Very high residual Sum-of-Squares - r

I'm having a problem with the square sum-of-residues of an fitting. The square sum of the residues is too high which indicates that the fit is not very good. However, visually it looks fine to have this very high residual value ... Can anyone help me to know what's going on?
My data:
x=c(0.017359, 0.019206, 0.020619, 0.021022, 0.021793, 0.022366, 0.025691, 0.025780, 0.026355, 0.028858, 0.029766, 0.029967, 0.030241, 0.032216, 0.033657,
0.036250, 0.039145, 0.040682, 0.042334, 0.043747, 0.044165, 0.044630, 0.046045, 0.048138, 0.050813, 0.050955, 0.051910, 0.053042, 0.054853, 0.056886,
0.058651, 0.059472, 0.063770,0.064567, 0.067415, 0.067802, 0.068995, 0.070742,0.073486, 0.074085 ,0.074452, 0.075224, 0.075853, 0.076192, 0.077002,
0.078273, 0.079376, 0.083269, 0.085902, 0.087619, 0.089867, 0.092606, 0.095944, 0.096327, 0.097019, 0.098444, 0.098868, 0.098874, 0.102027, 0.103296,
0.107682, 0.108392, 0.108719, 0.109184, 0.109623, 0.118844, 0.124023, 0.124244, 0.129600, 0.130892, 0.136721, 0.137456, 0.147343, 0.149027, 0.152818,
0.155706,0.157650, 0.161060, 0.162594, 0.162950, 0.165031, 0.165408, 0.166680, 0.167727, 0.172882, 0.173264, 0.174552,0.176073, 0.185649, 0.194492,
0.196429, 0.200050, 0.208890, 0.209826, 0.213685, 0.219189, 0.221417, 0.222662, 0.230860, 0.234654, 0.235211, 0.241819, 0.247527, 0.251528, 0.253664,
0.256740, 0.261723, 0.274585, 0.278340, 0.281521, 0.282332, 0.286166, 0.288103, 0.292959, 0.295201, 0.309456, 0.312158, 0.314132, 0.319906, 0.319924,
0.322073, 0.325427, 0.328132, 0.333029, 0.334915, 0.342098, 0.345899, 0.345936, 0.350355, 0.355015, 0.355123, 0.356335, 0.364257, 0.371180, 0.375171,
0.377743, 0.383944, 0.388606, 0.390111, 0.395080, 0.398209, 0.409784, 0.410324, 0.424782 )
y= c(34843.40, 30362.66, 27991.80 ,28511.38, 28004.74, 27987.13, 22272.41, 23171.71, 23180.03, 20173.79, 19751.84, 20266.26, 20666.72, 18884.42, 17920.78, 15980.99, 14161.08, 13534.40, 12889.18, 12436.11,
12560.56, 12651.65, 12216.11, 11479.18, 10573.22, 10783.99, 10650.71, 10449.87, 10003.68, 9517.94, 9157.04, 9104.01, 8090.20, 8059.60, 7547.20, 7613.51, 7499.47, 7273.46, 6870.20, 6887.01,
6945.55, 6927.43, 6934.73, 6993.73, 6965.39, 6855.37, 6777.16, 6259.28, 5976.27, 5835.58, 5633.88, 5387.19, 5094.94, 5129.89, 5131.42, 5056.08, 5084.47, 5155.40, 4909.01, 4854.71,
4527.62, 4528.10, 4560.14, 4580.10, 4601.70, 3964.90, 3686.20, 3718.46, 3459.13, 3432.05, 3183.09, 3186.18, 2805.15, 2773.65, 2667.73, 2598.55, 2563.02, 2482.63, 2462.49, 2478.10,
2441.70, 2456.16, 2444.00, 2438.47, 2318.64, 2331.75, 2320.43, 2303.10, 2091.95, 1924.55, 1904.91, 1854.07, 1716.52, 1717.12, 1671.00, 1602.70, 1584.89, 1581.34, 1484.16, 1449.26,
1455.06, 1388.60, 1336.71, 1305.60, 1294.58, 1274.36, 1236.51, 1132.67, 1111.35, 1095.21, 1097.71, 1077.05, 1071.04, 1043.99, 1036.22, 950.26, 941.06, 936.37, 909.72, 916.45,
911.01, 898.94, 890.68, 870.99, 867.45, 837.39, 824.93, 830.61, 815.49, 799.77, 804.84, 804.88, 775.53, 751.95, 741.01, 735.86, 717.03, 704.57, 703.74, 690.63,
684.24, 650.30, 652.74, 612.95 )
Then make fit using the nlsLM function (minpack.lm package):
library(magicaxis)
library(minpack.lm)
sig.backg=3*10^(-3)
mod <- nlsLM(y ~ a *( 1 + (x/b)^2 )^c+sig.backg,
start = c(a = 0, b = 1, c = 0),
trace = TRUE)
## plot data
magplot(x, y, main = "data", log = "xy", pch=16)
## plot fitted values
lines(x, fitted(mod), col = 2, lwd = 4 )
This value is the residue:
> print(mod)
Nonlinear regression model
model: y ~ a * (1 + (x/b)^2)^c + sig.backg
data: parent.frame()
a b c
68504.2013 0.0122 -0.6324
residual sum-of-squares: 12641435
Number of iterations to convergence: 34
Achieved convergence tolerance: 0.0000000149
sum-of-squares residual is too high : 12641435 ...
Is that so or is something wrong with the adjustment? It is bad?


