# 1 Questions regarding distributions

## 1.1 How do I know the root name of a distribution?

The root name of a probability distribution is the name which is used the d, p, q, r functions. For base R distributions, the root names are given in R-intro : https://cran.r-project.org/doc/manuals/R-intro.html#Probability-distributions. For example, you must use "pois" for the Poisson distribution and not "poisson".

## 1.2 How do I find “non standard” distributions?

For non-standard distributions, you can either find a package implementing them or define by yourself. A comprehensive list of non-standard distributions is given in the Distributions task view https://cran.r-project.org/web/views/Distributions.html. Here are some two examples of user-defined distributions. A third example (shifted exponential) is given in FAQ 3.5.4.

• The Gumbel distribution
dgumbel <- function(x, a, b) 1/b*exp((a-x)/b)*exp(-exp((a-x)/b))
pgumbel <- function(q, a, b) exp(-exp((a-q)/b))
qgumbel <- function(p, a, b) a-b*log(-log(p))
data(groundbeef)
fitgumbel <- fitdist(groundbeef$serving, "gumbel", start=list(a=10, b=10)) • The zero-modified geometric distribution dzmgeom <- function(x, p1, p2) p1 * (x == 0) + (1-p1)*dgeom(x-1, p2) pzmgeom <- function(q, p1, p2) p1 * (q >= 0) + (1-p1)*pgeom(q-1, p2) rzmgeom <- function(n, p1, p2) { u <- rbinom(n, 1, 1-p1) #prob to get zero is p1 u[u != 0] <- rgeom(sum(u != 0), p2)+1 u } x2 <- rzmgeom(1000, 1/2, 1/10) fitdist(x2, "zmgeom", start=list(p1=1/2, p2=1/2)) ## 1.3 How do I set (or find) initial values for non standard distributions? As documented, we provide initial values for the following distributions: "norm", "lnorm", "exp", "pois", "cauchy", "gamma“, "logis", "nbinom", "geom", "beta", "weibull" from the stats package; "invgamma", "llogis", "invweibull", "pareto1", "pareto" from the actuar package. Look first at statistics and probability books such as • different volumes of N. L. Johnson, S. Kotz and N. Balakrishnan books, e.g. Continuous Univariate Distributions, Vol. 1, • Thesaurus of univariate discrete probability distributions by G. Wimmer and G. Altmann. • Statistical Distributions by M. Evans, N. Hastings, B. Peacock. • Distributional Analysis with L-moment Statistics using the R Environment for Statistical Computing by W. Asquith. If not available, find initial values by equalling theoretical and empirical quartiles. The graphical function plotdist() and plotdistcens() can also be used to assess the suitability of starting values : by an iterative manual process you can move parameter values so as to obtain a distribution that roughly fits the data and take these parameter values as starting values for the real fit. You may also consider the prefit() function to find initial values especially in case where parameters are constrained. ## 1.4 Is it possible to fit a distribution with at least 3 parameters? Yes, an example with the Burr distribution is detailed in the JSS paper. We reproduce it very quickly here. data("endosulfan") library("actuar") fendo.B <- fitdist(endosulfan$ATV, "burr", start = list(shape1 = 0.3, shape2 = 1, rate = 1))
summary(fendo.B)
## Fitting of the distribution ' burr ' by maximum likelihood
## Parameters :
##        estimate Std. Error
## shape1    0.206     0.0561
## shape2    1.540     0.3188
## rate      1.497     0.4683
## Loglikelihood:  -520   AIC:  1046   BIC:  1054
## Correlation matrix:
##        shape1 shape2   rate
## shape1  1.000 -0.900 -0.727
## shape2 -0.900  1.000  0.588
## rate   -0.727  0.588  1.000

## 1.5 Why there are differences between MLE and MME for the lognormal distribution?

We recall that the lognormal distribution function is given by $F_X(x) = \Phi\left(\frac{\log(x)-\mu}{\sigma} \right),$ where $$\Phi$$ denotes the distribution function of the standard normal distribution. We know that $$E(X) = \exp\left( \mu+\frac{1}{2} \sigma^2 \right)$$ and $$Var(X) = \exp\left( 2\mu+\sigma^2\right) (e^{\sigma^2} -1)$$. The MME is obtained by inverting the previous formulas, whereas the MLE has the following explicit solution $\hat\mu_{MLE} = \frac{1}{n}\sum_{i=1}^n \log(x_i),~~ \hat\sigma^2_{MLE} = \frac{1}{n}\sum_{i=1}^n (\log(x_i) - \hat\mu_{MLE})^2.$ Let us fit a sample by MLE and MME. The fit looks particularly good in both cases.

x3 <- rlnorm(1000)
f1 <- fitdist(x3, "lnorm", method="mle")
f2 <- fitdist(x3, "lnorm", method="mme")
par(mfrow=1:2)
cdfcomp(list(f1, f2), do.points=FALSE, xlogscale = TRUE, main = "CDF plot")
denscomp(list(f1, f2), demp=TRUE, main = "Density plot")

