In aursiber/fitdistrplus: Help to Fit of a Parametric Distribution to Non-Censored or Censored Data

knitr::opts_chunk$set(echo = TRUE)

We denote by the raw empirical moment by $$ m_j = \frac1n \sum_{i=1}^n x_i^j, $$ by the centered empirical moment by $$ \mu_j = \frac1n \sum_{i=1}^n (x_i^j-m_1). $$ Starting values are computed in R/util-startarg.R. We give below the starting values for discrete and continuous distributions and refer to the bibliograhy sections for details.

1. Discrete distributions

1.1. Base R distribution

1.1.1. Geometric distribution

The MME is used $\hat p=1/(1+m_1)$.

1.1.2. Negative binomial distribution

The MME is used $\hat n = m_1^2/(\mu_2-m_1)$.

1.1.3. Poisson distribution

Both the MME and the MLE is $\hat \lambda = m_1$.

1.1.4. Binomial distribution

The MME is used $$ Var[X]/E[X] = 1-p \Rightarrow \hat p = 1- \mu_2/m_1. $$ the size parameter is $$ \hat n = \lceil\max(\max_i x_i, m_1/\hat p)\rceil. $$

1.2. logarithmic distribution

The expectation simplifies for small values of $p$ $$ E[X] = -\frac{1}{\log(1-p)}\frac{p}{1-p} \approx -\frac{1}{-p}\frac{p}{1-p} =\frac{1}{1-p}. $$ So the initial estimate is $$ \hat p = 1-1/m_1. $$

1.3. Zero truncated distributions

This distribution are the distribution of $X\vert X>0$ when $X$ follows a particular discrete distributions. Hence the initial estimate are the one used for base R on sample $x-1$.

1.4. Zero modified distributions

The MLE of the probability parameter is the empirical mass at 0 $\hat p_0=\frac1n \sum_i 1_{x_i=0}$. For other estimators we use the classical estimator with probability parameter $1-\hat p_0$.

1.5. Poisson inverse Gaussian distribution

The first two moments are $$ E[X]=\mu, Var[X] = \mu+\phi\mu^3. $$ So the initial estimate are $$ \hat\mu=m_1, \hat\phi = (\mu_2 - m_1)/m_1^3. $$

2. Continuous distributions

2.1. Normal distribution

The MLE is the MME so we use the empirical mean and variance.

2.2. Lognormal distribution

The log sample follows a normal distribution, so same as normal on the log sample.

2.3. Beta distribution (of the first kind)

The density function for a beta $\mathcal Be(a,b)$ is $$ f_X(x) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} x^{a-1}(1-x)^{b-1}. $$ The initial estimate is the MME \begin{equation} \hat a = m_1 \delta, \hat b = (1-m_1)\delta, \delta = \frac{m_1(1-m_1)}{\mu_2}-1, (#eq:betaguessestimator) \end{equation}

2.4. Other continuous distribution in `actuar`

2.4.1. Log-gamma

Use the gamma initial values on the sample $\log(x)$

2.4.2. Gumbel

The distribution function is $$ F(x) = \exp(-\exp(-\frac{x-\alpha}{\theta})). $$ Let $q_1$ and $q_3$ the first and the third quartiles. $$ \left{\begin{array} -\theta\log(-\log(p_1)) = q_1-\alpha \ -\theta\log(-\log(p_3)) = q_3-\alpha \end{array}\right. \Leftrightarrow \left{\begin{array} -\theta\log(-\log(p_1))+\theta\log(-\log(p_3)) = q_1-q_3 \ \alpha= \theta\log(-\log(p_3)) + q_3 \end{array}\right. \Leftrightarrow \left{\begin{array} \theta= \frac{q_1-q_3}{\log(-\log(p_3)) - \log(-\log(p_1))} \ \alpha= \theta\log(-\log(p_3)) + q_3 \end{array}\right.. $$ Using the median for the location parameter $\alpha$ yields to initial estimate $$ \hat\theta= \frac{q_1-q_3}{\log(\log(4/3)) - \log(\log(4))}, \hat\alpha = \hat\theta\log(\log(2)) + q_2. $$

2.4.3. Inverse Gaussian distribution

The moments of this distribution are $$ E[X] = \mu, Var[X] = \mu^3\phi. $$ Hence the initial estimate are $\hat\mu=m_1$, $\hat\phi=\mu_2/m_1^3$.