It makes sense, since the squared mean of your response variable is 38110960. You can scale your data if you prefer to work with smaller numbers.


The residual sum of squares doesn't have much meaning without knowing the total sum of squares (from which R^2 can be calculated). Its value is going to increase if your data have large values or if you add more data points, regardless of how good your fit is. Also, you may want to look at a plot of your residuals versus fitted data, there is a clear pattern that should be explained by your model to ensure that your errors are Normally distributed.

Related

MLE of a distribution in R: fitdistrplus (SGT distribution), how do I do it?

For return data I am doing research about the importance of skewness and kurtosis for the cVaR calculation. We are comparing some distributions first, by estimating the parameters of the distribution using fitdist() in R using package "fitdistrplus". However, we want to do this for a various number of distributions (see picture: SGT, GT, SGED, GED, t, norm).
Below is a sample code for SGT, where there is a problem: it introduced NaN's for the standard errors for parameters p and q. I also don't really know how to choose the starting values exactly.
SGTstart <- list(mu=0,sigma=2, lambda = 0.5, p=2, q=8)
SGTfit_R <- fitdistrplus::fitdist(data = as.vector(coredata(R)), distr = "sgt", method = "mle", SGTstart)
summary(SGTfit_R)
Sample of the data to make it reproducable: return vector of my stock index
c("0", "-1,008599424", "0,73180187", "0,443174024", "-0,351935172", "-1,318784086", "-2,171323799", "1,270243431", "-0,761354019", "0,417350946", "0,906432976", "-0,066736422", "-0,867085373", "-0,119914361", "-0,300989601", "0,482518259", "0,787365385", "-1,443105439", "-0,318546686", "-3,467674998", "1,041540157", "1,371281289", "-1,176752782", "-1,116893343", "-0,127522915", "-0,658070287", "1,098348016", "0,296391358", "-0,810635352", "-0,041779322", "0,353974233", "0,120090141", "0,304927119", "-1,22772592", "0,040768364", "1,182218724", "0,123136685", "-0,682709972", "-0,174093506", "-0,539704174", "0,579080595", "0,326346169", "0,205503526", "-0,771928642", "1,490828799", "0,734822712", "-0,025733101", "0,246531452", "-0,695585736", "-0,732413919", "0,806417952", "0,396105099", "0,024558388", "-0,791232528", "0,730410255", "-1,438890702", "0,668400286", "1,440996039", "0,731823553", "0,177515522", "0,740085418", "0,926248628", "-0,63516084", "-0,89996829", "1,655117371", "0,501033581", "0,06526534", "1,320866692", "-0,496350734", "-0,10157668", "0,022333393", "-1,236934596", "-1,070586427", "0,661662029", "0,871334714", "0,758891429", "0,064748766", "-0,305132153", "-0,424033661", "1,223444774", "-0,441840866", "-0,661390655", "-2,148399329", "0,843067435", "0,601099664", "-0,329590349", "0,210791225", "-0,341341769", "-0,555892395", "0,624026986", "0,218851965", "-0,015859171", "0,524283138", "-0,855634719", "0,339281481", "0,038507713", "-1,943784688", "0,315857689", "-0,368982834", "-1,111684011", "-0,2409217", "0,421815833", "-0,079319721", "0,915338199", "0,537387704", "-0,023004636", "-0,331854888", "0,702733882", "-1,084343115", "0,16901282", "0,559404916", "-0,538587484", "0,153683523", "-0,336562411", "-0,274946953", "0,862901957", "0,117407383", "1,205205829", "0,633347347", "0,058712615", "-0,083562948", "1,343190727", "1,281380185", "0,750972389", "-1,538678151", "0,228222073", "0,635385022", "0,037379479", "-0,491444798", "-1,220272752", "1,093162287", "1,499512169", "0,041394336", "-0,113330512", "0,657485999", "-0,264647978", "0,115056075", "-0,009763771", "0,454629881", "0,322398317", "0,347112494", "0,948127411", "0,461194301", "-0,407013048", "-0,469481931", "-0,536045151", "0,114726251", "0,396772868", "0,525885581")
Best, Enjo