Let us compare the theoretical moments (mean and variance) given the fitted values ($$\hat\mu,\hat\sigma$$), that is $E(X) = \exp\left( \hat\mu+\frac{1}{2} \hat\sigma^2 \right), Var(X) = \exp\left( 2\hat\mu+\hat\sigma^2\right) (e^{\hat\sigma^2} -1).$

c("E(X) by MME"=as.numeric(exp(f2$estimate["meanlog"]+f2$estimate["sdlog"]^2/2)),
"E(X) by MLE"=as.numeric(exp(f1$estimate["meanlog"]+f1$estimate["sdlog"]^2/2)),
"empirical"=mean(x3))
## E(X) by MME E(X) by MLE   empirical
##        1.61        1.60        1.61
c("Var(X) by MME"=as.numeric(exp(2*f2$estimate["meanlog"]+f2$estimate["sdlog"]^2)*(exp(f2$estimate["sdlog"]^2)-1)), "Var(X) by MLE"=as.numeric(exp(2*f1$estimate["meanlog"]+f1$estimate["sdlog"]^2)*(exp(f1$estimate["sdlog"]^2)-1)),
"empirical"=var(x3))
## Var(X) by MME Var(X) by MLE     empirical
##          4.30          4.36          4.30

From a MLE point of view, a lognormal sample $$x_1,\dots,x_n$$ is equivalent to handle a normal sample $$\log(x_1),\dots,\log(x_n)$$. However, it is well know by the Jensen inequality that $$E(X) = E(\exp(\log(X))) \geq \exp(E(\log(X)))$$ implying the MME estimates provides better moment estimates than with MLE.

## 1.6 Can I fit a distribution with positive support when data contains negative values?

The answer is no: you cannot fit a distribution with positive support (say gamma distribution) when data contains negative values.

set.seed(1234)
x <- rnorm(100, mean = 1, sd = 0.5)
(try(fitdist(x, "exp")))
## [1] "Error in start.arg.default(data10, distr = distname) : \n  values must be positive to fit an exponential  distribution\n"
## attr(,"class")
## [1] "try-error"
## attr(,"condition")
## <simpleError in start.arg.default(data10, distr = distname): values must be positive to fit an exponential  distribution>

It is irrelevant to do such fit. If you really need to use that distribution, you have two options: either to remove negative values (not recommended) or to shift the data.

fitdist(x[x >= 0], "exp")
## Fitting of the distribution ' exp ' by maximum likelihood
## Parameters:
##      estimate Std. Error
## rate     1.06      0.107
fitdist(x - min(x), "exp")
## Fitting of the distribution ' exp ' by maximum likelihood
## Parameters:
##      estimate Std. Error
## rate    0.914     0.0914

## 1.7 Can I fit a finite-support distribution when data is outside that support?

The answer is no: you cannot fit a distribution with finite-support (say beta distribution) when data is outside $$[0,1]$$.

set.seed(1234)
x <- rnorm(100, mean = 0.5, sd = 0.25)
(try(fitdist(x, "beta")))
## [1] "Error in start.arg.default(data10, distr = distname) : \n  values must be in [0-1] to fit a beta distribution\n"
## attr(,"class")
## [1] "try-error"
## attr(,"condition")
## <simpleError in start.arg.default(data10, distr = distname): values must be in [0-1] to fit a beta distribution>

It is irrelevant to do such fit. If you really need to use that distribution, you have two options: either to remove impossible values (not recommended) or to shift/scale the data.

fitdist(x[x > 0 & x < 1], "beta")
## Fitting of the distribution ' beta ' by maximum likelihood
## Parameters:
##        estimate Std. Error
## shape1     2.08      0.288
## shape2     2.50      0.352
fitdist((x - min(x)*1.01) / (max(x) * 1.01 - min(x) * 1.01), "beta")
## Fitting of the distribution ' beta ' by maximum likelihood
## Parameters:
##        estimate Std. Error
## shape1     1.77      0.236
## shape2     2.17      0.296

# 2 Questions regarding goodness-of-fit tests and statistics

## 2.1 Where can we find the results of goodness-of-fit tests ?

Results of goodness-of-fit tests are not printed but are given in the object returned by gofstat() and you can have access to them as described in the example below. Nevertheless, p-values are not given for every test. For Anderson-Darling (ad), Cramer von Mises (cvm) and Kolomogorov (ks), only the decision (rejection of H0 or not) is given, when available (see FAQ 2.3 for more details).