2.4.4. Generalized beta

This is the distribution of $\theta X^{1/\tau}$ when $X$ is beta distributed $\mathcal Be(a,b)$ The moments are $$ E[X] = \theta \beta(a+1/\tau, b)/\beta(a,b) = \theta \frac{\Gamma(a+1/\tau)}{\Gamma(a)}\frac{\Gamma(a+b)}{\Gamma(a+b+1/\tau)}, $$ $$ E[X^2] = \theta^2 \frac{\Gamma(a+2/\tau)}{\Gamma(a)}\frac{\Gamma(a+b)}{\Gamma(a+b+2/\tau)}. $$ Hence for large value of $\tau$, we have $$ E[X^2] /E[X] = \theta \frac{\Gamma(a+2/\tau)}{\Gamma(a+b+2/\tau)} \frac{\Gamma(a+b+1/\tau)}{\Gamma(a+1/\tau)} \approx \theta. $$ Note that the MLE of $\theta$ is the maximum We use $$ \hat\tau=3, \hat\theta = \frac{m_2}{m_1}\max_i x_i 1_{m_2>m_1} +\frac{m_1}{m_2}\max_i x_i 1_{m_2\geq m_1}. $$ then we use beta initial estimate on sample $(\frac{x_i}{\hat\theta})^{\hat\tau}$.

2.5. Feller-Pareto family

The Feller-Pareto distribution is the distribution $X=\mu+\theta(1/B-1)^{1/\gamma}$ when $B$ follows a beta distribution with shape parameters $\alpha$ and $\tau$. See details at https://doi.org/10.18637/jss.v103.i06 Hence let $Y = (X-\mu)/\theta$, we have $$ \frac{Y}{1+Y} = \frac{X-\mu}{\theta+X-\mu} = (1-B)^{1/\gamma}. $$ For $\gamma$ close to 1, $\frac{Y}{1+Y}$ is approximately beta distributed $\tau$ and $\alpha$.

The log-likelihood is \begin{equation} \mathcal L(\mu, \theta, \alpha, \gamma, \tau) = (\tau \gamma - 1) \sum_{i} \log(\frac{x_i-\mu}\theta) - (\alpha+\tau)\sum_i \log(1+(\frac{x_i-\mu}\theta)^\gamma) + n\log(\gamma) - n\log(\theta) -n \log(\beta(\alpha,\tau)). (#eq:fellerparetologlik). \end{equation} The MLE of $\mu$ is the minimum.

