{
  "index": "2021-B-6",
  "type": "COMB",
  "tag": [
    "COMB",
    "ANA",
    "NT"
  ],
  "difficulty": "",
  "question": "Given an ordered list of $3N$ real numbers, we can \\emph{trim} it to form a list of $N$ numbers as follows: We divide the list into $N$ groups of $3$ consecutive numbers, and within each group, discard the highest and lowest numbers, keeping only the median.\n\nConsider generating a random number $X$ by the following procedure: Start with a list of $3^{2021}$ numbers, drawn independently and uniformly at random between 0 and 1. Then trim this list as defined above, leaving a list of $3^{2020}$ numbers. Then trim again repeatedly until just one number remains; let $X$ be this number. Let $\\mu$ be the expected value of $|X - \\frac{1}{2}|$. Show that\n\\[\n\\mu \\geq \\frac{1}{4} \\left( \\frac{2}{3} \\right)^{2021}.\n\\]\n\n\\end{itemize}\n\n\\end{document}",
  "solution": "\\noindent\n\\textbf{First solution.}\n(based on a suggestion of Noam Elkies)\nLet $f_k(x)$ be the probability distribution of $X_k$, the last number remaining when one repeatedly trims a list of $3^k$ random variables chosen with respect to the uniform distribution on $[0,1]$; note that $f_0(x) = 1$ for $x \\in [0,1]$.\nLet $F_k(x)=\\int_0^x f_k(t)\\,dt$ be the cumulative distribution function; by symmetry,\n$F_k(\\frac{1}{2}) = \\frac{1}{2}$.\nLet $\\mu_k$ be the expected value of $X_k - \\frac{1}{2}$; then $\\mu_0 = \\frac{1}{4}$, so it will suffice to prove that $\\mu_{k} \\geq \\frac{2}{3} \\mu_{k-1}$ for $k > 0$.\n\nBy integration by parts and symmetry, we have\n\\[\n\\mu_k = 2 \\int_0^{1/2} \\left( \\frac{1}{2} - x \\right) f_k(x)\\,dx  = 2 \\int_0^{1/2} F_k(x)\\,dx;\n\\]\nthat is, $\\mu_k$ computes twice the area under the curve $y = F_k(x)$ for $0 \\leq x \\leq\\frac{1}{2}$. Since $F_k$ is a monotone function from $[0, \\frac{1}{2}]$ \nwith $F_k(0) = 0$ and $F_k(\\frac{1}{2}) = \\frac{1}{2}$, we may transpose the axes to obtain\n\\begin{equation} \\label{eq:2021B6 eq4}\n\\mu_k = 2 \\int_0^{1/2} \\left( \\frac{1}{2} - F_k^{-1}(y) \\right)\\,dy.\n\\end{equation}\n\nSince $f_k(x)$ is the probability distribution of the median of three random variables chosen with respect to the distribution $f_{k-1}(x)$,\n\\begin{equation} \\label{eq:2021B6 eq1}\nf_k(x) = 6 f_{k-1}(x) F_{k-1}(x) ( 1-F_{k-1}(x) )\n\\end{equation}\nor equivalently\n\\begin{equation} \\label{eq:2021B6 eq2}\nF_k(x) = 3 F_{k-1}(x)^2 - 2 F_{k-1}(x)^3.\n\\end{equation}\nBy induction, $F_k$ is the $k$-th iterate of $F_1(x) = 3x^2 -2x^3$, so\n\\begin{equation} \\label{eq:2021B6 eq5}\nF_k(x) = F_{k-1}(F_1(x)).\n\\end{equation}\nSince $f_1(t) = 6t(1-t) \\leq \\frac{3}{2}$ for $t \\in [0,\\frac{1}{2}]$,\n\\[\n\\frac{1}{2} - F_1(x) = \\int_x^{1/2} 6t(1-t)\\,dt \\leq \\frac{3}{2}\\left(\\frac{1}{2}-x\\right);\n\\]\nfor $y \\in [0, \\frac{1}{2}]$, we may take $x = F_{k}^{-1}(y)$ to obtain\n\\begin{equation} \\label{eq:2021B6 eq3}\n\\frac{1}{2} - F_k^{-1}(y) \\geq \\frac{2}{3} \\left( \\frac{1}{2} - F_{k-1}^{-1}(y) \\right).\n\\end{equation}\nUsing \\eqref{eq:2021B6 eq5} and \\eqref{eq:2021B6 eq3}, we obtain\n\\begin{align*}\n\\mu_k &= 2 \\int_0^{1/2} \\left( \\frac{1}{2} - F_k^{-1}(y) \\right) \\,dy \\\\\n&\\geq \\frac{4}{3} \\int_0^{1/2} \\left( \\frac{1}{2} - F_{k-1}^{-1}(y) \\right) \\,dy = \\frac{2}{3}\\mu_{k-1}\n\\end{align*}\nas desired.