The answer was using package sgt

Double summation in function which will be used for log likelihood estimation

I am trying to get the estimation parameters for the following equation:
First, I need to write the function with double summation. Then, I will use any optimization method to get the estimated parameters (can be nlm, optim etc). Here is my attempt (though, I am not sure how to add the double summation part in the end, two for loops):
There are 5 parameters: alpha_x, sigma_x, lambda_x, theta_x, phi_x >>> step_3[1] == alpha_x
for (a in 1:nrow(returns[[4]][,1])){
abc <- 0
for (b in 0:7){
abc <- abc + (step_3[3]^b*(factorial(b))^-1*(sqrt(step_3[2]^2+b*step_3[5]^2))^-1*exp(-1*(as.numeric(returns[[4]][a,1])-(step_3[1]+b*step_3[4]))^2*(2*(step_3[2]^2+b*step_3[5]^2))^-1))
}
abcd <- abcd + log(abc)
}
abcd <- -abcd + 0.5*nrow(returns[[4]][,1])*log(2*pi) + nrow(returns[[4]][,1])*step_3[3]
return(abcd)
}
nlm(s3, step_3 <- c(0.01,0.01,0.01,0.01,0.01), hessian = TRUE )
I am getting the estimation results, but the estimates are way off:
My question is how to implement (efficiently, any suggestions for sapply etc. would be appreciated) the double summation (sum_0_T) > ln (sum_1_l) into a function. The function will be used in optimization.
(l=0:7, T=number of observations)
return1 = some vector with price returns (all returns between -1,1).
After the estimation sample mean and variance should be roughly equal to:
sample_mean = alpha_x + lambda_x * theta_x
sample_variance = sigma_x^2 + lambda_x * (theta_x^2 + phi_x^2)
Thanks.
Here is a sample that can be used for the analysis:
c(0.02423747 ,-0.00419738, -0.03153278, 0.05343888, -0.0175492, 0.00848472, 0.01043673, -0.02123556, 0.01954968, -0.06462473, 0.02679873, 0.07971938, 0.00050474, -0.01768566,
-0.05206762, -0.00413691, 0.06390499, 0.00269576, 0.01520837, 0.00091051, 0.03499043, -0.00121999, -0.00123521, -0.01961684, 0.03360355, 0.01803711, 0.01772631, 0.036523
-0.00038927, 0.00905013, 0.01150976, 0.00480223, 0.01916402, 0.00054628, 0.01911904, 0.02194556, 0.00371314, 0.03376601, 0.0546574, -0.03972611, -0.0272525, 0.00271509,
0.02137819, 0.00483075, 0.03538795, 0.02981431, 0.00428509, -0.07192935, 0.01770175, -0.09626522, 0.07574215, 0.02929555, 0.01776551, 0.0385604, -0.06804089, 0.0666583,
0.01304272, -0.01825728, 0.01703525, 0.02022584, 0.03348027, 0.02818876, -0.00162942, -0.08785954, -0.13366772, 0.10243928)

How to calculate the root mean squared deviation in R?

