fixed typo in 8.3

_book/search.json

@@ -533,7 +533,7 @@
     "href": "r-session-03.html",
     "title": "\n8  R Session 03\n",
     "section": "",
-    "text": "8.1 Setup\nCodelibrary(here)\nlibrary(rio)\nlibrary(deSolve)\nlibrary(tidyverse)\nlibrary(ggtext)\nlibrary(gt)\nCodetheme_set(theme_minimal())\nCode# Loads the datasets: flu, measles, niamey, plauge\nflu <- rio::import(\n  \"https://raw.githubusercontent.com/arnold-c/SISMID-Module-02_2023/main/data/flu.csv\"\n)\nniamey <- rio::import(\n  \"https://raw.githubusercontent.com/arnold-c/SISMID-Module-02_2023/main/data/niamey.csv\"\n)\n# flu <- rio::import(here::here(\"data\", \"flu.csv\"))\n# niamey <- rio::import(here::here(\"data\", \"niamey.csv\"))",
+    "text": "8.1 Setup\nCodelibrary(here)\nlibrary(rio)\nlibrary(deSolve)\nlibrary(tidyverse)\nlibrary(ggtext)\nlibrary(gt)\nCodetheme_set(theme_minimal())\nCode# Loads the datasets: flu, measles, niamey, plauge\nflu <- rio::import(\n  \"https://raw.githubusercontent.com/arnold-c/SISMID-Module-02_2025/main/data/flu.csv\"\n)\nniamey <- rio::import(\n  \"https://raw.githubusercontent.com/arnold-c/SISMID-Module-02_2025/main/data/niamey.csv\"\n)\n# flu <- rio::import(here::here(\"data\", \"flu.csv\"))\n# niamey <- rio::import(here::here(\"data\", \"niamey.csv\"))",
     "crumbs": [
       "Day 2",
       "<span class='chapter-number'>8</span>  <span class='chapter-title'>R Session 03</span>"
@@ -555,7 +555,7 @@
     "href": "r-session-03.html#estimating-r_0-from-the-final-outbreak-size",
     "title": "\n8  R Session 03\n",
     "section": "\n8.3 Estimating \\(R_0\\) From The Final Outbreak Size",
-    "text": "8.3 Estimating \\(R_0\\) From The Final Outbreak Size\nOur first approach is to estimate \\(R_0\\) from the final outbreak size. Although unhelpful at the early stages of an epidemic (before the final epidemic size is observed), this method is nonetheless a useful tool for post hoc analysis. The method is general and can be motivated by the argument listed in (Keeling and Rohani 2008):\nFirst, we assume that the epidemic is started by a single infectious individual in a completely susceptible population. On average, this individual infects \\(R_0\\) others. The probability a particular individual escaped infection is therefore \\(e^{-R_0 / N}\\).\nIf \\(Z\\) individuals have been infected, the probability of an individual escaping infection from all potential sources is \\(e^{-Z R_0 / N}\\). It follows that at the end of the epidemic a proportion \\(R(\\infty) = Z / N\\) have been infected and the fraction remaining susceptible is \\(S(\\infty) = e^{-R(\\infty) R_0}\\), which is equal to \\(2 - R(\\infty)\\).\n\n\n\n\n\n\nNote\n\n\n\n\n\n\\(S(\\infty) = e^{-R(\\infty) R_0}\\) can be calculated by acknowledging that at equilibrium (\\(t = \\infty\\)), \\(S(\\infty) = 1 - R(\\infty) = Z / N\\), so substituting \\(R(\\infty)\\) into \\(1 - e^{-Z R_0 / N}\\) gives the desired result.\nIt could also be calculated by dividing \\(\\frac{\\dd{S}}{\\dd{t}}\\) by \\(\\frac{\\dd{R}}{\\dd{t}}\\):\n\\[\\begin{aligned}\n\\frac{\\dd{S}}{\\dd{R}} &= - \\frac{\\beta S}{\\gamma} \\\\\n&= - R_0 S\n\\end{aligned}\\]\nwhich is a separable differential equation, so can be integrated as follows:\n\\[\\begin{aligned}\n- \\int_{0}^{t} \\frac{1}{R_0 S} \\dd{S} &= \\int_{0}^{t} \\dd{R} \\\\\n- \\frac{1}{R_0} \\left(\\ln{S(t)} - \\ln{S(0)} \\right) &= R(t) - \\cancelto{0}{R(0)} \\\\\n\\ln{S(t)} &= \\ln{S(0)} - R_0 R(t) \\\\\nS(t) &= S(0) e^{-R_0 R(t)}\n\n\\end{aligned}\\]\n\n\n\nPutting this together, we get:\n\\[\n1 - R(\\infty) - e^{-R(\\infty) R_0} = 0\n\\]\nRearranging, we have the estimator\n\\[\n  \\hat{R_0} = \\frac{\\ln(1 - Z / N)}{-Z / N},\n\\]\nwhich, in this case, evaluates to \\(\\frac{\\ln(1 - 512 / 764)}{-512 / 764} = 1.655\\).\n\n8.3.1 Exercise 1\nThis equation shows the important one-to-one relationship between \\(R_0\\) and the final epidemic size. Plot the relationship between the total epidemic size and \\(R_0\\) for the complete range of values between 0 and 1.",