\n\n\\noindent\n\\textbf{Second solution.}\nRetain notation as in the first solution. Again $F_k(\\frac{1}{2}) = \\frac{1}{2}$, so \\eqref{eq:2021B6 eq1} implies\n\\[\nf_k\\left( \\frac{1}{2} \\right) = 6 f_{k-1} \\left( \\frac{1}{2} \\right) \\times \\frac{1}{2} \\times \\frac{1}{2}.\n\\]\nBy induction on $k$, we deduce that %$f_k(x)$ is a polynomial in $x$,\n$f_k(\\frac{1}{2}) = (\\frac{3}{2})^k$\nand $f_k(x)$ is nondecreasing on $[0,\\frac{1}{2}]$.\n(More precisely, besides \\eqref{eq:2021B6 eq1}, the second assertion uses that $F_{k-1}(x)$ increases from $0$ to $1/2$\nand $y \\mapsto y - y^2$ is nondecreasing on $[0, 1/2]$.)\n\nThe expected value of $|X_k-\\frac{1}{2}|$ equals\n\\begin{align*}\n\\mu_k &= 2 \\int_0^{1/2} \\left( \\frac{1}{2} - x \\right) f_k(x)\\,dx \\\\\n&= 2 \\int_0^{1/2} x f_k\\left( \\frac{1}{2} - x \\right)\\,dx.% \\\\\n%&= \\int_0^{1/2} \\left( \\frac{1}{2} - F_k\\left( \\frac{1}{2} - x \\right)\\right)\\,dx \\\\\n\\end{align*}\n%where the last step is integration by parts. Define the function\nDefine the function\n\\[\ng_k(x) = \\begin{cases} \\left( \\frac{3}{2} \\right)^k & x \\in \\left[ 0, \\frac{1}{2} \\left( \\frac{2}{3} \\right)^k \\right] \\\\ 0 & \\mbox{otherwise}.\n\\end{cases}\n\\]\nNote that for $x \\in [0, 1/2]$ we have\n\\[\n\\int_0^x (g_k(t) - f_k(1/2-t))\\,dt \\geq 0\n\\]\nwith equality at $x=0$ or $x=1/2$. (On the interval $[0, (1/2)(2/3)^k]$ the integrand is nonnegative, so the function increases from 0; on the interval $[(1/2)(2/3)^k, 1/2]$ the integrand is nonpositive, so the function decreases to 0.)\nHence by integration by parts,\n\\begin{align*}\n&\\mu_k - 2 \\int_0^{1/2} x g_k(x) \\,dx \\\\\n&\\quad = \\int_0^{1/2} 2x (f_k\\left( \\frac{1}{2} - x \\right) - g_k(x)) \\,dx \\\\\n&\\quad = \\int_0^{1/2} x^2 \\left( \\int_0^x g_k(t) - \\int_0^x f_k\\left( \\frac{1}{2} - t \\right)\\,dt \\,dt \\right)\\,dx \\geq 0. \n\\end{align*}\n(This can also be interpreted as an instance of the \\emph{rearrangement inequality}.)\n\nWe now see that\n\\begin{align*}\n\\mu_k &\\geq 2\\int_0^{1/2} x g_k(x)\\,dx \\\\\n&\\quad \\geq 2 \\left( \\frac{3}{2} \\right)^k \\int_0^{(1/2)(2/3)^k} x\\,dx\\\\\n&\\quad = 2 \\left( \\frac{3}{2} \\right)^k \\left. \\frac{1}{2} x^2 \\right|_0^{(1/2)(2/3)^k} \\\\\n&\\quad = 2 \\left( \\frac{3}{2} \\right)^k \\frac{1}{8} \\left( \\frac{2}{3} \\right)^{2k}  = \\frac{1}{4} \\left( \\frac{2}{3} \\right)^k\n\\end{align*}\nas desired.\n\n\n\n\\noindent\n\\textbf{Remark.}\nFor comparison, if we instead take the median of a list of $n$ numbers, the probability distribution is given by\n\\[\nP_{2n+1}(x) = \\frac{(2n+1)!}{n!n!} x^n (1-x)^n.\n\\]\nThe expected value of the absolute difference between $1/2$ and the median is \n\\[\n2 \\int_0^{1/2} (1/2 - x) P_{2n+1}(x) dx = 2^{-2n-2}{{2n+1}\\choose n}.\n\\]\nFor $n = 3^{2021}$, using Stirling's approximation this can be estimated as\n$1.13 (0.577)^{2021} < 0.25 (0.667)^{2021}$. This shows that the trimming procedure produces a quantity that is on average further away from 1/2 than the median.\n\n\\end{itemize}\n\\end{document}",
  "vars": [
    "x",
    "y",
    "t"
  ],
  "params": [
    "N",
    "k",
    "n",
    "X",
    "X_k",
    "f_k",
    "f_k-1",
    "F_k",
    "F_k-1",
    "F_1",
    "g_k",
    "P_2n+1",
    "\\\\mu"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "randomx",
        "y": "variabley",
        "t": "variablet",