I state that I never used the root mean squared deviation before.
I'm just tryng to reproduce what I found in an article.
Pratically, I have to quantify the noise of a "methodology" (that is the result of different noises, because of the coupling of two instruments), measuring the noise of the methodology at three different points out of the normal operation, where we know that it measures only the noise.
At the end, in the process that I'm following, you have to calculate the standard deviation among this three points and multiply it for the factor of 1.96 for the 95th percentile confidence interval (and in this way get a Limit of Detection of the method).
The time resolution is 30 minutes, then a standard deviation among three points, then the following three points, and so on. I already have a dataset organized in this way and I have calculated the standard deviation.
Just because I'm following the method of an article, they compare the limit of detection calculated using the standard deviation and the one using the root mean squared deviation.
Because at the end I have to use this limit of detection to filter my data by noise, I want to compare which is the best method for my situation, as they did.
How can I calculate the root mean squared deviation, as well as I calculated the standard deviation? Each three points (three differente columns), then the three following points (same three columns but the following row) and so on?
I already tried with the rmse function of the Metrics package, but the problem is the it require only two values: an actual and a predicted values.
Of course, as well as I did for the standard deviation calculation, I should use aggregate to iterate the function each three columns per each row.
EDIT
I put a piece of the article that I'm following for this method, to try to make you understand better what I would like
"For time lags much different from the true time lag, the standard deviation of the flux provides a measure of the random error affecting the flux"......."Multiplying this measure of the random error by α gives an estimate of the measurement precision at a given confidence interval (α =1.96 for the 95th percentile; α =3 for the 99th percentile) which can be used as the flux limit of detection (LoD) (i.e. LoDσ = α ×REσ )......"A modification of the LoDσ approach is to calculate the random error based on the root mean squared deviation (RMSE) of flux from zero ,which reflects the variability in the crosscovariance function in these regions, but also its offset from zero
I share with you part of the dataset and the code the I tried to use.
Dataset
structure(list(`101_LOD` = c(-0.00647656063436054, 0.00645714072316343,
0.00174533523902105, -0.000354643362187957, -0.000599093190801188,
0.00086188829059792), `101_LOD.1` = c(0.00380625456526623, -0.00398115037246045,
0.00158673927930099, -0.00537583996746438, -0.00280048350643599,
0.00348232298529063), `101_LOD.2` = c(-0.00281100080425964, -0.00335537844222041,
0.00611652518452308, -0.000738139825060029, 0.00485039477849737,
0.00412428118507656), `107_LOD` = c(0.00264717678436649, 0.00339296025595841,
0.00392733001719888, 0.0106686039973083, 0.00886643251752075,
0.0426091484273961), `107_LOD.1` = c(0.000242380702002215, -0.00116108069669281,
0.0119784744970561, 0.00380805756323248, 0.00190407945251567,
0.00199684331869391), `107_LOD.2` = c(-0.0102716279438754, -0.00706528150567528,
-0.0108745954674186, -0.0122962259781756, -0.00590383880635847,
-0.00166664119985051), `111_LOD` = c(-0.00174374098054644, 0.00383270191075735,
-0.00118363208946644, 0.00107908760333878, -9.30127551375776e-05,
-0.00141500588842743), `111_LOD.1` = c(0.000769378300959002,
0.00253820252869653, 0.00110643824418424, -0.000338050323261079,
-0.00313666295753596, 0.0043919374295125), `111_LOD.2` = c(0.000177265973907964,
0.00199829884609846, -0.000490950219515303, -0.00100263695578483,
0.00122606902671889, 0.00934018452187161), `113_LOD` = c(0.000997977666838309,
0.0062400770296875, -0.00153620247996209, 0.00136849054508488,
-0.00145700847633675, -0.000591288575933268), `113_LOD.1` = c(-0.00114161441697546,
0.00152607521404826, 0.000811193628975422, -0.000799514037634276,
-0.000319008435039752, -0.0010086036089075), `113_LOD.2` = c(-0.000722312098377764,
0.00364767954707251, 0.000547744649351312, 0.000352509651080838,
-0.000852173274761947, 0.00360487150682726), `135_LOD` = c(-0.00634051802134062,
0.00426062889500736, 0.00484049067127332, 0.00216220020394825,
0.00165634168942681, -0.00537970105199375), `135_LOD.1` = c(-0.00209301968088832,
0.00535855274344209, -0.00119679744329422, 0.0041216882161451,
0.00512978202611836, 0.0014048506490567), `135_LOD.2` = c(0.00022377545723911,
0.00400550696583795, 0.00198972253447825, 0.00301341644871015,
0.00256802839330668, 0.00946109288597202), `137_LOD` = c(-0.0108508893475138,
-0.0231919072487789, -0.00346546003410657, -0.00154066625155414,
0.0247266017774909, -0.0254464953061609), `137_LOD.1` = c(-0.00363025194918789,
-0.00291104074373261, 0.0024998477144967, 0.000877707284759669,
0.0095477003599792, 0.0501795740749602), `137_LOD.2` = c(0.00930498343499501,
-0.011839104725282, 0.000274929503053888, 0.000715665078729413,
0.0145503185102915, 0.0890428314632625), `149_LOD` = c(-0.000194406250680231,
0.000355157226357547, -0.000353931679163222, 0.000101471293242973,
-0.000429409422518444, 0.000344585379249552), `149_LOD.1` = c(-0.000494386150759807,
0.000384907974061922, 0.000582537329068263, -0.000173285705433721,
-6.92758935962043e-05, 0.00237942557324254), `149_LOD.2` = c(0.000368606958615297,
0.000432568466833549, 3.33092313366271e-05, 0.000715304544370804,
-0.000656902381786168, 0.000855422043674721), `155_LOD` = c(-0.000696168382693618,
-0.000917607266525328, 4.77049670728094e-06, 0.000140297660927979,
-5.99898679530658e-06, 6.71169142984434e-06), `155_LOD.1` = c(-0.000213644203677328,
-3.44396001911029e-07, -0.000524232671878577, -0.000830180665933627,
1.47799998238307e-06, -5.97640014667251e-05), `155_LOD.2` = c(-0.000749882784933487,
0.000345737159390042, -0.00076916001239521, -0.000135205762575321,
-2.55352420251723e-06, -3.07199008030628e-05), `31_LOD` = c(-0.00212014938530172,
0.0247411322547065, -0.00107990654365844, -0.000409195814154659,
-0.00768439381433953, 0.001860128524035), `31_LOD.1` = c(-0.00248488588195854,
-0.011146734518705, -0.000167943850441196, -0.0021998906531997,
0.0166775965182051, -0.0156939303287719), `31_LOD.2` = c(0.00210626277375321,
-0.00327815351414411, -0.00271043947479133, 0.00118991079627845,
-0.00838520090692615, 0.0255825346347586), `33_LOD` = c(0.0335175783154054,
0.0130192144768818, 0.0890608024914352, -0.0142431454793663,
0.00961009674973182, -0.0429774973256228), `33_LOD.1` = c(0.018600175159935,