set.seed(1234)
x <- rgamma(n = 100, shape = 2, scale = 1)
# fit of the good distribution
fgamma <- fitdist(x, "gamma")
# fit of a bad distribution
fexp <- fitdist(x, "exp")
g <- gofstat(list(fgamma, fexp), fitnames = c("gamma", "exp"))
denscomp(list(fgamma, fexp), legendtext = c("gamma", "exp"))

# results of the tests
## chi square test (with corresponding table with theoretical and observed counts)
g$chisqpvalue ## gamma exp ## 1.89e-01 7.73e-05 g$chisqtable
##           obscounts theo gamma theo exp
## <= 0.5483         9      10.06    23.66
## <= 0.8122         9       8.82     9.30
## <= 0.9592         9       5.27     4.68
## <= 1.368          9      14.63    11.37
## <= 1.523          9       5.24     3.74
## <= 1.701          9       5.73     3.97
## <= 1.94           9       7.09     4.82
## <= 2.381          9      11.08     7.50
## <= 2.842          9       9.00     6.29
## <= 3.801          9      11.93     9.28
## > 3.801          10      11.16    15.40
## Anderson-Darling test
g$adtest ## gamma exp ## "not rejected" "rejected" ## Cramer von Mises test g$cvmtest
##          gamma            exp
## "not rejected"     "rejected"
## Kolmogorov-Smirnov test
g$kstest ## gamma exp ## "not rejected" "rejected" ## 2.2 Is it reasonable to use goodness-of-fit tests to validate the fit of a distribution ? In the first versions of fitdistrplus, when they were available, the results of the GOF tests (AD, KS, CvM) were automatically printed. We decided to suppress this automatic printing when we realized that some users had some difficulties to interpret the results of those tests and sometimes misused them. Goodness-of-fit tests often appear as objective tools to decide wether a fitted distribution well describes a data set. But they are not ! It would not be reasonable at all to reject a distribution just because a goodness-of-fit test rejects it (see FAQ 2.2.1). And it would not be reasonable at all any more to validate a distribution because goodness-of-fit tests do not reject it (see FAQ 2.2.2). A fitted distribution should be evaluated using graphical methods (goodness-of-fit graphs automatically provided in our package by plotting the result of the fit (output of fitdist() or fitdistcens() and the complementary graphs that help to compare different fits - see ?graphcomp). We really think it is the most appropriate way to evaluate the adequacy of a fit and we are not the only ones to recommend it. You can find the same type of recommendations in reference books : • Probabilistic techniques in exposure assessment - a handbook dealing with variability and uncertainty in models and inputs by A.C. Cullen and H.C. Frey. • Application of uncertainty analysis to ecological risks of pesticides by W.J. Warren-Hicks and A. Hart. • Statistical inference by G. Casella and R.L. Berger • Loss models: from data to decision by S.A. Klugman and H.H. Panjer and G.E. Willmot Moreover, the selection of a distribution should also be driven by knowledge of underlying processes when available. For example when a variable cannot be negative, one would be very cautious while fitting a normal distribution, that potentially gives negative values, even if the observed data of this variable seem well fitted by a normal distribution. ### 2.2.1 Should I reject a distribution because a goodness-of-fit test rejects it ? No it would not be reasonable at all to reject a distribution just because a goodness-of-fit test rejects it, especially in the case of big samples. In the real life, as soon as you have a sufficient amount of data, you will reject the fitted distribution. We know that a model cannot perfectly describe real data, and generally the true question is to find the better distribution among a pool of simple parametric distributions to describe the data, so to compare different models (see FAQ 2.4 and 2.5 for corresponding questions). To illustre this point let us comment the example