        "N": "triplesize",
        "k": "levelindex",
        "n": "samplecount",
        "X": "trimvalue",
        "X_k": "trimvaluelevel",
        "f_k": "densitylevel",
        "f_k-1": "densityprev",
        "F_k": "cumullevel",
        "F_k-1": "cumulprev",
        "F_1": "cumulfirst",
        "g_k": "auxilevel",
        "P_2n+1": "medidistrib",
        "\\mu": "meanabdev"
      },
      "question": "Given an ordered list of $3triplesize$ real numbers, we can \\emph{trim} it to form a list of $triplesize$ numbers as follows: We divide the list into $triplesize$ groups of $3$ consecutive numbers, and within each group, discard the highest and lowest numbers, keeping only the median.\n\nConsider generating a random number $trimvalue$ by the following procedure: Start with a list of $3^{2021}$ numbers, drawn independently and uniformly at random between 0 and 1. Then trim this list as defined above, leaving a list of $3^{2020}$ numbers. Then trim again repeatedly until just one number remains; let $trimvalue$ be this number. Let $meanabdev$ be the expected value of $|trimvalue - \\frac{1}{2}|$. Show that\n\\[\nmeanabdev \\geq \\frac{1}{4} \\left( \\frac{2}{3} \\right)^{2021}.\n\\]\n\n\\end{itemize}\n\n\\end{document}",
      "solution": "\\noindent\n\\textbf{First solution.}\n(based on a suggestion of Noam Elkies)\nLet $densitylevel(randomx)$ be the probability distribution of $trimvaluelevel$, the last number remaining when one repeatedly trims a list of $3^{levelindex}$ random variables chosen with respect to the uniform distribution on $[0,1]$; note that $f_0(randomx) = 1$ for $randomx \\in [0,1]$.\nLet $cumullevel(randomx)=\\int_0^{randomx} densitylevel(variablet)\\,d variablet$ be the cumulative distribution function; by symmetry,\n$cumullevel\\!\\left(\\frac{1}{2}\\right) = \\frac{1}{2}$.\nLet $meanabdev_{levelindex}$ be the expected value of $trimvaluelevel - \\frac{1}{2}$; then $meanabdev_{0} = \\frac{1}{4}$, so it will suffice to prove that $meanabdev_{levelindex} \\geq \\frac{2}{3} meanabdev_{levelindex-1}$ for $levelindex > 0$.\n\nBy integration by parts and symmetry, we have\n\\[\nmeanabdev_{levelindex} = 2 \\int_0^{1/2} \\left( \\frac{1}{2} - randomx \\right) densitylevel(randomx)\\,d randomx  = 2 \\int_0^{1/2} cumullevel(randomx)\\,d randomx;\n\\]\nthat is, $meanabdev_{levelindex}$ computes twice the area under the curve $variabley = cumullevel(randomx)$ for $0 \\leq randomx \\leq\\frac{1}{2}$. Since $cumullevel$ is a monotone function from $[0, \\frac{1}{2}]$ \nwith $cumullevel(0) = 0$ and $cumullevel(\\frac{1}{2}) = \\frac{1}{2}$, we may transpose the axes to obtain\n\\begin{equation} \\label{eq:2021B6 eq4}\nmeanabdev_{levelindex} = 2 \\int_0^{1/2} \\left( \\frac{1}{2} - cumullevel^{-1}(variabley) \\right)\\,d variabley.\n\\end{equation}\n\nSince $densitylevel(randomx)$ is the probability distribution of the median of three random variables chosen with respect to the distribution $densityprev(randomx)$,\n\\begin{equation} \\label{eq:2021B6 eq1}\ndensitylevel(randomx) = 6\\, densityprev(randomx)\\, cumulprev(randomx)\\, \\bigl( 1-cumulprev(randomx) \\bigr)\n\\end{equation}\nor equivalently\n\\begin{equation} \\label{eq:2021B6 eq2}\ncumullevel(randomx) = 3\\, cumulprev(randomx)^{2} - 2\\, cumulprev(randomx)^{3}.