The gradient with respect to $\theta, \alpha, \gamma, \tau$ is \begin{equation} \nabla \mathcal L(\mu, \theta, \alpha, \gamma, \tau) = \begin{pmatrix} -(\tau \gamma - 1) \sum_{i} \frac{x_i}{\theta(x_i-\mu)} + (\alpha+\tau)\sum_i \frac{x_i\gamma(\frac{x_i-\mu}\theta)^{\gamma-1}}{\theta^2(1+(\frac{x_i-\mu}\theta)^\gamma)} - n/\theta \ - \sum_i \log(1+(\frac{x_i-\mu}\theta)^\gamma) -n(\psi(\tau) - \psi(\alpha+\tau)) \ (\tau - 1) \sum_{i} \log(\frac{x_i-\mu}\theta) - (\alpha+\tau)\sum_i \frac{(\frac{x_i-\mu}\theta)^\gamma}{ 1+(\frac{x_i-\mu}\theta)^\gamma}\log(\frac{x_i-\mu}\theta) + n/\gamma \ (\gamma - 1) \sum_{i} \log(\frac{x_i-\mu}\theta) - \sum_i \log(1+(\frac{x_i-\mu}\theta)^\gamma) -n (\psi(\tau) - \psi(\alpha+\tau)) \end{pmatrix}. (#eq:fellerparetogradient) \end{equation} Cancelling the first component of score for $\gamma=\alpha=2$, we get $$ -(2\tau - 1) \sum_{i} \frac{x_i}{\theta(x_i-\mu)} + (2+\tau)\sum_i \frac{x_i 2(x_i-\mu)}{\theta^3(1+(\frac{x_i-\mu}\theta)^2)} = \frac{n}{\theta} \Leftrightarrow -(2\tau - 1)\theta^2\frac1n \sum_{i} \frac{x_i}{x_i-\mu} + (2+\tau) \frac1n\sum_i \frac{x_i 2(x_i-\mu)}{(1+(\frac{x_i-\mu}\theta)^2)} = \theta^2 $$ $$ \Leftrightarrow (2+\tau) \frac1n\sum_i \frac{x_i 2(x_i-\mu)}{1+(\frac{x_i-\mu}\theta)^2} = (2\tau - 1)\theta^2\left(\frac1n \sum_{i} \frac{x_i}{x_i-\mu} -1\right) \Leftrightarrow \sqrt{ \frac{(2+\tau) \frac1n\sum_i \frac{x_i 2(x_i-\mu)}{1+(\frac{x_i-\mu}\theta)^2} }{(2\tau - 1)\left(\frac1n \sum_{i} \frac{x_i}{x_i-\mu} -1\right)} } = \theta. $$ Neglecting unknown value of $\tau$ and the denominator in $\theta$, we get with $\hat\mu$ set with (\@ref(eq:pareto4muinit)) \begin{equation} \hat\theta = \sqrt{ \frac{ \frac1n\sum_i \frac{x_i 2(x_i-\hat\mu)}{1+(x_i-\hat\mu)^2} }{\left(\frac1n \sum_{i} \frac{x_i}{x_i-\hat\mu} -1\right)} }. (#eq:fellerparetothetahat) \end{equation} Initial value of $\tau,\alpha$ are obtained on the sample $(z_i)_i$ $$ z_i = y_i/(1+y_i), y_i = (x_i - \hat\mu)/\hat\theta, $$ with initial values of a beta distribution which is based on MME (\@ref(eq:betaguessestimator)).

Cancelling the last component of the gradient leads to $$ (\gamma - 1) \frac1n\sum_{i} \log(\frac{x_i-\mu}\theta) - \frac1n\sum_i \log(1+(\frac{x_i-\mu}\theta)^\gamma) = \psi(\tau) - \psi(\alpha+\tau) \Leftrightarrow (\gamma - 1) \frac1n\sum_{i} \log(\frac{x_i-\mu}\theta) = \psi(\tau) - \psi(\alpha+\tau) +\frac1n\sum_i \log(1+(\frac{x_i-\mu}\theta)^\gamma) . $$ Neglecting the value $\gamma$ on the right-hand side we obtain \begin{equation} \hat\gamma = 1+ \frac{ \psi(\tau) - \psi(\alpha+\tau) +\frac1n\sum_i \log(1+(\frac{x_i-\mu}\theta)) }{ \frac1n\sum_{i} \log(\frac{x_i-\mu}\theta) }. (#eq:fellerparetogammahat) \end{equation}

2.5.1. Transformed beta

This is the Feller-Pareto with $\mu=0$. So the first component of \@ref(eq:fellerparetogradient) simplifies to with $\gamma=\alpha=2$ $$ -(2\tau - 1) \sum_{i} \frac{x_i}{\theta(x_i)} + (2+\tau)\sum_i \frac{2x_i^2}{\theta^3(1+(\frac{x_i}\theta)^2)} = \frac{n}{\theta} \Leftrightarrow -(2\tau - 1) \theta^2 + (2+\tau)\frac1n\sum_i \frac{2x_i^2}{1+(\frac{x_i}\theta)^2} = \theta^2 $$ $$ \theta^2=\frac{2+\tau}{2\tau}\frac1n\sum_i \frac{2x_i^2}{1+(\frac{x_i}\theta)^2}. $$ Neglecting unknown value of $\tau$ in the denominator in $\theta$, we get \begin{equation} \hat\theta = \sqrt{ \frac1n\sum_i \frac{2x_i^2}{1+x_i^2} }. (#eq:trbetathetahat) \end{equation} Initial value of $\tau,\alpha$ are obtained on the sample $(z_i)i$ $$ z_i = y_i/(1+y_i), y_i = x_i/\hat\theta, $$ with initial values of a beta distribution which is based on MME (\@ref(eq:betaguessestimator)). Similar to Feller-Pareto, we set \begin{equation} \hat\gamma = 1+ \frac{ \psi(\tau) - \psi(\alpha+\tau) +\frac1n\sum_i \log(1+\frac{x_i}\theta) }{ \frac1n\sum{i} \log(\frac{x_i}\theta) }. (#eq:fellerparetogammahat) \end{equation}

