=500, rel.tol=1e-11) integrate(function(x) dnorm(x, -700,50), -Inf, Inf, subdivisions=500, rel.tol=1e-11) The problem is that the quadrature algorithm is not smart enough to recognize that the center of mass is at `mean'. The QUADPACK algorithm (Gauss-Kronrod quadratutre) is adaptive, but it does not look to identify regions of high mass. Algorithms which can locate regions of highest density, such as those developed by Alan Genz, will do much better in problems like this. Genz and Kass (1997). J Computational Graphics and Statistics. There is a FORTRAN routine DCHURE that you might want to try for infinite regions. I don't think this has been ported to R (although I may be wrong). There may be other more recent ones. A simple solution is to locate the mode of the integrand, which should be quite easy to do, and then do a coordinate shift to that point and then integrate the mean-shifted integrand using `integrate'. Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410
Re: [R] Integral of PDF
Yes, Albyn. I do not think that this is a dangerous behavior of the tool (`integrate'). It is certainly a dangerous use of the tool. One will be hard-pressed to find a numerical algorithm/software that is fool-proof in the sense that it always gives you either the correct results or warns you that the results could be wrong for your problem. We can always devise a problem to defeat even the most robust algorithm. So, I would argue that the ultimate responsibility lies with the user, via careful thinking/prior analysis/ and multiple types of checks, to ensure that the results are reasonably accurate. This, of course, does not mean that `integrate' gets away scot free. I am sure it could be improved, but I am also quite certain that there are better quadrature algorithms. So, one needs to choose appropriate algorithms to suit their problem needs. Perhaps, the simplest thing to do for `integrate' might be to print a generic warning like: The error estimate is not always reliable. Of course, this does not help the user since they would not know when it is NOT reliable. The next thing to try might be to examine how it picks the initial interval [0, x], and see if this procedure can be improved. Best, Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu -Original Message- From: Albyn Jones [mailto:jo...@reed.edu] Sent: Thursday, December 02, 2010 6:41 PM To: Ravi Varadhan Cc: 'Hans W Borchers'; r-help@r-project.org Subject: Re: [R] Integral of PDF On Thu, Dec 02, 2010 at 06:23:45PM -0500, Ravi Varadhan wrote: A simple solution is to locate the mode of the integrand, which should be quite easy to do, and then do a coordinate shift to that point and then integrate the mean-shifted integrand using `integrate'. Ravi. Translation: think before you compute! albyn __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Integral of PDF
The integral of any probability density from -Inf to Inf should equal 1, correct? I don't understand last result below. integrate(function(x) dnorm(x, 0,1), -Inf, Inf) 1 with absolute error 9.4e-05 integrate(function(x) dnorm(x, 100,10), -Inf, Inf) 1 with absolute error 0.00012 integrate(function(x) dnorm(x, 500,50), -Inf, Inf) 8.410947e-11 with absolute error 1.6e-10 all.equal(integrate(function(x) dnorm(x, 500,50), -Inf, Inf)$value, 0) [1] TRUE sessionInfo() R version 2.10.1 (2009-12-14) i386-pc-mingw32 locale: [1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] statmod_1.4.6 mlmRev_0.99875-1 lme4_0.999375-35 Matrix_0.999375-33 lattice_0.17-26 loaded via a namespace (and not attached): [1] grid_2.10.1 nlme_3.1-96 stats4_2.10.1 tools_2.10.1 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Integral of PDF
You can dive into the thread puzzle with integrate over infinite range from September this year. The short answer appears to be: Increase the error tolerance. integrate(function(x) dnorm(x, 500,50), -Inf, Inf, subdivisions=500, rel.tol=1e-11) # 1 with absolute error 1.1e-12 Hans Werner Doran, Harold wrote: The integral of any probability density from -Inf to Inf should equal 1, correct? I don't understand last result below. integrate(function(x) dnorm(x, 0,1), -Inf, Inf) 1 with absolute error 9.4e-05 integrate(function(x) dnorm(x, 100,10), -Inf, Inf) 1 with absolute error 0.00012 integrate(function(x) dnorm(x, 500,50), -Inf, Inf) 8.410947e-11 with absolute error 1.6e-10 all.equal(integrate(function(x) dnorm(x, 500,50), -Inf, Inf)$value, 0) [1] TRUE sessionInfo() R version 2.10.1 (2009-12-14) i386-pc-mingw32 locale: [1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] statmod_1.4.6 mlmRev_0.99875-1 lme4_0.999375-35 Matrix_0.999375-33 lattice_0.17-26 loaded via a namespace (and not attached): [1] grid_2.10.1 nlme_3.1-96 stats4_2.10.1 tools_2.10.1 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://r.789695.n4.nabble.com/Integral-of-PDF-tp3070243p3070315.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Integral of PDF
To really understaand it you will have to look at the fortran code
underlying integrate. I tracked it back through a couple of layers
(dqagi, dqagie, ... just use google, these are old netlib
subroutines) then decided I ought to get back to grading papers :-)
It looks like the integral is split into the sum of two integrals,
(-Inf,0] and [0,Inf).