\n\\end{equation}\nBy induction, $cumullevel$ is the $levelindex$-th iterate of $cumulfirst(randomx) = 3randomx^{2} -2randomx^{3}$, so\n\\begin{equation} \\label{eq:2021B6 eq5}\ncumullevel(randomx) = cumulprev\\!\\bigl(cumulfirst(randomx)\\bigr).\n\\end{equation}\nSince $f_1(variablet) = 6variablet(1-variablet) \\leq \\frac{3}{2}$ for $variablet \\in [0,\\frac{1}{2}]$,\n\\[\n\\frac{1}{2} - cumulfirst(randomx) = \\int_{randomx}^{1/2} 6variablet(1-variablet)\\,d variablet \\leq \\frac{3}{2}\\left(\\frac{1}{2}-randomx\\right);\n\\]\nfor $variabley \\in [0, \\frac{1}{2}]$, we may take $randomx = cumullevel^{-1}(variabley)$ to obtain\n\\begin{equation} \\label{eq:2021B6 eq3}\n\\frac{1}{2} - cumullevel^{-1}(variabley) \\geq \\frac{2}{3} \\left( \\frac{1}{2} - cumulprev^{-1}(variabley) \\right).\n\\end{equation}\nUsing \\eqref{eq:2021B6 eq5} and \\eqref{eq:2021B6 eq3}, we obtain\n\\begin{align*}\nmeanabdev_{levelindex} &= 2 \\int_0^{1/2} \\left( \\frac{1}{2} - cumullevel^{-1}(variabley) \\right) \\,d variabley \\\\\n&\\geq \\frac{4}{3} \\int_0^{1/2} \\left( \\frac{1}{2} - cumulprev^{-1}(variabley) \\right) \\,d variabley = \\frac{2}{3}meanabdev_{levelindex-1}\n\\end{align*}\nas desired.\n\n\\noindent\n\\textbf{Second solution.}\nRetain notation as in the first solution. Again $cumullevel\\!\\left( \\frac{1}{2} \\right) = \\frac{1}{2}$, so \\eqref{eq:2021B6 eq1} implies\n\\[\ndensitylevel\\!\\left( \\frac{1}{2} \\right) = 6\\, densityprev \\!\\left( \\frac{1}{2} \\right) \\times \\frac{1}{2} \\times \\frac{1}{2}.\n\\]\nBy induction on $levelindex$, we deduce that\ndensitylevel\\!\\left(\\frac{1}{2}\\right) = \\left(\\frac{3}{2}\\right)^{levelindex}$\nand $densitylevel(randomx)$ is nondecreasing on $[0,\\frac{1}{2}]$.\n(More precisely, besides \\eqref{eq:2021B6 eq1}, the second assertion uses that $cumulprev(randomx)$ increases from $0$ to $1/2$\nand $y \\mapsto y - y^{2}$ is nondecreasing on $[0, 1/2]$.)\n\nThe expected value of $|trimvaluelevel-\\frac{1}{2}|$ equals\n\\begin{align*}\nmeanabdev_{levelindex} &= 2 \\int_0^{1/2} \\left( \\frac{1}{2} - randomx \\right) densitylevel(randomx)\\,d randomx \\\\\n&= 2 \\int_0^{1/2} randomx \\, densitylevel\\!\\left( \\frac{1}{2} - randomx \\right)\\,d randomx.\n\\end{align*}\nDefine the function\n\\[\nauxilevel(randomx) = \\begin{cases} \\left( \\frac{3}{2} \\right)^{levelindex} & randomx \\in \\left[ 0, \\frac{1}{2} \\left( \\frac{2}{3} \\right)^{levelindex} \\right] \\\\ 0 & \\mbox{otherwise}.\n\\end{cases}\n\\]\nNote that for $randomx \\in [0, 1/2]$ we have\n\\[\n\\int_0^{randomx} \\bigl(auxilevel(variablet) - densitylevel(1/2 - variablet)\\bigr)\\,d variablet \\ge 0\n\\]\nwith equality at $randomx=0$ or $randomx=1/2$. (On the interval $[0, (1/2)(2/3)^{levelindex}]$ the integrand is nonnegative, so the function increases from 0; on the interval $[(1/2)(2/3)^{levelindex}, 1/2]$ the integrand is nonpositive, so the function decreases to 0.)\nHence by integration by parts,\n\\begin{align*}\n&meanabdev_{levelindex} - 2 \\int_0^{1/2} randomx\\, auxilevel(randomx) \\,d randomx \\\\\n&\\quad = \\int_0^{1/2} 2\\, randomx \\bigl( densitylevel\\!\\left( \\tfrac{1}{2} - randomx \\right) - auxilevel(randomx) \\bigr) \\,d randomx \\\\\n&\\quad = \\int_0^{1/2} randomx^{2} \\left( \\int_0^{randomx} auxilevel(variablet)\\,d variablet - \\int_0^{randomx} densitylevel\\!\\left( \\tfrac{1}{2} - variablet \\right)\\,d variablet \\right)\\,d randomx \\ge 0. \n\\end{align*}\n(This can also be interpreted as an instance of the \\emph{rearrangement inequality}.)