確率を $P$ で表す。 $x \gt 0$ について、 \begin{align} P \left( X \leq x \right) &= P \left( Y \leq \ln x \right) \\ &= \frac{1}{\sqrt{2 \pi} \zeta} \int_{-\infty}^{\ln x} \exp \left[ - \frac{(y-\lambda)^2}{2 \zeta^2} \right] dy \\ \therefore \ \ f_X(x) &= \frac{d}{dx} P \left( X \leq x \right) \\ &= \frac{1}{\sqrt{2 \pi} \zeta x} \exp \left[ - \frac{(\ln x - \lambda)^2}{2 \zeta^2} \right] \end{align} である。
\begin{align} \mu_X &= E \left[ X \right] \\ &= E \left[ e^Y \right] \\ &= \frac{1}{\sqrt{2 \pi} \zeta} \int_{-\infty}^\infty \exp \left[ y - \frac{(y-\lambda)^2}{2 \zeta^2} \right] dy \\ &= \frac{1}{\sqrt{2 \pi} \zeta} \int_{-\infty}^\infty \exp \left[ - \frac{y^2-2(\lambda + \zeta^2) y + \lambda^2} {2 \zeta^2} \right] dy \\ &= \frac{1}{\sqrt{2 \pi} \zeta} \int_{-\infty}^\infty \exp \left[ - \frac{(y-(\lambda+\zeta^2))^2 - 2 \lambda \zeta^2 - \zeta^4} {2 \zeta^2} \right] dy \\ &= \exp \left[ \lambda + \frac{\zeta^2}{2} \right] \\ E \left[ X^2 \right] &= E \left[ e^{2Y} \right] \\ &= \frac{1}{\sqrt{2 \pi} \zeta} \int_{-\infty}^\infty \exp \left[ 2y - \frac{(y-\lambda)^2}{2 \zeta^2} \right] dy \\ &= \frac{1}{\sqrt{2 \pi} \zeta} \int_{-\infty}^\infty \exp \left[ - \frac{y^2-2(\lambda + 2 \zeta^2) y + \lambda^2} {2 \zeta^2} \right] dy \\ &= \frac{1}{\sqrt{2 \pi} \zeta} \int_{-\infty}^\infty \exp \left[ - \frac{(y-(\lambda+2 \zeta^2))^2 - 4 \lambda \zeta^2 - 4 \zeta^4} {2 \zeta^2} \right] dy \\ &= \exp \left[ 2 \lambda + 2 \zeta^2 \right] \\ \sigma_X^2 &= E \left[ X^2 \right] - E \left[ X \right]^2 \\ &= \exp \left[ 2 \lambda + 2 \zeta^2 \right] - \exp \left[ 2 \lambda + \zeta^2 \right] \\ &= \left( e^{\zeta^2} - 1 \right) \exp \left[ 2 \lambda + \zeta^2 \right] \end{align}
\begin{align} L (x_1, x_2, \cdots, x_n ; \lambda, \zeta) &= \prod_{i=1}^n \frac{1}{\sqrt{2 \pi} \zeta x_i} \exp \left[ - \frac{(\ln x_i - \lambda)^2}{2 \zeta^2} \right] \\ &= \left( \sqrt{2 \pi} \zeta \right)^{-n} \left( \prod_{i=1}^n x_i \right)^{-1} \exp \left[ - \frac{1}{2 \zeta^2} \sum_{j=1}^n (\ln x_j - \lambda)^2 \right] \end{align}
\begin{align} \ln L (x_1, x_2, \cdots, x_n ; \lambda, \zeta) &= - \frac{n}{2} \ln (2 \pi) - n \ln \zeta - \sum_{i=1}^n \ln x_i - \frac{1}{2 \zeta^2} \sum_{j=1}^n (\ln x_j - \lambda)^2 \\ \frac{\partial}{\partial \lambda} \ln L (x_1, x_2, \cdots, x_n ; \lambda, \zeta) &= \frac{1}{\zeta^2} \sum_{j=1}^n (\ln x_j - \lambda) \\ &= \frac{n}{\zeta^2} \left( \frac{1}{n} \sum_{j=1}^n \ln x_j - \lambda \right) \\ \frac{\partial}{\partial \zeta} \ln L (x_1, x_2, \cdots, x_n ; \lambda, \zeta) &= - \frac{n}{\zeta} + \frac{1}{\zeta^3} \sum_{j=1}^n (\ln x_j - \lambda)^2 \\ &= - \frac{n}{\zeta^3} \left( \zeta^2 - \frac{1}{n} \sum_{j=1}^n (\ln x_j - \lambda)^2 \right) \end{align} なので、 \begin{align} \hat{\lambda} &= \frac{1}{n} \sum_{j=1}^n \ln x_j \\ \hat{\zeta} &= \sqrt{ \frac{1}{n} \sum_{j=1}^n \left( \ln x_j - \hat{\lambda} \right)^2 } \end{align} がわかる。