presented below. We drew two samples from the same Poisson distribution with a mean parameter equal to 100. In many applications, for this value of its parameter, the Poisson distribution would be considered to be well approximated by a normal distribution. Testing the fit (here using a Kolmogorov-Smirnov test ) of the normal distribution on a sample of 100 observations would not reject the normal fit, while testing it on a sample of 10000 observations would reject it, while both samples come from the same distribution. set.seed(1234) x1 <- rpois(n = 100, lambda = 100) f1 <- fitdist(x1, "norm") g1 <- gofstat(f1) g1$kstest
##     1-mle-norm
## "not rejected"
x2 <- rpois(n = 10000, lambda = 100)
f2 <- fitdist(x2, "norm")
g2 <- gofstat(f2)
g2$kstest ## 1-mle-norm ## "rejected" par(mfrow=1:2) denscomp(f1, demp = TRUE, addlegend = FALSE, main = "small sample") denscomp(f2, demp = TRUE, addlegend = FALSE, main = "big sample") ### 2.2.2 Should I accept a distribution because goodness-of-fit tests do not reject it ? No, it would not be reasonable at all to validate a distribution because goodness-of-fit tests do not reject it. Like all the other hypothesis tests, goodness-of-fit tests lack of statistical power when the sample size is not so high. And the different goodness-of-fit tests are not equally sensitive to different types of deviation between empirical and fitted distributions. For example the Kolmogorov-Smirnov test is sensitive when distributions differ in a global fashion near the centre of the distribution. The Anderson-Darling test will be more sensitive when distributions differ in their tails, and the Cramer von Mises will be more sensitive when there are small but repetitive differences between empirical and theoretical distribution functions. The sensitivity of a chi square test will depend on the definition of classes, and even if we propose a default definition of classes when the user does not provide classes, this choice is not obvious and could impact the results of the test. This test is more appropriate when data are discrete, even if they are modelled by a continuous distribution, as in the following example. Two samples of respective sizes 500 and 50 are drawn from a Poisson distribution of mean parameter equal to 1 (not a sufficiently high value to consider that the Poisson distribution could be approximated by a normal one). Using a Kolmogorov-Smirnov test, for the small sample the normal fit is rejected only for the bigger sample. It is not rejected with the smaller sample even if the fit could be rejected after a simple visual confrontation of the distributions. In that particular case, the chi square test with classes defined by default would have rejected te normal fit for both samples. set.seed(1234) x3 <- rpois(n = 500, lambda = 1) f3 <- fitdist(x3, "norm") g3 <- gofstat(f3) g3$kstest
## 1-mle-norm
## "rejected"
x4 <- rpois(n = 50, lambda = 1)
f4 <- fitdist(x4, "norm")
g4 <- gofstat(f4)
g4$kstest ## 1-mle-norm ## "not rejected" par(mfrow=1:2) denscomp(f3, addlegend = FALSE, main = "big sample") denscomp(f4, addlegend = FALSE, main = "small sample") g3$chisqtable
##      obscounts theocounts
## <= 0     180.0       80.3
## <= 1     187.0      163.5
## <= 2      87.0      168.1
## <= 3      32.0       73.4
## > 3       14.0       14.7
g3$chisqpvalue ## [1] 7.11e-42 g4$chisqtable
##      obscounts theocounts
## <= 0     14.00       5.46
## <= 1     15.00      14.23
## <= 2     15.00      18.09
## > 2       6.00      12.22
L0 <- logLik(f0)