\n\nWe now see that\n\\begin{align*}\nmeanabdev_{levelindex} &\\ge 2\\int_0^{1/2} randomx\\, auxilevel(randomx)\\,d randomx \\\\\n&\\quad \\ge 2 \\left( \\frac{3}{2} \\right)^{levelindex} \\int_0^{(1/2)(2/3)^{levelindex}} randomx\\,d randomx\\\\\n&\\quad = 2 \\left( \\frac{3}{2} \\right)^{levelindex} \\left. \\frac{1}{2} randomx^{2} \\right|_0^{(1/2)(2/3)^{levelindex}} \\\\\n&\\quad = 2 \\left( \\frac{3}{2} \\right)^{levelindex} \\frac{1}{8} \\left( \\frac{2}{3} \\right)^{2\\, levelindex}  = \\frac{1}{4} \\left( \\frac{2}{3} \\right)^{levelindex}\n\\end{align*}\nas desired.\n\n\n\n\\noindent\n\\textbf{Remark.}\nFor comparison, if we instead take the median of a list of $samplecount$ numbers, the probability distribution is given by\n\\[\nmedidistrib(randomx) = \\frac{(2samplecount+1)!}{samplecount!\\,samplecount!} randomx^{samplecount} (1-randomx)^{samplecount}.\n\\]\nThe expected value of the absolute difference between $1/2$ and the median is \n\\[\n2 \\int_0^{1/2} \\left( \\frac{1}{2} - randomx \\right) medidistrib(randomx) \\,d randomx = 2^{-2samplecount-2}\\binom{2samplecount+1}{samplecount}.\n\\]\nFor $samplecount = 3^{2021}$, using Stirling's approximation this can be estimated as\n$1.13 (0.577)^{2021} < 0.25 (0.667)^{2021}$. This shows that the trimming procedure produces a quantity that is on average further away from 1/2 than the median.\n\n\\end{itemize}\n\\end{document}"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "butterscotch",
        "y": "nightingale",
        "t": "drumsticks",
        "N": "gingerbread",
        "k": "marigolds",
        "n": "harmonica",
        "X": "bluewhale",
        "X_k": "blacksmith",
        "f_k": "seashells",
        "f_k-1": "chessboard",
        "F_k": "riverbank",
        "F_k-1": "featherbed",
        "F_1": "raincloud",
        "g_k": "starlight",
        "P_2n+1": "skateboard",
        "\\\\mu": "wanderlust"
      },
      "question": "Given an ordered list of $3gingerbread$ real numbers, we can \\emph{trim} it to form a list of $gingerbread$ numbers as follows: We divide the list into $gingerbread$ groups of $3$ consecutive numbers, and within each group, discard the highest and lowest numbers, keeping only the median.\n\nConsider generating a random number $bluewhale$ by the following procedure: Start with a list of $3^{2021}$ numbers, drawn independently and uniformly at random between 0 and 1. Then trim this list as defined above, leaving a list of $3^{2020}$ numbers. Then trim again repeatedly until just one number remains; let $bluewhale$ be this number. Let $wanderlust$ be the expected value of $|bluewhale - \\frac{1}{2}|$. Show that\n\\[\nwanderlust \\geq \\frac{1}{4} \\left( \\frac{2}{3} \\right)^{2021}.\n\\]\n\n\\end{itemize}\n\n\\end{document}",
      "solution": "\\noindent\n\\textbf{First solution.}\n(based on a suggestion of Noam Elkies)\nLet $seashells(butterscotch)$ be the probability distribution of $blacksmith$, the last number remaining when one repeatedly trims a list of $3^{marigolds}$ random variables chosen with respect to the uniform distribution on $[0,1]$; note that $seashells_0(butterscotch) = 1$ for $butterscotch \\in [0,1]$.\nLet $riverbank(butterscotch)=\\int_0^{butterscotch} seashells(drumsticks)\\,d drumsticks$ be the cumulative distribution function; by symmetry,\n$riverbank(\\frac{1}{2}) = \\frac{1}{2}$.