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  "body": "\n## Counterexample\n\nLet $f$ be the density of the standard [[log-normal distribution|4W40NHWPQT]]:\n$$f(x) = \\frac{1}{x \\sqrt{2\\pi}} \\exp\\left(-\\frac{(\\ln x)^2}{2}\\right) \\quad (x \u003e 0), \\qquad f(x) = 0 \\quad (x \\le 0).$$\nIts [[moments|01XAHPADBA]] are $\\int_0^\\infty x^n f(x)\\, dx = e^{n^2/2}$ for every integer $n \\ge 0$.\nFor a real number $a$ with $-1 \\le a \\le 1$, put\n$$f_a(x) = f(x)\\bigl(1 + a \\sin(2\\pi \\ln x)\\bigr) \\quad (x \u003e 0), \\qquad f_a(x) = 0 \\quad (x \\le 0).$$\nEach $f_a$ is a [[probability density|04YG690N5V]], because $f_a \\ge 0$ and it has the same total mass as $f$ by the computation below.\nCheck that $f_a$ has the same moments as $f$.\nIt suffices to show $\\int_0^\\infty x^n f(x) \\sin(2\\pi \\ln x)\\, dx = 0$ for every integer $n \\ge 0$.\nThe substitution $x = e^y$ turns this integral into $\\int_\\R e^{ny} \\varphi(y) \\sin(2\\pi y)\\, dy$, where $\\varphi(y) = e^{-y^2/2}/\\sqrt{2\\pi}$ is the [[standard normal|05AQE0QR0D]] density.\nSince $e^{ny} \\varphi(y) = e^{n^2/2} \\varphi(y - n)$, the substitution $y = u + n$ gives\n$$e^{n^2/2} \\int_\\R \\varphi(u) \\sin(2\\pi u + 2\\pi n)\\, du = e^{n^2/2} \\int_\\R \\varphi(u) \\sin(2\\pi u)\\, du = 0,$$\nbecause $n$ is an integer and the last integrand is odd.\nSo the densities $f_a$, $-1 \\le a \\le 1$, are pairwise different and all have the moments $e^{n^2/2}$.\nHence the moment problem for this moment sequence is not [[determinate|H0GY2J14CA]]: it is [[indeterminate|GCTN4BZ1ET]].\n\n## Remarks\n\nThe log-normal law has all moments finite, yet its moments do not determine it.\nThe moments grow like $e^{n^2/2}$, so fast that [[Carleman's condition|RVZ831CNYW]] fails.\nThe normal distribution, whose moments grow much more slowly, is determined by its moments.\n\n## Explanations\n\n### L1 human\n\nOne might hope that knowing all the averages $E[X], E[X^2], E[X^3], \\dots$ pins down a random quantity.\nThe log-normal law shows that this can fail.\nAdding a small wiggle $a \\sin(2\\pi \\ln x)$ to its density changes the shape, but every moment stays exactly the same.\nThe wiggles cancel out in each moment because, after taking logarithms, they become an odd function against a bell curve.\n",
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