2.5.2. Generalized Pareto

This is the Feller-Pareto with $\mu=0$ $\gamma=1$. So the first component of \@ref(eq:fellerparetogradient) simplifies to with $\gamma=2$ $$ -(\tau - 1) \frac{n}{\theta} + (2+\tau)\sum_i \frac{x_i}{\theta^2(1+\frac{x_i}\theta} = n/\theta \Leftrightarrow -(\tau - 1) \theta + (2+\tau)\frac1n\sum_i \frac{x_i}{(1+\frac{x_i}\theta} = \theta. $$ Neglecting unknown value of $\tau$ leads to \begin{equation} \hat\theta = \frac1n\sum_i \frac{x_i}{1+x_i} (#eq:generalizedparetotheta) \end{equation}

Initial value of $\tau,\alpha$ are obtained on the sample $(z_i)_i$ $$ z_i = y_i/(1+y_i), y_i = x_i/\hat\theta, $$ with initial values of a beta distribution which is based on MME (\@ref(eq:betaguessestimator)).

2.5.3. Burr

Burr is a Feller-Pareto distribution with $\mu=0$, $\tau=1$.

The survival function is $$ 1-F(x) = (1+(x/\theta)^\gamma)^{-\alpha}. $$ Using the median $q_2$, we have $$ \log(1/2) = - \alpha \log(1+(q_2/\theta)^\gamma). $$ The initial value is \begin{equation} \alpha = \frac{\log(2)}{\log(1+(q_2/\theta)^\gamma)}, (#eq:burralpharelation) \end{equation}

So the first component of \@ref(eq:fellerparetogradient) simplifies to with $\gamma=\alpha=2$, $\tau=1$, $\mu=0$. $$ - n/\theta + 3\sum_i \frac{2x_i(\frac{x_i}\theta)}{\theta^2(1+(\frac{x_i}\theta)^2)} = n/\theta \Leftrightarrow \theta^2\frac1n\sum_i \frac{2x_i(\frac{x_i}\theta)}{(1+(\frac{x_i}\theta)^2)} = 2/3. $$ Neglecting unknown value in the denominator in $\theta$, we get \begin{equation} \hat\theta = \sqrt{ \frac{2}{3 \frac1n\sum_i \frac{2x_i^2}{1+(x_i)^2} } }. (#eq:trbetathetahat) \end{equation} We use for $\hat\gamma$ \@ref(eq:fellerparetogammahat) with $\tau=1$ and $\alpha=2$ and previous $\hat\theta$.

2.5.4. Loglogistic

Loglogistic is a Feller-Pareto distribution with $\mu=0$, $\tau=1$, $\alpha=1$. The survival function is $$ 1-F(x) = (1+(x/\theta)^\gamma)^{-1}. $$ So $$ \frac1{1-F(x)}-1 = (x/\theta)^\gamma \Leftrightarrow \log(\frac{F(x)}{1-F(x)}) = \gamma\log(x/\theta). $$ Let $q_1$ and $q_3$ be the first and the third quartile. $$ \log(\frac{1/3}{2/3})= \gamma\log(q_1/\theta), \log(\frac{2/3}{1/3})= \gamma\log(q_3/\theta) \Leftrightarrow -\log(2)= \gamma\log(q_1/\theta), \log(2)= \gamma\log(q_3/\theta). $$ The difference of previous equations simplifies to $$ \hat\gamma=\frac{2\log(2)}{\log(q_3/q_1)}. $$ The sum of previous equations $$ 0 = \gamma\log(q_1)+\gamma\log(q_3) - 2\gamma\log(\theta). $$ \begin{equation} \hat\theta = \frac12 e^{\log(q_1q_3)}. (#eq:llogisthetahat) \end{equation}