sigma=as.numeric(f0$estimate["sdlog"]), shift=D) # works with BFGS, but not Nelder-Mead f <- fitdist(y, "shiftlnorm", start=start, optim.method="BFGS") summary(f) ## Fitting of the distribution ' shiftlnorm ' by maximum likelihood ## Parameters : ## estimate Std. Error ## mean -1.386 0.02401 ## sigma 0.071 0.00192 ## shift 0.248 0.00598 ## Loglikelihood: 8299 AIC: -16591 BIC: -16573 ## Correlation matrix: ## mean sigma shift ## mean 1.000 -0.885 0.999 ## sigma -0.885 1.000 -0.886 ## shift 0.999 -0.886 1.000 If we don’t use the log argument, the algorithms stalls. f2 <- try(fitdist(y, "shiftlnorm_no", start=start, optim.method="BFGS")) ## <simpleError in optim(par = vstart, fn = fnobj, fix.arg = fix.arg, obs = data, gr = gradient, ddistnam = ddistname, hessian = TRUE, method = meth, lower = lower, upper = upper, ...): non-finite finite-difference value [2]> print(attr(f2, "condition")) ## <simpleError in fitdist(y, "shiftlnorm_no", start = start, optim.method = "BFGS"): the function mle failed to estimate the parameters, ## with the error code 100 ## > Indeed the algorithm stops because at the following value, the log-likelihood is infinite. sum(log(dshiftlnorm_no(y, 0.16383978, 0.01679231, 1.17586600 ))) ## [1] -Inf log(prod(dshiftlnorm_no(y, 0.16383978, 0.01679231, 1.17586600 ))) ## [1] -Inf sum(dshiftlnorm(y, 0.16383978, 0.01679231, 1.17586600, TRUE )) ## [1] 7761 There is something wrong in the computation. Only the R-base implementation using log argument seems reliable. This happens the C-base implementation of dlnorm takes care of the log value. In the file ../src/nmath/dlnorm.c in the R sources, we find the C code for dlnorm double dlnorm(double x, double meanlog, double sdlog, int give_log) { double y; #ifdef IEEE_754 if (ISNAN(x) || ISNAN(meanlog) || ISNAN(sdlog)) return x + meanlog + sdlog; #endif if(sdlog <= 0) { if(sdlog < 0) ML_ERR_return_NAN; // sdlog == 0 : return (log(x) == meanlog) ? ML_POSINF : R_D__0; } if(x <= 0) return R_D__0; y = (log(x) - meanlog) / sdlog; return (give_log ? -(M_LN_SQRT_2PI + 0.5 * y * y + log(x * sdlog)) : M_1_SQRT_2PI * exp(-0.5 * y * y) / (x * sdlog)); /* M_1_SQRT_2PI = 1 / sqrt(2 * pi) */ } In the last four lines with the logical condtion give_log?, we see how the log argument is handled: • when log=TRUE, we use $$-(\log(\sqrt{2\pi}) + y^2/2+\log(x\sigma))$$ -(M_LN_SQRT_2PI + 0.5 * y * y + log(x * sdlog)) • when log=FALSE, we use $$\sqrt{2\pi} *\exp( y^2/2)/(x\sigma))$$ (and then the logarithm outside dlnorm) M_1_SQRT_2PI * exp(-0.5 * y * y) / (x * sdlog)) Note that the constant $$\log(\sqrt{2\pi})$$ is pre-computed in the C macro M_LN_SQRT_2PI. In order to sort out this problem, we use the constrOptim wrapping optim to take into account linear constraints. This allows also to use other optimization methods than L-BFGS-B (low-memory BFGS bounded) used in optim. f2 <- fitdist(y, "shiftlnorm", start=start, lower=c(-Inf, 0, -min(y)), optim.method="Nelder-Mead") summary(f2) ## Fitting of the distribution ' shiftlnorm ' by maximum likelihood ## Parameters : ## estimate Std. Error ## mean -1.3873 NA ## sigma 0.0711 NA ## shift 0.2481 NA ## Loglikelihood: 8299 AIC: -16591 BIC: -16573 ## Correlation matrix: ## [1] NA print(cbind(BFGS=f$estimate, NelderMead=f2$estimate)) ## BFGS NelderMead ## mean -1.386 -1.3873 ## sigma 0.071 0.0711 ## shift 0.248 0.2481 Another possible would be to perform all computations with higher precision arithmetics as implemented in the package Rmpfr using the MPFR library. ## 3.4 What to do when there is a scaling issue? Let us consider a dataset which has particular small values. data(dataFAQscale1) head(dataFAQscale1) ## [1] -0.007077 -0.000947 -0.001898 -0.000475 -0.001902 -0.000476 summary(dataFAQscale1) ## Min. 1st Qu. Median Mean 3rd Qu. Max. ## -0.00708 -0.00143 -0.00047 -0.00031 0.00096 0.00428 The only way to sort out is to multiply the dataset by a large value. for(i in 6:0) cat(10^i, try(mledist(dataFAQscale1*10^i, "cauchy")$estimate), "\n")
## 1e+06 -290 1194
## 1e+05 -29 119
## 10000 -2.9 11.9
## 1000 -0.29 1.19
## 100 -0.029 0.119
## 10 -0.0029 0.0119
## <simpleError in optim(par = vstart, fn = fnobj, fix.arg = fix.arg, obs = data,     gr = gradient, ddistnam = ddistname, hessian = TRUE, method = meth,     lower = lower, upper = upper, ...): non-finite finite-difference value [2]>
## 1 NA NA