0.04588362587764, 0.0517479021554752, 0.0453766081395813, -0.0483559729403664,
0.123771869764484), `33_LOD.2` = c(0.01906507758481, -0.00984821669825455,
0.134177176083007, -0.00544320457445977, 0.0516083894733814,
-0.0941500564321804), `39_LOD` = c(-0.148517395684098, -0.21311281527214,
0.112875846920874, -0.134256453140454, 0.0429030528286934, -0.0115143877745049
), `39_LOD.1` = c(-0.0431568202849291, -0.159003698955288, 0.0429009071238143,
-0.126060096927082, -0.078848020069061, -0.0788748111534866),
`39_LOD.2` = c(-0.16276833960171, 0.0236589399437796, 0.0828435027244962,
-0.50219849047847, -0.105196237549017, -0.161206838628339
), `42_LOD` = c(-0.00643926654994104, -0.0069253267922805,
7.63419856289838e-05, -0.0185223126108671, 0.00120855708103566,
-0.00275288147011515), `42_LOD.1` = c(-0.000866169150506504,
-0.00147791175852563, -0.000670310173141084, -0.00757733007180311,
0.0151353172950393, -0.00114193461500327), `42_LOD.2` = c(0.00719928454572906,
0.00311615354837406, 0.00270759483782046, -0.0108062423259522,
0.00158765505419478, -0.0034831499672973), `45_LOD` = c(0.00557787518897268,
0.022337270533665, 0.00657118689440082, -0.00247269227623608,
0.0191646343214611, 0.0233090596023039), `45_LOD.1` = c(-0.0305395220788143,
0.077105031761457, -0.00101713990356452, 0.0147500116150713,
-5.43009569586179e-05, -0.0235006181977403), `45_LOD.2` = c(-0.0216498682456909,
-0.0413426968184435, -0.0210779895848601, -0.0147549519865421,
0.00305229143870313, -0.0483293292336662), `47_LOD` = c(-0.00467568767221499,
-0.0199796182799552, 0.00985966068611855, -0.031010117051163,
0.0319279109813341, 0.0350743318265918), `47_LOD.1` = c(0.00820166533285921,
-0.00748186905620154, -0.010483251821707, -0.00921919551377505,
0.0129546148757833, 0.000223462281435923), `47_LOD.2` = c(0.00172469728530889,
0.0181683409295075, 0.00264937907258855, -0.0569837400476351,
0.00514558635349483, 0.0963339573489031), `59_LOD` = c(-0.00664210061621158,
-0.062069664217766, 0.0104345353700492, 0.0115323589989968,
-0.000701276829098035, -0.0397759501000331), `59_LOD.1` = c(-0.00844888486350536,
0.0207426674766074, -0.0227755432761471, -0.00370561240222376,
0.0152046240483297, -0.0127327412801225), `59_LOD.2` = c(-0.000546590647534814,
0.0178115310450356, 0.00776130696191998, 0.00162470375408126,
-0.036140754156005, 0.0197791914089296), `61_LOD` = c(0.00797528044191513,
-0.00358928087671818, 0.000662870138322471, -0.0412142836466128,
-0.00571822580078707, -0.0333870884803465), `61_LOD.1` = c(0.000105849888219735,
-0.00694734283847093, -0.00656216592134899, 0.00161225110022219,
0.0125744958934939, -0.0178560868664668), `61_LOD.2` = c(0.0049288443167774,
0.0059411543659837, -0.00165857112209555, -0.0093669075333705,
0.00655185371925189, 0.00516436591134869), `69_LOD` = c(0.0140014747729604,
0.0119645827116724, 0.0059880663080946, -0.00339119330845176,
0.00406436116298777, 0.00374425148741196), `69_LOD.1` = c(0.00465076983995792,
0.00664902297016735, -0.00183936649215524, 0.00496509351837152,
-0.0224812403463345, -0.0193087796456654), `69_LOD.2` = c(-0.00934638876711703,
-0.00802183076602164, 0.00406752039394799, -0.000421337136630527,
-0.00406768983408334, -0.0046016148041856), `71_LOD` = c(-0.00206064862123214,
0.0058604630066848, -0.00353440181333921, -0.000305197461077327,
0.00266085011303462, -0.00105635261106644), `71_LOD.1` = c(3.66652318354654e-06,
0.00542612739642576, 0.000860385212430484, 0.00157520645492044,
-0.00280256517377998, -0.00474358065422048), `71_LOD.2` = c(-0.00167098030843413,
0.0059622082597603, -0.00121597491543965, -0.000791592953383716,
-0.0022790991468459, 0.00508978650148816), `75_LOD` = c(NA,
-0.00562613898652477, -0.000103076958936504, -3.76628574664693e-05,
-0.000325767611573817, 0.000117404893823389), `75_LOD.1` = c(NA,
NA, -0.000496324358203359, -0.000517476831074487, -0.00213096062838051,
-0.00111202867609916), `75_LOD.2` = c(NA, NA, -0.000169651845347418,
-4.72864955070539e-05, -0.00144880109085214, 0.00421635976535877
), `79_LOD` = c(-0.0011901810540199, 0.00731686066269579,
0.00538551997145174, -0.00578723012473479, -0.0030246805255648,
0.00146141135533218), `79_LOD.1` = c(-0.00424278455960268,
-0.010593752642875, 0.0065136497427927, -0.00427355522802769,
0.000539975609490915, -0.0206849687839064), `79_LOD.2` = c(-0.00366739576561779,
-0.00374066839898667, -0.00132764684703939, -0.00534145222725701,
0.00920940542227595, -0.0101871763957068), `85_LOD` = c(-0.0120254177480422,
0.00369546541331518, -0.00420718877886963, 0.00414911885475517,
-0.00130381692844529, -0.00812757789798261), `85_LOD.1` = c(-0.00302024868281014,
0.00537704163310547, 0.00184264538884543, -0.00159032685888543,
-0.0062127769817834, 0.00349476605688194), `85_LOD.2` = c(0.0122689407380797,
-0.00509605601025503, -0.00641413996554198, 0.000592176121486696,
0.00131237912317341, -0.00535018996837309), `87_LOD` = c(0.00613621268007298,
0.000410268892659307, -0.00239014321624482, -0.00171179729894864,
-0.00107159765522861, -0.00708388174601732), `87_LOD.1` = c(0.00144787264098156,
-0.0025946273860992, -0.00194897899110034, 0.00157863310440493,
-0.0048913305554607, -0.000585669821053749), `87_LOD.2` = c(-0.00224691693198253,
-0.00277315666829267, 0.00166487067514155, -0.00173757960229744,
-0.00362252480121682, -0.0101992979591839), `93_LOD` = c(-0.0234225447373586,
0.0390095666365413, 0.00606244490932179, 0.0264258422783391,
0.0161211132913951, -0.0617678157059), `93_LOD.1` = c(-0.0124876313221369,
-0.0309636779639578, 0.00610883313140442, -0.0192442672220773,
0.0129557286224975, -0.00869066964782635), `93_LOD.2` = c(-0.0219837540560547,
-0.00521242297372905, 0.0179965615561871, 0.0081370991723329,
1.45427765512579e-06, -0.0111199632179688), `99_LOD` = c(0.00412086456443205,
-0.00259940538393106, 0.00742537463584133, -0.00302091572866969,
-0.00320466045653491, -0.00168702410433936), `99_LOD.1` = c(0.00280546156134205,
-0.00472591065687533, 0.00518402193979284, -0.00130887074314965,
0.00148769905391341, 0.00366250488078969), `99_LOD.2` = c(-0.00240469207099292,
-9.57307699040024e-05, -0.000145493235845501, 0.000667454164326723,
-0.0057445759245933, 0.00433464631989088), H_LOD = c(-6248.9128518109,
-10081.9540490064, -6696.91582671427, -5414.20614601348,
-3933.64339240365, -13153.7509294302), H_LOD.1 = c(-6.2489128518109,
-10.0819540490064, -6.69691582671427, -5.41420614601348,
-3.93364339240365, -13.1537509294302), H_LOD.2 = c(-6248.9128518109,
-10081.9540490064, -6696.91582671427, -5414.20614601348,
-3933.64339240365, -13153.7509294302)), row.names = c(NA,
6L), class = "data.frame")
Code
LOD_rdu=sapply(split.default(LOD_ut, rep(seq((ncol(LOD_ut) / 3)), each = 3)), function(i)
apply(i, 1, rmse))
And I get this error Error in mse(actual, predicted) :
argument "predicted" is missing, with no default