+    "text": "8.3 Estimating \\(R_0\\) From The Final Outbreak Size\nOur first approach is to estimate \\(R_0\\) from the final outbreak size. Although unhelpful at the early stages of an epidemic (before the final epidemic size is observed), this method is nonetheless a useful tool for post hoc analysis. The method is general and can be motivated by the argument listed in (Keeling and Rohani 2008):\nFirst, we assume that the epidemic is started by a single infectious individual in a completely susceptible population. On average, this individual infects \\(R_0\\) others. The probability a particular individual escaped infection is therefore \\(e^{-R_0 / N}\\).\nIf \\(Z\\) individuals have been infected, the probability of an individual escaping infection from all potential sources is \\(e^{-Z R_0 / N}\\). It follows that at the end of the epidemic a proportion \\(R(\\infty) = Z / N\\) have been infected and the fraction remaining susceptible is \\(S(\\infty) = e^{-R(\\infty) R_0}\\), which is equal to \\(1 - R(\\infty)\\).\n\n\n\n\n\n\nNote\n\n\n\n\n\n\\(S(\\infty) = e^{-R(\\infty) R_0}\\) can be calculated by acknowledging that at equilibrium (\\(t = \\infty\\)), \\(S(\\infty) = 1 - R(\\infty) = Z / N\\), so substituting \\(R(\\infty)\\) into \\(1 - e^{-Z R_0 / N}\\) gives the desired result.\nIt could also be calculated by dividing \\(\\frac{\\dd{S}}{\\dd{t}}\\) by \\(\\frac{\\dd{R}}{\\dd{t}}\\):\n\\[\\begin{aligned}\n\\frac{\\dd{S}}{\\dd{R}} &= - \\frac{\\beta S}{\\gamma} \\\\\n&= - R_0 S\n\\end{aligned}\\]\nwhich is a separable differential equation, so can be integrated as follows:\n\\[\\begin{aligned}\n- \\int_{0}^{t} \\frac{1}{R_0 S} \\dd{S} &= \\int_{0}^{t} \\dd{R} \\\\\n- \\frac{1}{R_0} \\left(\\ln{S(t)} - \\ln{S(0)} \\right) &= R(t) - \\cancelto{0}{R(0)} \\\\\n\\ln{S(t)} &= \\ln{S(0)} - R_0 R(t) \\\\\nS(t) &= S(0) e^{-R_0 R(t)}\n\n\\end{aligned}\\]\n\n\n\nPutting this together, we get:\n\\[\n1 - R(\\infty) - e^{-R(\\infty) R_0} = 0\n\\]\nRearranging, we have the estimator\n\\[\n  \\hat{R_0} = \\frac{\\ln(1 - Z / N)}{-Z / N},\n\\]\nwhich, in this case, evaluates to \\(\\frac{\\ln(1 - 512 / 764)}{-512 / 764} = 1.655\\).\n\n8.3.1 Exercise 1\nThis equation shows the important one-to-one relationship between \\(R_0\\) and the final epidemic size. Plot the relationship between the total epidemic size and \\(R_0\\) for the complete range of values between 0 and 1.",
     "crumbs": [
       "Day 2",
       "<span class='chapter-number'>8</span>  <span class='chapter-title'>R Session 03</span>"

r-session-03.qmd

@@ -70,7 +70,7 @@ On average, this individual infects $R_0$ others.
 The probability a particular individual escaped infection is therefore $e^{-R_0 / N}$.
 
 If $Z$ individuals have been infected, the probability of an individual escaping infection from all potential sources is $e^{-Z R_0 / N}$.
-It follows that at the end of the epidemic a proportion $R(\infty) = Z / N$ have been infected and the fraction remaining susceptible is $S(\infty) = e^{-R(\infty) R_0}$, which is equal to $2 - R(\infty)$.
+It follows that at the end of the epidemic a proportion $R(\infty) = Z / N$ have been infected and the fraction remaining susceptible is $S(\infty) = e^{-R(\infty) R_0}$, which is equal to $1 - R(\infty)$.
 
 ::: {.callout-note collapse="true"}
 $S(\infty) = e^{-R(\infty) R_0}$ can be calculated by acknowledging that at equilibrium ($t = \infty$), $S(\infty) = 1 - R(\infty) = Z / N$, so substituting $R(\infty)$ into $1 - e^{-Z R_0 / N}$ gives the desired result.