integrate(function(x) dnorm(x, 350,50), 0, Inf)
1 with absolute error 1.5e-05
integrate(function(x) dnorm(x, 400,50), 0, Inf)
1.068444e-05 with absolute error 2.1e-05
integrate(function(x) dnorm(x, 500,50), 0, Inf)
8.410947e-11 with absolute error 1.6e-10
integrate(function(x) dnorm(x, 500,50), 0, 1000)
1 with absolute error 7.4e-05
The integral from 0 to Inf is the lim_{x - Inf} of the integral
over [0,x]. I suspect that the algorithm is picking an interval
[0,x], evaluating the integral over that interval, and then performing
some test to decide whether to expand the interval. When the initial
interval doesn't contain much, expanding a little won't add enough to
trigger another iteration.
albyn
On Thu, Dec 02, 2010 at 04:21:34PM -0500, Doran, Harold wrote:
The integral of any probability density from -Inf to Inf should equal 1,
correct? I don't understand last result below.
integrate(function(x) dnorm(x, 0,1), -Inf, Inf)
1 with absolute error 9.4e-05
integrate(function(x) dnorm(x, 100,10), -Inf, Inf)
1 with absolute error 0.00012
integrate(function(x) dnorm(x, 500,50), -Inf, Inf)
8.410947e-11 with absolute error 1.6e-10
all.equal(integrate(function(x) dnorm(x, 500,50), -Inf, Inf)$value, 0)
[1] TRUE
sessionInfo()
R version 2.10.1 (2009-12-14)
i386-pc-mingw32
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] statmod_1.4.6 mlmRev_0.99875-1 lme4_0.999375-35
Matrix_0.999375-33 lattice_0.17-26
loaded via a namespace (and not attached):
[1] grid_2.10.1 nlme_3.1-96 stats4_2.10.1 tools_2.10.1
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Albyn Jones
Reed College
jo...@reed.edu
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Integral of PDF
You can use trace() to see what is happening trace(dnorm, quote(print(round(x integrate(function(x) dnorm(x, 500,50), -Inf, Inf) Tracing dnorm(x, 500, 50) on entry [1] 1 233 0 38 0 14 0 7 0 4 0 2 0 2 1 Tracing dnorm(x, 500, 50) on entry [1] -1 -2330 -380 -140 -70 -40 -20 -2 -1 Tracing dnorm(x, 500, 50) on entry [1] 3 467 1 78 1 29 1 14 1 9 2 6 2 4 2 Tracing dnorm(x, 500, 50) on entry [1] -3 -467 -1 -78 -1 -29 -1 -14 -1 -9 -2 -6 -2 -4 -2 Tracing dnorm(x, 500, 50) on entry [1] 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0 Tracing dnorm(x, 500, 50) on entry [1] 0 -1 0 -1 0 -1 0 -1 0 -1 0 -1 0 0 0 Tracing dnorm(x, 500, 50) on entry [1] 7 935 3 156 3 58 3 30 4 18 4 12 5 9 6 Tracing dnorm(x, 500, 50) on entry [1] -7 -935 -3 -156 -3 -58 -3 -30 -4 -18 -4 -12 -5 -9 -6 Tracing dnorm(x, 500, 50) on entry [1] 2 3 1 3 1 3 1 3 1 2 1 2 1 2 1 Tracing dnorm(x, 500, 50) on entry [1] -2 -3 -1 -3 -1 -3 -1 -3 -1 -2 -1 -2 -1 -2 -1 8.410947e-11 with absolute error 1.6e-10 You are asking integrate to find the relatively tiny portion of the the floating point real line (from c. -10^300 to 10^300) on which dnorm(x) is bigger than c. 10^-300. It found a few points where it was that big, but apparently not enough to home on the answer. You need to give it better hints abouts dnorm's support region. E.g., integrate(function(x) dnorm(x, 500,50), -100, 900) Tracing dnorm(x, 500, 50) on entry [1] 400 -87 887 -33 833 60 740 183 617 326 474 -98 898 -65 865 10 790 119 681 [20] 253 547 Tracing dnorm(x, 500, 50) on entry [1] 150 -93 393 -66 366 -20 320 42 258 113 187 -99 399 -83 383 -45 345 9 291 [20] 76 224 Tracing dnorm(x, 500, 50) on entry [1] 650 407 893 434 866 480 820 542 758 613 687 401 899 417 883 455 845 509 791 [20] 576 724 Tracing dnorm(x, 500, 50) on entry [1] 525 403 647 417 633 440 610 471 579 506 544 401 649 409 641 427 623 455 595 [20] 488 562 Tracing dnorm(x, 500, 50) on entry [1] 775 653 897 667 883 690 860 721 829 756 794 651 899 659 891 677 873 705 845 [20] 738 812 1 with absolute error 4.0e-07 Making the limits +-10^4 works ok also, but +-Inf is apparently asking for too much. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Doran, Harold Sent: Thursday, December 02, 2010 1:22 PM To: r-help@r-project.org Subject: [R] Integral of PDF The integral of any probability density from -Inf to Inf should equal 1, correct? I don't understand last result below. integrate(function(x) dnorm(x, 0,1), -Inf, Inf) 1 with absolute error 9.4e-05 integrate(function(x) dnorm(x, 100,10), -Inf, Inf) 1 with absolute error 0.00012 integrate(function(x) dnorm(x, 500,50), -Inf, Inf) 8.410947e-11 with absolute error 1.6e-10 all.equal(integrate(function(x) dnorm(x, 500,50), -Inf, Inf)$value, 0) [1] TRUE sessionInfo() R version 2.10.1 (2009-12-14) i386-pc-mingw32 locale: [1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] statmod_1.4.6 mlmRev_0.99875-1 lme4_0.999375-35 Matrix_0.999375-33 lattice_0.17-26 loaded via a namespace (and not attached): [1] grid_2.10.1 nlme_3.1-96 stats4_2.10.1 tools_2.10.1 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Integral of PDF
But that only postpones the misery, Hans Werner! You can always make the mean large enough to get a wrong answer from `integrate'. integrate(function(x) dnorm(x, 700,50), -Inf, Inf, subdivisions=500, rel.tol=1e-11) integrate(function(x) dnorm(x, -700,50), -Inf, Inf, subdivisions=500, rel.tol=1e-11) The problem is that the quadrature algorithm is not smart enough to recognize that the center of mass is at `mean'. The QUADPACK algorithm (Gauss-Kronrod quadratutre) is adaptive, but it does not look to identify regions of high mass. Algorithms which can locate regions of highest density, such as those developed by Alan Genz, will do much better in problems like this. Genz and Kass (1997). J Computational Graphics and Statistics. There is a FORTRAN routine DCHURE that you might want to try for infinite regions. I don't think this has been ported to R (although I may be wrong). There may be other more recent ones. A simple solution is to locate the mode of the integrand, which should be quite easy to do, and then do a coordinate shift to that point and then integrate the mean-shifted integrand using `integrate'. Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University Ph. (410) 502-2619 email: rvarad...@jhmi.edu -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Hans W Borchers Sent: Thursday, December 02, 2010 5:16 PM To: r-help@r-project.org Subject: Re: [R] Integral of PDF You can dive into the thread puzzle with integrate over infinite range from September this year. The short answer appears to be: Increase the error tolerance. integrate(function(x) dnorm(x, 500,50), -Inf, Inf, subdivisions=500, rel.tol=1e-11) # 1 with absolute error 1.1e-12 Hans Werner Doran, Harold wrote: The integral of any probability density from -Inf to Inf should equal 1, correct? I don't understand last result below. integrate(function(x) dnorm(x, 0,1), -Inf, Inf) 1 with absolute error 9.4e-05 integrate(function(x) dnorm(x, 100,10), -Inf, Inf) 1 with absolute error 0.00012 integrate(function(x) dnorm(x, 500,50), -Inf, Inf) 8.410947e-11 with absolute error 1.6e-10 all.equal(integrate(function(x) dnorm(x, 500,50), -Inf, Inf)$value, 0) [1] TRUE sessionInfo() R version 2.10.1 (2009-12-14) i386-pc-mingw32 locale: [1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] statmod_1.4.6 mlmRev_0.99875-1 lme4_0.999375-35 Matrix_0.999375-33 lattice_0.17-26 loaded via a namespace (and not attached): [1] grid_2.10.1 nlme_3.1-96 stats4_2.10.1 tools_2.10.1 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://r.789695.n4.nabble.com/Integral-of-PDF-tp3070243p3070315.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Integral of PDF
Albyn's reply is in line with an hypothesis I was beginning to
formulate (without looking at the underlying FoRTRAN code),
prompted by the hint in '?integrate':
Details:
If one or both limits are infinite, the infinite range
is mapped onto a finite interval.