\nLet $wanderlust_{marigolds}$ be the expected value of $blacksmith - \\frac{1}{2}$; then $wanderlust_0 = \\frac{1}{4}$, so it will suffice to prove that $wanderlust_{marigolds} \\geq \\frac{2}{3} wanderlust_{marigolds-1}$ for $marigolds > 0$.\n\nBy integration by parts and symmetry, we have\n\\[\nwanderlust_{marigolds} = 2 \\int_0^{1/2} \\left( \\frac{1}{2} - butterscotch \\right) seashells(butterscotch)\\,d butterscotch  = 2 \\int_0^{1/2} riverbank(butterscotch)\\,d butterscotch;\n\\]\nthat is, $wanderlust_{marigolds}$ computes twice the area under the curve $y = riverbank(butterscotch)$ for $0 \\leq butterscotch \\leq\\frac{1}{2}$. Since $riverbank$ is a monotone function from $[0, \\frac{1}{2}]$ \nwith $riverbank(0) = 0$ and $riverbank(\\frac{1}{2}) = \\frac{1}{2}$, we may transpose the axes to obtain\n\\begin{equation} \\label{eq:2021B6 eq4}\nwanderlust_{marigolds} = 2 \\int_0^{1/2} \\left( \\frac{1}{2} - riverbank^{-1}(nightingale) \\right)\\,d nightingale.\n\\end{equation}\n\nSince $seashells(butterscotch)$ is the probability distribution of the median of three random variables chosen with respect to the distribution $seashells_{marigolds-1}(butterscotch)$,\n\\begin{equation} \\label{eq:2021B6 eq1}\nseashells(butterscotch) = 6 seashells_{marigolds-1}(butterscotch) riverbank_{marigolds-1}(butterscotch) ( 1-riverbank_{marigolds-1}(butterscotch) )\n\\end{equation}\nor equivalently\n\\begin{equation} \\label{eq:2021B6 eq2}\nriverbank(butterscotch) = 3 riverbank_{marigolds-1}(butterscotch)^2 - 2 riverbank_{marigolds-1}(butterscotch)^3.\n\\end{equation}\nBy induction, $riverbank$ is the $marigolds$-th iterate of $raincloud(butterscotch) = 3butterscotch^2 -2butterscotch^3$, so\n\\begin{equation} \\label{eq:2021B6 eq5}\nriverbank(butterscotch) = riverbank_{marigolds-1}(raincloud(butterscotch)).\n\\end{equation}\nSince $seashells_1(drumsticks) = 6drumsticks(1-drumsticks) \\leq \\frac{3}{2}$ for $drumsticks \\in [0,\\frac{1}{2}]$,\n\\[\n\\frac{1}{2} - raincloud(butterscotch) = \\int_{butterscotch}^{1/2} 6drumsticks(1-drumsticks)\\,d drumsticks \\leq \\frac{3}{2}\\left(\\frac{1}{2}-butterscotch\\right);\n\\]\nfor $nightingale \\in [0, \\frac{1}{2}]$, we may take $butterscotch = riverbank^{-1}(nightingale)$ to obtain\n\\begin{equation} \\label{eq:2021B6 eq3}\n\\frac{1}{2} - riverbank^{-1}(nightingale) \\geq \\frac{2}{3} \\left( \\frac{1}{2} - riverbank_{marigolds-1}^{-1}(nightingale) \\right).\n\\end{equation}\nUsing \\eqref{eq:2021B6 eq5} and \\eqref{eq:2021B6 eq3}, we obtain\n\\begin{align*}\nwanderlust_{marigolds} &= 2 \\int_0^{1/2} \\left( \\frac{1}{2} - riverbank^{-1}(nightingale) \\right) \\,d nightingale \\\\\n&\\geq \\frac{4}{3} \\int_0^{1/2} \\left( \\frac{1}{2} - riverbank_{marigolds-1}^{-1}(nightingale) \\right) \\,d nightingale = \\frac{2}{3}wanderlust_{marigolds-1}\n\\end{align*}\nas desired.\n\n\\noindent\n\\textbf{Second solution.}\nRetain notation as in the first solution. Again $riverbank(\\frac{1}{2}) = \\frac{1}{2}$, so \\eqref{eq:2021B6 eq1} implies\n\\[\nseashells\\left( \\frac{1}{2} \\right) = 6 seashells_{marigolds-1} \\left( \\frac{1}{2} \\right) \\times \\frac{1}{2} \\times \\frac{1}{2}.\n\\]\nBy induction on $marigolds$, we deduce that %$seashells(butterscotch)$ is a polynomial in $butterscotch$,\n$seashells(\\frac{1}{2}) = (\\frac{3}{2})^{marigolds}$\nand $seashells(butterscotch)$ is nondecreasing on $[0,\\frac{1}{2}]$.