Cumulative distribution

The MLP cumulative distribution function (${\displaystyle F(m)=\int _{-\infty }^{m}f(t)\,dt}$) is given by:

${\displaystyle {\begin{aligned}F(m)={\frac {1}{2}}{\text{erfc}}\left(-{\frac {\ln(m)-\mu _{0}}{{\sqrt {2}}\sigma _{0}}}\right)-{\frac {1}{2}}\exp \left(\alpha \mu _{0}+{\frac {\alpha ^{2}\sigma _{0}^{2}}{2}}\right)m^{-\alpha }{\text{erfc}}\left({\frac {\alpha \sigma _{0}}{\sqrt {2}}}\left(\alpha \sigma _{0}-{\frac {\ln(m)-\mu _{0}}{{\sqrt {2}}\sigma _{0}}}\right)\right)\end{aligned}}}$

We can see that as ${\displaystyle m\to 0,}$ that ${\displaystyle \textstyle F(m)\to {\frac {1}{2}}\operatorname {erfc} \left(-{\frac {\ln(m-\mu _{0})}{{\sqrt {2}}\sigma _{0}}}\right),}$ which is the cumulative distribution function for a lognormal distribution with parameters μ₀ and σ₀.

Mean, variance, raw moments

The expectation value of ${\displaystyle M}$^k gives the ${\displaystyle k}$^th raw moment of ${\displaystyle M}$,

${\displaystyle {\begin{aligned}\langle M^{k}\rangle =\int _{0}^{\infty }m^{k}f(m)\mathrm {d} m\end{aligned}}}$

This exists if and only if α > ${\displaystyle k}$, in which case it becomes:

${\displaystyle {\begin{aligned}\langle M^{k}\rangle ={\frac {\alpha }{\alpha -k}}\exp \left({\frac {\sigma _{0}^{2}k^{2}}{2}}+\mu _{0}k\right),\ \alpha >k\end{aligned}}}$

which is the ${\displaystyle k}$^th raw moment of the lognormal distribution with the parameters μ₀ and σ₀ scaled by α⁄α-${\displaystyle k}$ in the limit α→∞. This gives the mean and variance of the MLP distribution:

${\displaystyle {\begin{aligned}\langle M\rangle ={\frac {\alpha }{\alpha -1}}\exp \left({\frac {\sigma _{0}^{2}}{2}}+\mu _{0}\right),\ \alpha >1\end{aligned}}}$

${\displaystyle {\begin{aligned}\langle M^{2}\rangle ={\frac {\alpha }{\alpha -2}}\exp \left(2\left(\sigma _{0}^{2}+\mu _{0}\right)\right),\ \alpha >2\end{aligned}}}$

Var(${\displaystyle M}$) = ⟨${\displaystyle M}$²⟩-(⟨${\displaystyle M}$⟩)² = α exp(σ₀² + 2μ₀) (exp(σ₀²)/α-2 - α/(α-2)²), α > 2

Mode

The solution to the equation ${\displaystyle f'(m)}$ = 0 (equating the slope to zero at the point of maxima) for ${\displaystyle m}$ gives the mode of the MLP distribution.

${\displaystyle f'(m)=0\Leftrightarrow K\operatorname {erfc} (u)=\exp(-u^{2}),}$

where ${\displaystyle \textstyle u={\frac {1}{\sqrt {2}}}\left(\alpha \sigma _{0}-{\frac {\ln m-\mu _{0}}{\sigma _{0}}}\right)}$ and ${\displaystyle K=\sigma _{0}(\alpha +1){\tfrac {\sqrt {\pi }}{2}}.}$

Numerical methods are required to solve this transcendental equation. However, noting that if ${\displaystyle K}$≈1 then u = 0 gives us the mode ${\displaystyle m}$^*:

${\displaystyle m^{*}=\exp(\mu _{0}+\alpha \sigma _{0}^{2})}$

Random variate

The lognormal random variate is:

${\displaystyle {\begin{aligned}L(\mu ,\sigma )=\exp(\mu +\sigma N(0,1))\end{aligned}}}$

where ${\displaystyle N(0,1)}$ is standard normal random variate. The exponential random variate is :

${\displaystyle {\begin{aligned}E(\delta )=-\delta ^{-1}\ln(R(0,1))\end{aligned}}}$

where R(0,1) is the uniform random variate in the interval [0,1]. Using these two, we can derive the random variate for the MLP distribution to be:

${\displaystyle {\begin{aligned}M(\mu _{0},\sigma _{0},\alpha )=\exp(\mu _{0}+\sigma _{0}N(0,1)-\alpha ^{-1}\ln(R(0,1)))\end{aligned}}}$