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.
The MME is used \(\hat p=1/(1+m_1)\).
The MME is used \(\hat n = m_1^2/(\mu_2-m_1)\).
Both the MME and the MLE is \(\hat \lambda = m_1\).
The MME is used \[ Var[X]/E[X] = 1-p \Rightarrow \hat p = 1- \mu_2/m_1. \] the size parameter is \[ \hat n = \lceil\max(\max_i x_i, m_1/\hat p)\rceil. \]
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 \[ \hat p = 1-1/m_1. \]
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 \(x-1\).
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 \(1-\hat p_0\).
The first two moments are \[ E[X]=\mu, Var[X] = \mu+\phi\mu^3. \] So the initial estimate are \[ \hat\mu=m_1, \hat\phi = (\mu_2 - m_1)/m_1^3. \]
The MLE is the MME so we use the empirical mean and variance.
The log sample follows a normal distribution, so same as normal on the log sample.
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 \[\begin{equation} \hat a = m_1 \delta, \hat b = (1-m_1)\delta, \delta = \frac{m_1(1-m_1)}{\mu_2}-1, \tag{2.1} \end{equation}\]
actuarUse the gamma initial values on the sample \(\log(x)\)
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 \[ \hat\theta= \frac{q_1-q_3}{\log(\log(4/3)) - \log(\log(4))}, \hat\alpha = \hat\theta\log(\log(2)) + q_2. \]
The moments of this distribution are \[ E[X] = \mu, Var[X] = \mu^3\phi. \] Hence the initial estimate are \(\hat\mu=m_1\), \(\hat\phi=\mu_2/m_1^3\).
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 \((\frac{x_i}{\hat\theta})^{\hat\tau}\).
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 \[\begin{equation} \mathcal L(\mu, \theta, \alpha, \gamma, \tau) = (\tau \gamma - 1) \sum_{i} \log(\frac{x_i-\mu}\theta) - (\alpha+\tau)\sum_i \log(1+(\frac{x_i-\mu}\theta)^\gamma) + n\log(\gamma) - n\log(\theta) -n \log(\beta(\alpha,\tau)). \tag{2.2}. \end{equation}\] The MLE of \(\mu\) is the minimum.
The gradient with respect to \(\theta, \alpha, \gamma, \tau\) is \[\begin{equation} \nabla \mathcal L(\mu, \theta, \alpha, \gamma, \tau) = \begin{pmatrix} -(\tau \gamma - 1) \sum_{i} \frac{x_i}{\theta(x_i-\mu)} + (\alpha+\tau)\sum_i \frac{x_i\gamma(\frac{x_i-\mu}\theta)^{\gamma-1}}{\theta^2(1+(\frac{x_i-\mu}\theta)^\gamma)} - n/\theta \\ - \sum_i \log(1+(\frac{x_i-\mu}\theta)^\gamma) -n(\psi(\tau) - \psi(\alpha+\tau)) \\ (\tau - 1) \sum_{i} \log(\frac{x_i-\mu}\theta) - (\alpha+\tau)\sum_i \frac{(\frac{x_i-\mu}\theta)^\gamma}{ 1+(\frac{x_i-\mu}\theta)^\gamma}\log(\frac{x_i-\mu}\theta) + n/\gamma \\ (\gamma - 1) \sum_{i} \log(\frac{x_i-\mu}\theta) - \sum_i \log(1+(\frac{x_i-\mu}\theta)^\gamma) -n (\psi(\tau) - \psi(\alpha+\tau)) \end{pmatrix}. \tag{2.3} \end{equation}\] 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 ((2.16)) \[\begin{equation} \hat\theta = \sqrt{ \frac{ \frac1n\sum_i \frac{x_i 2(x_i-\hat\mu)}{1+(x_i-\hat\mu)^2} }{\left(\frac1n \sum_{i} \frac{x_i}{x_i-\hat\mu} -1\right)} }. \tag{2.4} \end{equation}\] 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 ((2.1)).
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 \[\begin{equation} \hat\gamma = 1+ \frac{ \psi(\tau) - \psi(\alpha+\tau) +\frac1n\sum_i \log(1+(\frac{x_i-\mu}\theta)) }{ \frac1n\sum_{i} \log(\frac{x_i-\mu}\theta) }. \tag{2.5} \end{equation}\]
This is the Feller-Pareto with \(\mu=0\). So the first component of (2.3) 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 \[\begin{equation} \hat\theta = \sqrt{ \frac1n\sum_i \frac{2x_i^2}{1+x_i^2} }. \tag{2.6} \end{equation}\] 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 ((2.1)). Similar to Feller-Pareto, we set \[\begin{equation} \hat\gamma = 1+ \frac{ \psi(\tau) - \psi(\alpha+\tau) +\frac1n\sum_i \log(1+\frac{x_i}\theta) }{ \frac1n\sum_{i} \log(\frac{x_i}\theta) }. \tag{2.5} \end{equation}\]
This is the Feller-Pareto with \(\mu=0\) \(\gamma=1\). So the first component of (2.3) 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 \[\begin{equation} \hat\theta = \frac1n\sum_i \frac{x_i}{1+x_i} \tag{2.7} \end{equation}\]
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 ((2.1)).