\n(More precisely, besides \\eqref{eq:2021B6 eq1}, the second assertion uses that $riverbank_{marigolds-1}(butterscotch)$ increases from $0$ to $1/2$\nand $nightingale \\mapsto nightingale - nightingale^2$ is nondecreasing on $[0, 1/2]$.)\n\nThe expected value of $|blacksmith-\\frac{1}{2}|$ equals\n\\begin{align*}\nwanderlust_{marigolds} &= 2 \\int_0^{1/2} \\left( \\frac{1}{2} - butterscotch \\right) seashells(butterscotch)\\,d butterscotch \\\\\n&= 2 \\int_0^{1/2} butterscotch seashells\\left( \\frac{1}{2} - butterscotch \\right)\\,d butterscotch.% \\\\\n%&= \\int_0^{1/2} \\left( \\frac{1}{2} - riverbank\\left( \\frac{1}{2} - butterscotch \\right)\\right)\\,d butterscotch \\\\\n\\end{align*}\n%where the last step is integration by parts. Define the function\nDefine the function\n\\[\nstarlight(butterscotch) = \\begin{cases} \\left( \\frac{3}{2} \\right)^{marigolds} & butterscotch \\in \\left[ 0, \\frac{1}{2} \\left( \\frac{2}{3} \\right)^{marigolds} \\right] \\\\ 0 & \\mbox{otherwise}.\n\\end{cases}\n\\]\nNote that for $butterscotch \\in [0, 1/2]$ we have\n\\[\n\\int_0^{butterscotch} (starlight(drumsticks) - seashells(1/2-drumsticks))\\,d drumsticks \\geq 0\n\\]\nwith equality at $butterscotch=0$ or $butterscotch=1/2$. (On the interval $[0, (1/2)(2/3)^{marigolds}]$ the integrand is nonnegative, so the function increases from 0; on the interval $[(1/2)(2/3)^{marigolds}, 1/2]$ the integrand is nonpositive, so the function decreases to 0.)\nHence by integration by parts,\n\\begin{align*}\n&wanderlust_{marigolds} - 2 \\int_0^{1/2} butterscotch starlight(butterscotch) \\,d butterscotch \\\\\n&\\quad = \\int_0^{1/2} 2butterscotch (seashells\\left( \\frac{1}{2} - butterscotch \\right) - starlight(butterscotch)) \\,d butterscotch \\\\\n&\\quad = \\int_0^{1/2} butterscotch^2 \\left( \\int_0^{butterscotch} starlight(drumsticks) - \\int_0^{butterscotch} seashells\\left( \\frac{1}{2} - drumsticks \\right)\\,d drumsticks \\,d drumsticks \\right)\\,d butterscotch \\geq 0. \n\\end{align*}\n(This can also be interpreted as an instance of the \\emph{rearrangement inequality}.)\n\nWe now see that\n\\begin{align*}\nwanderlust_{marigolds} &\\geq 2\\int_0^{1/2} butterscotch starlight(butterscotch)\\,d butterscotch \\\\\n&\\quad \\geq 2 \\left( \\frac{3}{2} \\right)^{marigolds} \\int_0^{(1/2)(2/3)^{marigolds}} butterscotch\\,d butterscotch\\\\\n&\\quad = 2 \\left( \\frac{3}{2} \\right)^{marigolds} \\left. \\frac{1}{2} butterscotch^2 \\right|_0^{(1/2)(2/3)^{marigolds}} \\\\\n&\\quad = 2 \\left( \\frac{3}{2} \\right)^{marigolds} \\frac{1}{8} \\left( \\frac{2}{3} \\right)^{2marigolds}  = \\frac{1}{4} \\left( \\frac{2}{3} \\right)^{marigolds}\n\\end{align*}\nas desired.\n\n\n\n\\noindent\n\\textbf{Remark.}\nFor comparison, if we instead take the median of a list of $harmonica$ numbers, the probability distribution is given by\n\\[\nskateboard(butterscotch) = \\frac{(2harmonica+1)!}{harmonica!harmonica!} butterscotch^{harmonica} (1-butterscotch)^{harmonica}.\n\\]\nThe expected value of the absolute difference between $1/2$ and the median is \n\\[\n2 \\int_0^{1/2} (1/2 - butterscotch) skateboard(butterscotch) d butterscotch = 2^{-2harmonica-2}{{2harmonica+1}\\choose harmonica}.\n\\]\nFor $harmonica = 3^{2021}$, using Stirling's approximation this can be estimated as\n$1.13 (0.577)^{2021} < 0.25 (0.667)^{2021}$. This shows that the trimming procedure produces a quantity that is on average further away from 1/2 than the median.\n\n\\end{itemize}\n\\end{document}"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "knownpoint",
