$y = \log x$ とおくと、 $x = \exp(y), dx = \exp(y) dy$ である。
期待値を $E$ , 分散を $V$ で表して、次のように計算する： \begin{align} E(X) &= \int_0^\infty x f(x) dx \\ &= \frac{1}{\sqrt{2 \pi}} \int_0^\infty \exp \left[ - \frac{1}{2} \left( \log x - \mu \right)^2 \right] dx \\ &= \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^\infty \exp \left[ - \frac{1}{2} \left( y - \mu \right)^2 \right] \exp(y) dy \\ &= \frac{\exp \left( \mu + \frac{1}{2} \right) }{\sqrt{2 \pi}} \int_{- \infty}^\infty \exp \left[ - \frac{1}{2} \left\{ y - (\mu + 1) \right\}^2 \right] dy \\ &= \exp \left( \mu + \frac{1}{2} \right) \\ E \left( X^2 \right) &= \int_0^\infty x^2 f(x) dx \\ &= \frac{1}{\sqrt{2 \pi}} \int_0^\infty x \exp \left[ - \frac{1}{2} \left( \log x - \mu \right)^2 \right] dx \\ &= \frac{1}{\sqrt{2 \pi}} \int_{- \infty}^\infty \exp \left[ - \frac{1}{2} \left( y - \mu \right)^2 \right] \exp(2y) dy \\ &= \frac{\exp \left( 2 \mu + 2 \right) }{\sqrt{2 \pi}} \int_{- \infty}^\infty \exp \left[ - \frac{1}{2} \left\{ y - (\mu + 2) \right\}^2 \right] dy \\ &= \exp \left( 2 \mu + 2 \right) \\ V(X) &= E \left( X^2 \right) - E(X)^2 \\ &= \exp \left( 2 \mu + 2 \right) - \exp \left( 2 \mu + 1 \right) \\ &= \exp \left( 2 \mu + 1 \right) (e-1) \end{align}