It is difficult to understand precisely what you need, I will try to answer you,
From wikipedia, RMSD serves to compare a dataset generated by a model (the model in your article i guess) to an observed distribution.
From CRAN, RMSE function in modelr package has two arguments, model and data:
modelr::rmse(model = ,data = )
This function will give you the fit of your model to the data. The first argument is a model, meaning you will probably use a function like lm() to generate it. Because you don't detail the model I can't help you more.
The second argument is the dataset, the one you provide is quite disturbing to me. R will expect a tidy set with two columns x the time of the observation and y the value.

you can first what groups your columns:
GRP = sub("[.][0-9]*","",colnames(LOD_ut))
head(GRP)
[1] "101_LOD" "101_LOD" "101_LOD" "107_LOD" "107_LOD" "107_LOD"
you can also use 1:(ncol(LOD_ut)/3), just that the above gives you the groups back.
We can use the above like this:
LOD_ut[,GRP=="101_LOD"]
101_LOD 101_LOD.1 101_LOD.2
1 -0.0064765606 0.003806255 -0.0028110008
2 0.0064571407 -0.003981150 -0.0033553784
3 0.0017453352 0.001586739 0.0061165252
4 -0.0003546434 -0.005375840 -0.0007381398
5 -0.0005990932 -0.002800484 0.0048503948
6 0.0008618883 0.003482323 0.0041242812
This calls out your first three grouped columns. Now if you do apply(..,1,sd) you get the standard deviation, now we just do it over all groups
SD = sapply(unique(GRP),function(i)apply(LOD_ut[,GRP==i],1,sd,na.rm=T))
If you must do RMSE, use predicted to be the mean:
rmse_func = function(x){
Metrics:::rmse(mean(x,na.rm=T),x[!is.na(x)])
}
RMSE = sapply(unique(GRP),function(i){
apply(LOD_ut[,GRP==i],1,rmse_func)
})