Let us consider a dataset which has particular large values.

data(dataFAQscale2)
head(dataFAQscale2)
## [1] 1.40e+09 1.41e+09 1.43e+09 1.44e+09 1.49e+09 1.57e+09
summary(dataFAQscale2)
##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
## 1.40e+09 1.58e+09 2.24e+09 2.55e+09 3.39e+09 4.49e+09

The only way to sort out is to multiply the dataset by a small value.

for(i in 0:5)
cat(10^(-2*i), try(mledist(dataFAQscale2*10^(-2*i), "cauchy")$estimate), "\n") ## 1 2.03e+09 6.59e+08 ## 0.01 20283641 6594932 ## 1e-04 202836 65949 ## 1e-06 2028 659 ## 1e-08 20.3 6.59 ## 1e-10 0.203 0.0659 ## 3.5 How do I set bounds on parameters when optimizing? ### 3.5.1 Setting bounds for scale parameters Consider the normal distribution $$\mathcal{N}(\mu, \sigma^2)$$ defined by the density $f(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right), x\in\mathbb{R},$ where $$\mu$$ is a location parameter such that $$\mu\in\mathbb{R}$$, $$\sigma^2$$ is a scale parameter such that $$\sigma^2>0$$. Therefore when optimizing the log-likelihood or the squared differences or the GoF statistics. Setting a lower bound for the scale parameter is easy with fitdist: just use the lower argument. set.seed(1234) x <- rnorm(1000, 1, 2) fitdist(x, "norm", lower=c(-Inf, 0)) ## Fitting of the distribution ' norm ' by maximum likelihood ## Parameters: ## estimate Std. Error ## mean 0.947 NA ## sd 1.994 NA ### 3.5.2 Setting bounds for shape parameters Consider the Burr distribution $$\mathcal B(\mu, \sigma^2)$$ defined by the density $f(x) = \frac{a b (x/s)^b}{x [1 + (x/s)^b]^{a + 1}}, x\in\mathbb{R},$ where $$a,b$$ are shape parameters such that $$a,b>0$$, $$s$$ is a scale parameter such that $$s>0$$. x <- rburr(1000, 1, 2, 3) fitdist(x, "burr", lower=c(0, 0, 0), start=list(shape1 = 1, shape2 = 1, rate = 1)) ## Fitting of the distribution ' burr ' by maximum likelihood ## Parameters: ## estimate Std. Error ## shape1 0.969 NA ## shape2 2.051 NA ## rate 3.180 NA ### 3.5.3 Setting bounds for probability parameters Consider the geometric distribution $$\mathcal G(p)$$ defined by the mass probability function $f(x) = p(1-p)^x, x\in\mathbb{N},$ where $$p$$ is a probability parameter such that $$p\in[0,1]$$. x <- rgeom(1000, 1/4) fitdist(x, "geom", lower=0, upper=1) ## Fitting of the distribution ' geom ' by maximum likelihood ## Parameters: ## estimate Std. Error ## prob 0.242 0.00666 ### 3.5.4 Setting bounds for boundary parameters Consider the shifted exponential distribution $$\mathcal E(\mu,\lambda)$$ defined by the mass probability function $f(x) = \lambda \exp(-\lambda(x-\mu)), x>\mu,$ where $$\lambda$$ is a scale parameter such that $$\lambda>0$$, $$\mu$$ is a boundary (or shift) parameter such that $$\mu\in\mathbb{R}$$. When optimizing the log-likelihood, the boundary constraint is $\forall i=1,\dots,n, x_i>\mu \Rightarrow \min_{i=1,\dots,n} x_i > \mu \Leftrightarrow \mu > -\min_{i=1,\dots,n} x_i.$ Note that when optimizing the squared differences or the GoF statistics, this constraint may not be necessary. Let us do it in R. dsexp <- function(x, rate, shift) dexp(x-shift, rate=rate) psexp <- function(x, rate, shift) pexp(x-shift, rate=rate) rsexp <- function(n, rate, shift) rexp(n, rate=rate)+shift x <- rsexp(1000, 1/4, 1) fitdist(x, "sexp", start=list(rate=1, shift=0), lower= c(0, -min(x))) ## Fitting of the distribution ' sexp ' by maximum likelihood ## Parameters: ## estimate Std. Error ## rate 0.248 NA ## shift 1.005 NA ## 3.6 How works quantile matching estimation for discrete distributions? Let us consider the geometric distribution with values in $$\{0,1,2,3,\dots\}$$. The probability mass function, the cumulative distribution function and the quantile function are $P(X=x)= p (1-p)^{\lfloor x\rfloor}, F_X(x) = 1- (1-p)^{\lfloor x\rfloor}, F_X^{-1}(q) = \left\lfloor\frac{\log(1-q)}{\log(1-p)}\right\rfloor.$ Due to the integer part (floor function), both the distribution function and the quantile function are step functions. pgeom(0:3, prob=1/2) ## [1] 0.500 0.750 0.875 0.938 qgeom(c(0.3, 0.6, 0.9), prob=1/2) ## [1] 0 1 3 par(mar=c(4,4,2,1), mfrow=1:2) curve(pgeom(x, prob=1/2), 0, 10, n=301, main="c.d.f.") curve(qgeom(x, prob=1/2), 0, 1, n=301, main="q.f.") Now we study the QME for the geometric distribution. Since we have only one parameter, we choose one probabiliy, $$p=1/2$$. The theoretical median is the following integer $F_X^{-1}(1/2) = \left\lfloor\frac{\log(1/2)}{\log(1-p)}\right\rfloor.