2.5.5. Paralogistic

Paralogistic is a Feller-Pareto distribution with $\mu=0$, $\tau=1$, $\alpha=\gamma$. The survival function is $$ 1-F(x) = (1+(x/\theta)^\alpha)^{-\alpha}. $$ So $$ \log(1-F(x)) = -\alpha \log(1+(x/\theta)^\alpha). $$ The log-likelihood is \begin{equation} \mathcal L(\theta, \alpha) = ( \alpha - 1) \sum_{i} \log(\frac{x_i}\theta) - (\alpha+1)\sum_i \log(1+(\frac{x_i}\theta)^\alpha) + 2n\log(\alpha) - n\log(\theta). (#eq:paralogisloglik) \end{equation} The gradient with respect to $\theta$, $\alpha$ is $$ \begin{pmatrix} ( \alpha - 1)\frac{-n}{\theta} - (\alpha+1)\sum_i \frac{-x_i\alpha(x_i/\theta)^{\alpha-1}}{1+(\frac{x_i}\theta)^\alpha} - n/\theta \ \sum_{i} \log(\frac{ \frac{x_i}\theta}{1+(\frac{x_i}\theta)^\alpha }) - (\alpha+1)\sum_i \frac{(\frac{x_i}\theta)^\alpha \log(x_i/\theta)}{1+(\frac{x_i}\theta)^\alpha} + 2n/\alpha \ \end{pmatrix}. $$ The first component cancels when $$ - (\alpha+1)\sum_i \frac{-x_i\alpha(x_i/\theta)^{\alpha-1}}{1+(\frac{x_i}\theta)^\alpha} = \alpha n/\theta \Leftrightarrow (\alpha+1)\frac1n\sum_i \frac{ (x_i)^{\alpha+1}}{1+(\frac{x_i}\theta)^\alpha} = \theta^\alpha. $$ The second component cancels when $$ \frac1n\sum_{i} \log(\frac{ \frac{x_i}\theta}{1+(\frac{x_i}\theta)^\alpha }) = -2/\alpha +(\alpha+1)\frac1n\sum_i \frac{(\frac{x_i}\theta)^\alpha \log(x_i/\theta)}{1+(\frac{x_i}\theta)^\alpha}. $$ Choosing $\theta=1$, $\alpha=2$ in sums leads to $$ \frac1n\sum_{i} \log(\frac{ \frac{x_i}\theta}{1+x_i^2 }) - \frac1n\sum_i \frac{x_i^2\log(x_i)}{1+x_i^2} = -2/\alpha +(\alpha)\frac1n\sum_i \frac{x_i^2\log(x_i)}{1+x_i^2}. $$ Initial estimators are \begin{equation} \hat\alpha = \frac{ \frac1n\sum_{i} \log(\frac{ x_i}{1+x_i^2 }) - \frac1n\sum_i \frac{x_i^2\log(x_i)}{1+x_i^2} }{ \frac1n\sum_i \frac{x_i^2\log(x_i)}{1+x_i^2} - 2 }, (#eq:paralogisalphahat) \end{equation} \begin{equation} \hat\theta = (\hat\alpha+1)\frac1n\sum_i \frac{ (x_i)^{\hat\alpha+1}}{1+(x_i)^{\hat\alpha}}. (#eq:paralogisthetahat) \end{equation}

2.5.6. Inverse Burr

Use Burr estimate on the sample $1/x$

2.5.7. Inverse paralogistic

Use paralogistic estimate on the sample $1/x$

2.5.8. Inverse pareto

Use pareto estimate on the sample $1/x$

2.5.9. Pareto IV

The survival function is $$ 1-F(x) = \left(1+ \left(\frac{x-\mu}{\theta}\right)^{\gamma} \right)^{-\alpha}, $$ see ?Pareto4 in actuar.

