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  "body": "\n## Example\n\nLet $F$ be the distribution function of a nonnegative random variable with $F(0) = 0$, and let $g$ be a locally bounded function on $[0, \\infty)$.\nThe *renewal equation* for an unknown locally bounded function $u$ on $[0, \\infty)$ is\n$$u(t) = g(t) + \\int_0^t u(t - s) \\, dF(s), \\quad t \\ge 0.$$\nWhen $F$ has a density $k = F'$, the integral is $\\int_0^t k(t - s) \\, u(s) \\, ds$, and the equation is a [[Volterra equation of convolution type|SNFYTC1TTA]] of the second kind.\n\nIn [[renewal theory|45V0W838BQ]], let $X_1, X_2, \\dots$ be independent waiting times with distribution function $F$, and let $N(t)$ be the number of $n \\ge 1$ with $X_1 + \\cdots + X_n \\le t$.\nConditioning on the first waiting time $X_1$ shows that the renewal function $m(t) = \\mathbb{E}[N(t)]$ solves the renewal equation with $g = F$:\n$$m(t) = F(t) + \\int_0^t m(t - s) \\, dF(s).$$\n\nCheck on the exponential case: let $F(t) = 1 - e^{-\\lambda t}$ with $\\lambda \u003e 0$, so $dF(s) = \\lambda e^{-\\lambda s} \\, ds$.\nThen $N$ is a Poisson process of rate $\\lambda$, and $m(t) = \\lambda t$.\nIndeed $\\int_0^t \\lambda (t - s) \\lambda e^{-\\lambda s} \\, ds = \\lambda t - 1 + e^{-\\lambda t}$, and adding $F(t) = 1 - e^{-\\lambda t}$ gives $\\lambda t$.\n\n## Remarks\n\nThe renewal equation has a unique locally bounded solution, $u = g + m * g$, that is, $u(t) = g(t) + \\int_0^t g(t - s) \\, dm(s)$.\nLet $\\varphi_F(p) = \\int_{[0, \\infty)} e^{-ps} \\, dF(s)$.\nTaking Laplace transforms gives $\\hat u = \\hat g / (1 - \\varphi_F)$, as in the [[Laplace transform method|EHGECREYB4]].\nFor the exponential case, $\\hat F(p) = \\frac{\\lambda}{p(p + \\lambda)}$ and $\\varphi_F(p) = \\frac{\\lambda}{p + \\lambda}$, so $\\hat m(p) = \\frac{\\lambda}{p(p + \\lambda)} \\cdot \\frac{p + \\lambda}{p} = \\frac{\\lambda}{p^2}$, the transform of $\\lambda t$.\n\n## Explanations\n\n### L1 human\n\nThink of a light bulb that is replaced each time it fails.\nThe expected number of replacements by time $t$ is the chance of a first failure by $t$, plus what happens after a first failure at time $s$.\nAfter that failure, the same problem starts again with $t - s$ time left.\nThis \"restart\" argument gives the renewal equation, an integral equation in which the unknown appears with a time shift.\n",
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