Search for optimal parameter value with curve fitting (difference equations)

I have two experimentally determined variables (y1_exp, y2_exp) which change with x, where y2_exp affects y1_exp with changing x (however y1_exp does not affect y2_exp). Here is the data:
library(tidyverse)
library(minpack.lm)
data <- read_csv("x,y1_exp,y2_exp
0,0.00553214,0.006103046
5,0.005526549,0.006079372
10,0.005521163,0.006092119
15,0.005528608,0.006065714
20,0.005526002,0.006062165
25,0.005523214,0.006057619
30,0.005523214,0.006048626
35,0.005523214,0.006044087
40,0.005523214,0.006044087
45,0.005520607,0.006048626
50,0.005512257,0.006025025
55,0.005520607,0.006013361
60,0.005518752,0.006007923
65,0.005524318,0.006025025
70,0.005520607,0.006016856
75,0.005513185,0.006002485
80,0.005515968,0.005995235
85,0.005788141,0.006016856
90,0.005733146,0.006033983
95,0.005683015,0.006028386
100,0.005644958,0.006013984
105,0.005618042,0.006004894
110,0.005604063,0.006010348
115,0.005590122,0.006010348
120,0.005572465,0.005997623
125,0.005572465,0.006000349
130,0.005567818,0.00599944
135,0.005567818,0.006000349
140,0.005569677,0.005997623
145,0.005567818,0.005990351
150,0.005567818,0.00598308
155,0.00555401,0.005977626
160,0.005569677,0.00598308
165,0.005569677,0.005977626
170,0.00556503,0.00598287
175,0.005562242,0.00598287
180,0.005572465,0.005974678
185,0.005578041,0.00598287
190,0.005586308,0.005971947
195,0.005585377,0.005966486
200,0.005588169,0.00598287
205,0.005602975,0.005978088
210,0.005628144,0.005991762
215,0.005642127,0.005992449
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225,0.00565609,0.005984232
230,0.00565609,0.005994952
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240,0.005672896,0.006000438
245,0.005684107,0.006001111
250,0.005696264,0.006009352
255,0.005707486,0.005998364
260,0.005710292,0.006012776
265,0.005724372,0.006003606
270,0.005732801,0.006018033
275,0.005758179,0.006018033
280,0.005769437,0.006018033
285,0.005780694,0.006026969
290,0.005791952,0.006028809
295,0.00581447,0.006026969
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305,0.005845434,0.006037761
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395,0.006018524,0.006112613
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435,0.006067108,0.00613115
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3320,0.006360987,0.00659698")
Using first order difference equation (recursively calculated function) I calculated modeled value of y2_mod at each x.
Parameter that I wanted to fit is a (parameter b is fixed).
I used nls.lm function for fitting. This is how the code looks:
fitparms <- c(a = .008)
fn_residuals <- function(fitparms, data) {
b <- .00111
t <- data$x
dT <- t[2] - t[1]
N <- numeric(length(t))
y1_exp <- data$y1_exp
y2_mod <- numeric(length(t))
y2_mod[1] <- data$y2_exp[1]
for (h in seq_len(length(t)-1)) {
y2_mod[h+1] <- (fitparms[1] * (y1_exp[h+1] - y2_mod[h]) + b * (y1_exp[h+1] - y2_mod[h])) * dT + y2_mod[h]
}
data_2 <<- data %>%
add_column(y2_mod = y2_mod)
# residuals
ssqres <- data_2$y2_mod - data_2$y2_exp
# # return predicted vs experimental residual
return(ssqres)
}
# Ploting data
data_2 %>%
gather(y1_exp, y2_exp, y2_mod, key = "y", value = "value") %>%
ggplot(aes(x = x)) +
geom_line(aes(y = value, color = y))
# Fitting
fitval <- nls.lm(par = fitparms,
fn = fn_residuals,
lower = rep(0, 1), #defined lower limit for parameter
control = nls.lm.control(nprint = 1),
data = data)
This is my output:
It. 0, RSS = 0.00020291, Par. = 0.008
It. 1, RSS = 8.10975e-05, Par. = 0.00326637
It. 2, RSS = 7.9177e-05, Par. = 0.00349277
It. 3, RSS = 7.90245e-05, Par. = 0.00356081
It. 4, RSS = 7.90137e-05, Par. = 0.00357927
It. 5, RSS = 7.90129e-05, Par. = 0.00358413
It. 6, RSS = 7.90129e-05, Par. = 0.00358539
It. 7, RSS = 7.90129e-05, Par. = 0.00358572
It. 8, RSS = 7.90129e-05, Par. = 0.0035858
# Summary of fit
summary(fitval)
Parameters:
Estimate Std. Error t value Pr(>|t|)
a 3.586e-03 6.921e-05 51.81 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.000345 on 664 degrees of freedom
Number of iterations to termination: 8
Reason for termination: Relative error in the sum of squares is at most `ftol'.
# Estimated parameter
parest <- as.list(coef(fitval))
parest
$a
[1] 0.003585804
I am new to this field, so my questions are:
if this is the proper (right) way to solve such problems (is using nls.lm correct)?
Is there another better way of tackling the problem of finding optimal value for such curve fitting?
Also, I did not find any package that I could use for implementing difference equations for my kind of problem, i.e. I used y1_exp data at each x for calculation of y2_mod.
For example deSolve package has a special-purpose solver (method = "iteration") for discrete time models, difference equations (page 30 of Vignette: Solving Initial Value Differential Equations in R), however with this solver you can only input initial values (e.g.: y1_exp[1]).
So, is there a better way of writing a code for my kind difference equation problem, or does some package already exists that deals with
such problems?
Thanks! Jernej