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 \[\begin{equation} \alpha = \frac{\log(2)}{\log(1+(q_2/\theta)^\gamma)}, \tag{2.8} \end{equation}\]
So the first component of (2.3) 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 \[\begin{equation} \hat\theta = \sqrt{ \frac{2}{3 \frac1n\sum_i \frac{2x_i^2}{1+(x_i)^2} } }. \tag{2.6} \end{equation}\] We use for \(\hat\gamma\) (2.5) with \(\tau=1\) and \(\alpha=2\) and previous \(\hat\theta\).
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 \[ 0 = \gamma\log(q_1)+\gamma\log(q_3) - 2\gamma\log(\theta). \] \[\begin{equation} \hat\theta = \frac12 e^{\log(q_1q_3)}. \tag{2.9} \end{equation}\]
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 \[\begin{equation} \mathcal L(\theta, \alpha) = ( \alpha - 1) \sum_{i} \log(\frac{x_i}\theta) - (\alpha+1)\sum_i \log(1+(\frac{x_i}\theta)^\alpha) + 2n\log(\alpha) - n\log(\theta). \tag{2.10} \end{equation}\] 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 \[\begin{equation} \hat\alpha = \frac{ \frac1n\sum_{i} \log(\frac{ x_i}{1+x_i^2 }) - \frac1n\sum_i \frac{x_i^2\log(x_i)}{1+x_i^2} }{ \frac1n\sum_i \frac{x_i^2\log(x_i)}{1+x_i^2} - 2 }, \tag{2.11} \end{equation}\] \[\begin{equation} \hat\theta = (\hat\alpha+1)\frac1n\sum_i \frac{ (x_i)^{\hat\alpha+1}}{1+(x_i)^{\hat\alpha}}. \tag{2.12} \end{equation}\]
Use Burr estimate on the sample \(1/x\)
Use paralogistic estimate on the sample \(1/x\)
Use pareto estimate on the sample \(1/x\)
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 \[\begin{equation} \gamma = \frac{ \log\left( \frac{ (\frac43)^{1/\alpha}-1 }{ (4)^{1/\alpha}-1 } \right) }{ \log\left(\frac{q_1-\mu}{q_3-\mu}\right) }, \tag{2.13} \end{equation}\] \[\begin{equation} \theta = \frac{q_1- q_3 }{ ((\frac43)^{1/\alpha}-1)^{1/\gamma} - ((4)^{1/\alpha}-1)^{1/\gamma} }. \tag{2.14} \end{equation}\]
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 \[\begin{equation} \alpha =n/\sum_i \log(1+ (\frac{x_i-\mu}{\theta})^{\gamma}). \tag{2.15} \end{equation}\]
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 \[\begin{equation} \hat\mu = \left\{ \begin{array}{ll} (1-\epsilon) \min_i x_i & \text{if } \min_i x_i <0 \\ (1+\epsilon)\min_i x_i & \text{if } \min_i x_i \geq 0 \\ \end{array} \right. . \tag{2.16} \end{equation}\] where \(\epsilon=0.05\).
Initial parameter estimation is \(\hat\mu\), \(\alpha^\star = 2\) , \(\hat\gamma\) from (2.13) with \(\alpha^\star\), \(\hat\theta\) from (2.14) with \(\alpha^\star\) and \(\hat\gamma\), \(\hat\alpha\) from (2.15) with \(\hat\mu\), \(\hat\theta\) and \(\hat\gamma\).
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 , \(\hat\theta\) from with \(\hat\gamma\).
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 with \(\alpha^\star\) and \(\gamma=1\), \(\hat\alpha\) from with \(\hat\mu\), \(\hat\theta\) and \(\gamma=1\),
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 .
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 with \(\alpha^\star\) and \(\gamma=1\), \(\hat\alpha\) from with \(\mu=0\), \(\hat\theta\) and \(\gamma=1\).
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 moment-estimator as in gamma \[ \hat\alpha = m_2^2/\mu_2, \hat\theta= \mu_2/m_1. \] Then cancelling the first component of the gradient we set \[ \hat\tau = \frac{\psi(\hat\alpha)}{\frac1n\sum_i \log(x_i/\hat\theta) }. \]
Transformed gamma with \(\tau=1\)
We compute moment-estimator as in gamma \[ \hat\alpha = m_2^2/\mu_2, \hat\theta= \mu_2/m_1. \]
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 \(\hat\tau\), \(\hat\theta\).
The MLE is the MME \(\hat\lambda = 1/m_1.\)
Same as transformed gamma distribution with \((1/x_i)_i\).
We compute moment-estimator as \[ \hat\alpha = (2m_2-m_1^2)/(m_2-m_1^2), \hat\theta= m_1m_2/(m_2-m_1^2). \]
We use the QME.
Same as transformed gamma distribution with \((1/x_i)_i\).