The first and third quartiles $q_1$ and $q_3$ verify $$ ((\frac34)^{-1/\alpha}-1)^{1/\gamma} = \frac{q_1-\mu}{\theta}, ((\frac14)^{-1/\alpha}-1)^{1/\gamma} = \frac{q_3-\mu}{\theta}. $$ Hence we get two useful relations \begin{equation} \gamma = \frac{ \log\left( \frac{ (\frac43)^{1/\alpha}-1 }{ (4)^{1/\alpha}-1 } \right) }{ \log\left(\frac{q_1-\mu}{q_3-\mu}\right) }, (#eq:pareto4gammarelation) \end{equation} \begin{equation} \theta = \frac{q_1- q_3 }{ ((\frac43)^{1/\alpha}-1)^{1/\gamma} - ((4)^{1/\alpha}-1)^{1/\gamma} }. (#eq:pareto4thetarelation) \end{equation}

The log-likelihood of a Pareto 4 sample (see Equation (5.2.94) of Arnold (2015) updated with Goulet et al. notation) is $$ \mathcal L(\mu,\theta,\gamma,\alpha) = (\gamma -1) \sum_i \log(\frac{x_i-\mu}{\theta}) -(\alpha+1)\sum_i \log(1+ (\frac{x_i-\mu}{\theta})^{\gamma}) +n\log(\gamma) -n\log(\theta)+n\log(\alpha). $$ Cancelling the derivate of $\mathcal L(\mu,\theta,\gamma,\alpha)$ with respect to $\alpha$ leads to \begin{equation} \alpha =n/\sum_i \log(1+ (\frac{x_i-\mu}{\theta})^{\gamma}). (#eq:pareto4alpharelation) \end{equation}

The MLE of the threshold parameter $\mu$ is the minimum. So the initial estimate is slightly under the minimum in order that all observations are strictly above it \begin{equation} \hat\mu = \left{ \begin{array}{ll} (1-\epsilon) \min_i x_i & \text{if } \min_i x_i <0 \ (1+\epsilon)\min_i x_i & \text{if } \min_i x_i \geq 0 \ \end{array} \right. . (#eq:pareto4muinit) \end{equation} where $\epsilon=0.05$.

Initial parameter estimation is $\hat\mu$, $\alpha^\star = 2$ , $\hat\gamma$ from \@ref(eq:pareto4gammarelation) with $\alpha^\star$, $\hat\theta$ from \@ref(eq:pareto4thetarelation) with $\alpha^\star$ and $\hat\gamma$, $\hat\alpha$ from \@ref(eq:pareto4alpharelation) with $\hat\mu$, $\hat\theta$ and $\hat\gamma$.

2.5.10. Pareto III

Pareto III corresponds to Pareto IV with $\alpha=1$. \begin{equation} \gamma = \frac{ \log\left( \frac{ \frac43-1 }{ 4-1 } \right) }{ \log\left(\frac{q_1-\mu}{q_3-\mu}\right) }, \label{eq:pareto3:gamma:relation} \end{equation}

\begin{equation} \theta = \frac{ (\frac13)^{1/\gamma} - (3)^{1/\gamma} }{q_1- q_3 }. \label{eq:pareto3:theta:relation} \end{equation}

Initial parameter estimation is $\hat\mu$, $\hat\gamma$ from \eqref{eq:pareto3:gamma:relation}, $\hat\theta$ from \eqref{eq:pareto3:theta:relation} with $\hat\gamma$.

2.5.11. Pareto II

Pareto II corresponds to Pareto IV with $\gamma=1$.

\begin{equation} \theta = \frac{ (\frac43)^{1/\alpha} - 4^{1/\alpha} }{q_1- q_3 }. \label{eq:pareto2:theta:relation} \end{equation}

Initial parameter estimation is $\hat\mu$, $\alpha^\star = 2$ , $\hat\theta$ from \eqref{eq:pareto4:theta:relation} with $\alpha^\star$ and $\gamma=1$, $\hat\alpha$ from \eqref{eq:pareto4:alpha:relation} with $\hat\mu$, $\hat\theta$ and $\gamma=1$,

2.5.12. Pareto I

Pareto I corresponds to Pareto IV with $\gamma=1$, $\mu=\theta$.

The MLE is \begin{equation} \hat\mu = \min_i X_i, \hat\alpha = \left(\frac1n \sum_{i=1}^n \log(X_i/\hat\mu) \right)^{-1}. \label{eq:pareto1:alpha:mu:relation} \end{equation}

This can be rewritten with the geometric mean of the sample $G_n = (\prod_{i=1}^n X_i)^{1/n}$ as $$ \hat\alpha = \log(G_n/\hat\mu). $$

Initial parameter estimation is $\hat\mu$, $\hat\alpha$ from \eqref{eq:pareto1:alpha:mu:relation}.