I have not checked if everything is correct, but nls.lm is indeed one suitable way to go. Another, more advanced package with more flexibility is package FME.
You are right, package deSolve can also handle difference equations. Depending on how the equations are specified, one can either use method="euler" (if the model returns the difference) or method="iteration", if the model returns the new value. The model starts of course with the (vector of) first initial value(s), but if you want to force your model by an external input, you can use a forcing function, see ?forcings.
A possible approach could be to combine forcing with "iteration".
Thomas

Nonlinear least square in r

Check please, there is a good fit or not.
I have two vectors here:
x=c(30,110)
y=c(0.000760289, 0.000800320, 0.000830345, 0.000840353, 0.000860370,
0.000910414, 0.000990490, 0.001090594, 0.001200721, 0.001350912,
0.001531172, 0.001751533, 0.001961923, 0.002192402, 0.002463031,
0.002793899, 0.003185067, 0.003636604, 0.004148594, 0.004721127,
0.005394524, 0.006168989, 0.007014544, 0.007870894, 0.008758242,
0.009656474, 0.010565620, 0.011485709, 0.012396520, 0.013308162,
0.014271353, 0.015326859, 0.016525802, 0.017889059, 0.019447890,
0.021223636, 0.023145810, 0.025174229, 0.027391752, 0.029645106,
0.032337259, 0.035347424, 0.039375125, 0.043575783, 0.048003973,
0.052926206, 0.058307309, 0.064581189, 0.071947231, 0.080494476,
0.089891885, 0.100671526, 0.111971207, 0.124237571, 0.137975539,
0.153629149, 0.171239194, 0.190712664, 0.212079979, 0.235026373,
0.259457493, 0.282867017, 0.307830359, 0.334773680, 0.364001719,
0.395742526, 0.425596389, 0.458391314, 0.494558651, 0.534657357,
0.579354317, 0.616075034, 0.656680256, 0.701804548, 0.752133146,
0.808558032, 0.872226001, 0.944664487, 1.027837007, 1.124484096,
1.238426232)
and this My codes:
a=0.000066332
b=0.091394349
k=50.2552812
fit = nls(y ~ (a*b*exp(b*(x+0.5))*k)/((k*b)+(a*(exp(b*(x+0.5))-1))),
start=list(a=a,b=b, k=k))
plot(x,y,col="red", xlab="Age",ylab=expression(paste(mu)))
lines(x,predict(fit), col="blue")
Is this a good fitting ? could I also use the maximum likelihood estimation for this x and y to estimate a, b, and k ? Could give me some codes for doing MLE in r ? I need for comparing between NLS and MLE.
Really appreaciate for the helps. Thank you

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