$ Note that the theoretical median for a discrete distribution is an integer. Empirically, the median may not be an integer. Indeed for an even length dataset, the empirical median is $q_{n,1/2} = \frac{x_{n/2}^\star + x_{n/2+1}^\star}{2},$ where $$x_{1}^\star<\dots<x_{n}^\star$$ is the sorted sample, which is not an integer value if $$x_{n/2}^\star + x_{n/2+1}^\star$$ is not an even number. However for an odd length dataset, the empirical median is an integer $$q_{n,1/2}=x_{(n+1)/2}^\star$$. x <- c(0, 0, 0, 0, 1, 1, 3, 2, 1, 0, 0) median(x[-1]) #sample size 10 ## [1] 0.5 median(x) #sample size 11 ## [1] 0 Therefore, a first issue arises: if the median is not an integer, it is impossible to match exactly the empirical median with the theoretical quantile. Furthermore, the second issue is the non-uniqueness of the solution. Admitting matching $$q_{n,1/2}$$ is an integer, QME aims to find some $$p$$ such that $\left\lfloor\frac{\log(1/2)}{\log(1-p)}\right\rfloor = q_{n,1/2} \Leftrightarrow q_{n,1/2} \leq \frac{\log(1/2)}{\log(1-p)} < q_{n,1/2} +1.$ Let us plot the squared differences $$(F_X^{-1}(1/2) - q_{n,1/2})^2$$. x <- rgeom(100, 1/3) L2 <- function(p) (qgeom(1/2, p) - median(x))^2 L2(1/3) #theoretical value ## [1] 0 curve(L2(x), 0.10, 0.95, xlab=expression(p), ylab=expression(L2(p)), main="squared differences", n=301) Any value between [1/3, 5/9] minimizes the squared differences. Therefore, fitdist() may be sensitive to the chosen initial value with deterministic optimization algorithm. fitdist(x, "geom", method="qme", probs=1/2, start=list(prob=1/2), control=list(trace=1, REPORT=1)) ## initial value 1.000000 ## iter 2 value 0.000000 ## iter 2 value 0.000000 ## iter 2 value 0.000000 ## final value 0.000000 ## converged ## Fitting of the distribution ' geom ' by matching quantiles ## Parameters: ## estimate ## prob 0.34 fitdist(x, "geom", method="qme", probs=1/2, start=list(prob=1/20), control=list(trace=1, REPORT=1)) ## initial value 144.000000 ## iter 1 value 144.000000 ## final value 144.000000 ## converged ## Fitting of the distribution ' geom ' by matching quantiles ## Parameters: ## estimate ## prob 0.05 The solution is to use a stochastic algorithm such as simulated annealing (SANN). fitdist(x, "geom", method="qme", probs=1/2, optim.method="SANN", start=list(prob=1/20)) ## Fitting of the distribution ' geom ' by matching quantiles ## Parameters: ## estimate ## prob 0.362 fitdist(x, "geom", method="qme", probs=1/2, optim.method="SANN", start=list(prob=1/2)) ## Fitting of the distribution ' geom ' by matching quantiles ## Parameters: ## estimate ## prob 0.473 Let us consider the Poisson distribution defined by the following mass probability and the cumulative distribution functions $P(X=k)=\frac{\lambda^k}{k!}\exp(-\lambda),~ F_X(x) = \exp(-\lambda)\sum_{k=0}^{\lfloor x \rfloor}\frac{\lambda^k}{k!},~ x\geq 0.$ The quantile function $$F_X^{-1}(p)=\inf(x, F_X(x)\geq p)$$ simplifies to $F_X^{-1}(p) = i \text{ such that } \sum_{k=0}^{i-1} P(X=k) \leq p < \sum_{k=0}^{i} P(X=k).$ Again, the quantile function is a step function $F_X^{-1}(p) = \left\{ \begin{array}{ll} 0 & \text{if } p < P(X=0) \\ 1 & \text{if } P(X=0) \leq p < P(X=0)+P(X=1) \\ 2 & \text{if } P(X=0)+P(X=1) \leq p < P(X=0)+P(X=1)+P(X=2) \\ \dots \\ i & \text{if } \sum_{k=0}^{i-1} P(X=k) \leq p < \sum_{k=0}^{i} P(X=k) \\ \dots \\ \end{array} \right.$ Again, the squared differences is a step function $$(F_X^{-1}(1/2) - q_{n,1/2})^2$$. par(mar=c(4,4,2,1)) x <- rpois(100, lambda=7.5) L2 <- function(lam) (qpois(1/2, lambda = lam) - median(x))^2 curve(L2(x), 6, 9, xlab=expression(lambda), ylab=expression(L2(lambda)), main="squared differences", n=201) Therefore, using fitdist() may be sensitive to the chosen initial value. fitdist(x, "pois", method="qme", probs=1/2, start=list(lambda=2)) ## Fitting of the distribution ' pois ' by matching quantiles ## Parameters: ## estimate ## lambda 2 fitdist(x, "pois", method="qme", probs=1/2, optim.method="SANN", start=list(lambda=2)) ## Fitting of the distribution ' pois ' by matching quantiles ## Parameters: ## estimate ## lambda 7.32 # 4 Questions regarding uncertainty ## 4.1 Can we compute marginal confidence intervals on parameter estimates from their reported standard error ? In statistics, deriving marginal confidence intervals on MLE parameter estimates using the approximation of their standard errors (calculated from the hessian) is a quite common procedure. It is based on the wald approximation which stands that when the sample size $$n$$ is sufficiently high, the marginal $$95\%$$ confidence on the ith component $$\theta_i$$ of a model parameter $$\theta$$ estimated by maximum likelihood (estimate denoted $$\hat \theta$$) can be approximated by : $$\hat \theta_i \pm 1.96 \times SE(\hat \theta_i )$$ with $$SE(\hat \theta_i )$$ the ith term of the diagonal of the covariance matrix of the estimates ($$V_{ii}$$). $$V$$ is generally approximated by the inverse of the Fisher information matrix ($$I(\hat \theta)$$). The Fisher information matrix corresponds to the opposite of the