2.5.13. Pareto

Pareto corresponds to Pareto IV with $\gamma=1$, $\mu=0$. \begin{equation} \theta = \frac{ (\frac43)^{1/\alpha} - 4^{1/\alpha} }{q_1- q_3 }. \label{eq:pareto:theta:relation} \end{equation}

Initial parameter estimation is $$ \alpha^\star = \max(2, 2(m_2-m_1^2)/(m_2-2m_1^2)), $$ with $m_i$ are empirical raw moment of order $i$, $\hat\theta$ from \eqref{eq:pareto4:theta:relation} with $\alpha^\star$ and $\gamma=1$, $\hat\alpha$ from \eqref{eq:pareto4:alpha:relation} with $\mu=0$, $\hat\theta$ and $\gamma=1$.

2.6. Transformed gamma family

2.6.1. Transformed gamma distribution

The log-likelihood is given by $$ \mathcal L(\alpha,\tau,\theta) = n\log(\tau) + \alpha\tau\sum_i \log(x_i/\theta) -\sum_i (x_i/\theta)^\tau - \sum_i\log(x_i) - n\log(Gamma(\alpha)). $$ The gradient with respect to $\alpha,\tau,\theta$ is given by $$ \begin{pmatrix} \tau- n\psi(\alpha)) \ n/\tau + \alpha\sum_i \log(x_i/\theta) -\sum_i (x_i/\theta)^{\tau} \log(x_i/\theta) \ -\alpha\tau /\theta +\sum_i \tau \frac{x_i}{\theta^2}(x_i/\theta)^{\tau-1} \end{pmatrix}. $$ We compute the moment-estimator as in gamma \eqref{eq:gamma:relation} $$ \hat\alpha = m_2^2/\mu_2, \hat\theta= \mu_2/m_1. $$ Then cancelling the first component of the gradient we set $$ \hat\tau = \frac{\psi(\hat\alpha)}{\frac1n\sum_i \log(x_i/\hat\theta) }. $$

2.6.2. gamma distribution

Transformed gamma with $\tau=1$

We compute the moment-estimator given by \begin{equation} \hat\alpha = m_2^2/\mu_2, \hat\theta= \mu_2/m_1. \label{eq:gamma:relation} \end{equation}

2.6.3. Weibull distribution

Transformed gamma with $\alpha=1$

Let $\tilde m=\frac1n\sum_i \log(x_i)$ and $\tilde v=\frac1n\sum_i (\log(x_i) - \tilde m)^2$. We use an approximate MME $$ \hat\tau = 1.2/sqrt(\tilde v), \hat\theta = exp(\tilde m + 0.572/\hat \tau). $$ Alternatively, we can use the distribution function $$ F(x) = 1 - e^{-(x/\sigma)^\tau} \Rightarrow \log(-\log(1-F(x))) = \tau\log(x) - \tau\log(\theta), $$ Hence the QME for Weibull is $$ \tilde\tau = \frac{ \log(-\log(1-p_1)) - \log(-\log(1-p_2)) }{ \log(x_1) - \log(x_2) }, \tilde\tau = x_3/(-\log(1-p_3))^{1/\tilde\tau} $$ with $p_1=1/4$, $p_2=3/4$, $p_3=1/2$, $x_i$ corresponding empirical quantiles.

Initial parameters are $\tilde\tau$ and $\tilde\theta$ unless the empirical quantiles $x_1=x_2$, in that case we use $\hat\tau$, $\hat\theta$.

2.6.4. Exponential distribution

The MLE is the MME $\hat\lambda = 1/m_1.$

2.7. Inverse transformed gamma family

2.7.1. Inverse transformed gamma distribution

Same as transformed gamma distribution with $(1/x_i)_i$.

2.7.2. Inverse gamma distribution

We compute moment-estimator as $$ \hat\alpha = (2m_2-m_1^2)/(m_2-m_1^2), \hat\theta= m_1m_2/(m_2-m_1^2). $$

2.7.3. Inverse Weibull distribution

We use the QME.

2.7.4. Inverse exponential