For a finite interval, globally adaptive interval
subdivision is used in connection with extrapolation
by the Epsilon algorithm.
as a result of some numerical experiments. Following up on
Harold Doran's original dnorm(x,mean.mean/10) pattern (and
quoting a small subset of the experiments ... ):
integrate(function(x) dnorm(x, 100,10), -Inf, Inf)
# 1 with absolute error 0.00012
integrate(function(x) dnorm(x, 200,20), -Inf, Inf)
# 1.429508e-08 with absolute error 2.8e-08
integrate(function(x) dnorm(x, 140,14), -Inf, Inf)
# 1 with absolute error 2.2e-06
integrate(function(x) dnorm(x, 150,15), -Inf, Inf)
# 2.261582e-05 with absolute error 4.4e-05
integrate(function(x) dnorm(x, 144,14.4), -Inf, Inf)
# 1 with absolute error 1.7e-06
integrate(function(x) dnorm(x, 145,14.5), -Inf, Inf)
# 5.447138e-05 with absolute error 0.00011
integrate(function(x) dnorm(x, 150,15), -33000, 33300)
# 1 with absolute error 0.00012
integrate(function(x) dnorm(x, 150,15), -34000, 34300)
# 5.239671e-05 with absolute error 0.00010
integrate(function(x) dnorm(x, 150,15),
-33768.260234277, 34068.2602334277)
# 0.5000307 with absolute error 6.1e-05
integrate(function(x) dnorm(x, 150,15),
-33768.260234278, 34068.2602334278)
# 6.139415e-05 with absolute error 0.00012
So it seems that, depending on how integrate() maps to a finite
interval, and on how the algorithm goes about its adaptive
interval subdivision, there are critical points where integrate()
switches from one to another of the following:
[A] The whole of a finite interval containing all but the extreme
outer tails of dnorm() is integrated over;
[B] The whole of a finite interval containing one half of the
distribution of dnorm() is integrated over;
[C] The whole of a finite interval lying in the extreme of one
tail of the dnorm distribution is integrated over.
with result: [A] close to 1; [B] close to 0.5; [C] close to 0.
This is compatible with Albyn's conclusion that the integral is
split into the sum of (-Inf,0) and (0,Inf), with the algorithm
ignoring one, or the other, or both, or neither!
This must be one of the nastiest exemplar's of rounding error ever!!!
Ted.
On 02-Dec-10 22:39:37, Albyn Jones wrote:
To really understaand it you will have to look at the fortran code
underlying integrate. I tracked it back through a couple of layers
(dqagi, dqagie, ... just use google, these are old netlib
subroutines) then decided I ought to get back to grading papers :-)
It looks like the integral is split into the sum of two integrals,
(-Inf,0] and [0,Inf).
integrate(function(x) dnorm(x, 350,50), 0, Inf)
1 with absolute error 1.5e-05
integrate(function(x) dnorm(x, 400,50), 0, Inf)
1.068444e-05 with absolute error 2.1e-05
integrate(function(x) dnorm(x, 500,50), 0, Inf)
8.410947e-11 with absolute error 1.6e-10
integrate(function(x) dnorm(x, 500,50), 0, 1000)
1 with absolute error 7.4e-05
The integral from 0 to Inf is the lim_{x - Inf} of the integral
over [0,x]. I suspect that the algorithm is picking an interval
[0,x], evaluating the integral over that interval, and then performing
some test to decide whether to expand the interval. When the initial
interval doesn't contain much, expanding a little won't add enough to
trigger another iteration.
albyn
On Thu, Dec 02, 2010 at 04:21:34PM -0500, Doran, Harold wrote:
The integral of any probability density from -Inf to Inf should equal
1, correct? I don't understand last result below.
integrate(function(x) dnorm(x, 0,1), -Inf, Inf)
1 with absolute error 9.4e-05
integrate(function(x) dnorm(x, 100,10), -Inf, Inf)
1 with absolute error 0.00012
integrate(function(x) dnorm(x, 500,50), -Inf, Inf)
8.410947e-11 with absolute error 1.6e-10
all.equal(integrate(function(x) dnorm(x, 500,50), -Inf, Inf)$value,
0)
[1] TRUE
sessionInfo()
R version 2.10.1 (2009-12-14)
i386-pc-mingw32
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United
States.1252
[3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] statmod_1.4.6 mlmRev_0.99875-1 lme4_0.999375-35
Matrix_0.999375-33 lattice_0.17-26
loaded via a namespace (and not attached):
[1] grid_2.10.1 nlme_3.1-96 stats4_2.10.1 tools_2.10.1
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Re: [R] Integral of PDF
On Thu, Dec 02, 2010 at 06:23:45PM -0500, Ravi Varadhan wrote: A simple solution is to locate the mode of the integrand, which should be quite easy to do, and then do a coordinate shift to that point and then integrate the mean-shifted integrand using `integrate'. Ravi. Translation: think before you compute! albyn __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