hessian matrix evaluated on the MLE estimate. Let us recall that the hessian matrix is defined by $$H_{ij}(y, \theta) = \frac{\partial^2 L(y, \theta)}{\partial \theta_i \partial \theta_j}$$ with $$L(y, \theta)$$ the loglikelihod function for data $$y$$ and parameter $$\theta$$. Before using this approximation, one must keep in mind that its validity does not only depend on the sample size. It also strongly depends on the data, of the distribution, and also on the parameterization of the distribution. For this reason we recommend potential users of the Wald approximation to compare the results to the ones obtained using the bootstrap procedure (see below) before using this approximation. A look at the loglikelihood contours is also interesting as the Wald approximation assumes elliptical contours. In a more general context, we recommend the use of bootstrap to compute confidence intervals on parameters or on any function of those parameters. Below you will find two examples, one for which Wald confidence intervals seem correct and one for which they give wrong results, with parameter values even outside their possible range (negative rate bound for the gamma distribution). set.seed(1234) n <- rnorm(30, mean = 10, sd = 2) fn <- fitdist(n, "norm") bn <- bootdist(fn) bn$CI
##      Median 2.5% 97.5%
## mean   9.41 8.78 10.02
## sd     1.73 1.33  2.15
fn$estimate + cbind("estimate"= 0, "2.5%"= -1.96*fn$sd, "97.5%"= 1.96*fn$sd) ## estimate 2.5% 97.5% ## mean 9.41 8.77 10.04 ## sd 1.78 1.33 2.22 llplot(fn, back.col = FALSE) set.seed(1234) g <- rgamma(30, shape = 0.1, rate = 10) fg <- fitdist(g, "gamma") bg <- bootdist(fg) bg$CI
##        Median   2.5%   97.5%
## shape  0.0923 0.0636   0.145
## rate  30.0782 9.6306 146.660
fg$estimate + cbind("estimate"= 0, "2.5%"= -1.96*fg$sd, "97.5%"= 1.96*fg$sd) ## estimate 2.5% 97.5% ## shape 0.0882 0.0553 0.121 ## rate 24.2965 -6.3504 54.944 llplot(fg, back.col = FALSE) ## 4.2 How can we compute confidence intervals on quantiles from the fit of a distribution ? The quantile() function can be used to calculate any quantile from a fitted distribution when it is called with an object of class fitdist or fitdistcens as the first argument. When called with an object of class bootdist or bootdistcens as the first argument, quantiles are returned accompanied with a confidence interval calculated using the bootstraped sample of parameters. Moreover, you can use the CIcdfplot() function to plot the fitted distribution as a CDF curve surrounded by a band corresponding to pointwise intervals on the quantiles. See an example below on censored data corresponding to 72-hour acute salinity tolerance (LC50values) of rivermarine invertebrates. data(salinity) log10LC50 <-log10(salinity) fit <- fitdistcens(log10LC50, "norm") # Bootstrap bootsample <- bootdistcens(fit, niter = 101) #### We used only 101 iterations in that example to limit the calculation time but #### in practice you should take at least 1001 bootstrap iterations # Calculation of the quantile of interest (here the 5 percent hazard concentration) (HC5 <- quantile(bootsample, probs = 0.05)) ## (original) estimated quantiles for each specified probability (censored data) ## p=0.05 ## estimate 1.12 ## Median of bootstrap estimates ## p=0.05 ## estimate 1.13 ## ## two-sided 95 % CI of each quantile ## p=0.05 ## 2.5 % 1.04 ## 97.5 % 1.19 # visualizing pointwise confidence intervals on other quantiles CIcdfplot(bootsample, CI.output = "quantile", CI.fill = "pink", xlim = c(0.5,2), main = "") ## 4.3 How can we compute confidence intervals on any function of the parameters of the fitted distribution ? The bootstrap sample of parameter estimates can be used to calculate a bootstrap sample of any variable defined as a function of the parameters of the fitted distribution. From such a bootstrap sample you can easily compute a conidence interval using percentiles. Below is an example that uses the bootstrap sample of parameters from the previous example (FAQ 4.2) to calculate a 95 percent confidence interval on the Potentially Affected Portion (PAF) of species at a given exposure to salinity (fixed to 1.2 in log10 in this example). exposure <- 1.2 # Bootstrap sample of the PAF at this exposure PAF <- pnorm(exposure, mean = bootsample$estim$mean, sd = bootsample$estim\$sd)
# confidence interval from 2.5 and 97.5 percentiles
quantile(PAF, probs = c(0.025, 0.975))
##   2.5%  97.5%
## 0.0541 0.1633

For more complex calculations especially to tranfer uncertainty within a quantitative risk assessment, we recommend the use of the package mc2d which aims at making such calculations easy and which gives extensive examples of use of such bootstrap samples of parameters estimated using functions of the package fitdistrplus.