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  "body": "\n## Example\n\nLet $\\mu \\in \\R$ and $\\sigma \u003e 0$, and let $Y$ be [[normal|05AQE0QR0D]] with mean $\\mu$ and variance $\\sigma^2$.\nThe law of $X = e^Y$ is the *log-normal distribution* with parameters $\\mu$ and $\\sigma^2$; so $X \u003e 0$ and $\\ln X$ is normal.\nIts [[density|0ZY9DX8FBF]] is\n$$f(x) = \\frac{1}{x \\sigma \\sqrt{2\\pi}} \\exp\\left(-\\frac{(\\ln x - \\mu)^2}{2 \\sigma^2}\\right) \\qquad (x \u003e 0),$$\nand $f(x) = 0$ for $x \\le 0$.\nCheck: for $0 \u003c a \u003c b$, $P(a \u003c X \u003c b) = P(\\ln a \u003c Y \u003c \\ln b)$, and the substitution $y = \\ln x$, $dy = dx / x$, turns the normal density of $Y$ into $f$.\nThe [[moments|01XAHPADBA]] are $E[X^n] = E[e^{nY}] = e^{n\\mu + n^2 \\sigma^2 / 2}$ for every real $n$, by the moment-generating function of the normal law.\nSo $E[X] = e^{\\mu + \\sigma^2/2}$ and $\\operatorname{Var}(X) = (e^{\\sigma^2} - 1) e^{2\\mu + \\sigma^2}$.\nThe median is $e^{\\mu}$, which is smaller than the mean.\n\n## Remarks\n\nAll moments are finite, but the [[moment-generating function|0WQVXP7PDB]] $E[e^{tX}]$ is infinite for every $t \u003e 0$.\nThe log-normal law is a standard example of a distribution that is not determined by its moments: other distributions have exactly the same moments.\nIf $X_1$ and $X_2$ are independent and log-normal, then $X_1 X_2$ is log-normal, because $\\ln X_1 + \\ln X_2$ is normal.\n\n## Explanations\n\n### L1 human\n\nA log-normal quantity is one whose logarithm follows a bell curve.\nIt arises when many independent factors multiply together, just as the normal law arises when many independent terms add.\nIts graph is skewed to the right: it is always positive, has a long tail of large values, and its mean lies above its median.\nSizes of particles, incomes and stock prices in the Black–Scholes model are often modelled with it.\n",
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