        "y": "finalval",
        "t": "steadystate",
        "N": "minisize",
        "k": "massiveindex",
        "n": "gigacount",
        "X": "consvalue",
        "X_k": "consstream",
        "f_k": "fixedfield",
        "f_k-1": "fixedfieldprev",
        "F_k": "voidcurve",
        "F_k-1": "voidcurveprev",
        "F_1": "voidcurveone",
        "g_k": "failfunc",
        "P_2n+1": "improbablepdf",
        "\\mu": "surprise"
      },
      "question": "Given an ordered list of $3minisize$ real numbers, we can \\emph{trim} it to form a list of $minisize$ numbers as follows: We divide the list into $minisize$ groups of $3$ consecutive numbers, and within each group, discard the highest and lowest numbers, keeping only the median.\n\nConsider generating a random number $consvalue$ by the following procedure: Start with a list of $3^{2021}$ numbers, drawn independently and uniformly at random between 0 and 1. Then trim this list as defined above, leaving a list of $3^{2020}$ numbers. Then trim again repeatedly until just one number remains; let $consvalue$ be this number. Let $surprise$ be the expected value of $|consvalue - \\frac{1}{2}|$. Show that\n\\[\nsurprise \\geq \\frac{1}{4} \\left( \\frac{2}{3} \\right)^{2021}.\n\\]",
      "solution": "\\noindent\\textbf{First solution.}\\newline\n(based on a suggestion of Noam Elkies)\nLet $fixedfield(knownpoint)$ be the probability distribution of $consstream$, the last number remaining when one repeatedly trims a list of $3^{massiveindex}$ random variables chosen with respect to the uniform distribution on $[0,1]$; note that $fixedfield_0(knownpoint)=1$ for $knownpoint\\in[0,1]$.  Let $voidcurve(knownpoint)=\\int_0^{knownpoint} fixedfield(steadystate)\\,dsteadystate$ be the cumulative distribution function; by symmetry, $voidcurve(\\frac12)=\\frac12$.  Let $surprise_{massiveindex}$ be the expected value of $consstream-\\frac12$; then $surprise_0=\\frac14$, so it will suffice to prove that $surprise_{massiveindex}\\ge\\frac23\\,surprise_{massiveindex-1}$ for $massiveindex>0$.\n\nBy integration by parts and symmetry,\n$$\nsurprise_{massiveindex}=2\\int_0^{1/2}\\!\\left(\\frac12-knownpoint\\right)fixedfield(knownpoint)\\,dknownpoint\n =2\\int_0^{1/2}\\!voidcurve(knownpoint)\\,dknownpoint,$$\nso $surprise_{massiveindex}$ is twice the area under $finalval=voidcurve(knownpoint)$ for $0\\le knownpoint\\le\\frac12$.  Transposing the axes yields\n$$\nsurprise_{massiveindex}=2\\int_0^{1/2}\\!\\left(\\frac12-voidcurve^{-1}(finalval)\\right)\\,dfinalval.$$\n\nBecause $fixedfield(knownpoint)$ is the density of the median of three independent variables having density $fixedfieldprev$, one has\n$$\nfixedfield(knownpoint)=6\\,fixedfieldprev(knownpoint)\\,voidcurveprev(knownpoint)\\bigl(1-voidcurveprev(knownpoint)\\bigr),\n$$\nand hence\n$$\nvoidcurve(knownpoint)=3\\,voidcurveprev(knownpoint)^2-2\\,voidcurveprev(knownpoint)^3.\n$$\nIterating gives $voidcurve=voidcurveprev\\circ voidcurveone$ with $voidcurveone(knownpoint)=3knownpoint^2-2knownpoint^3$.  From $fixedfield_1(steadystate)=6steadystate(1-steadystate)\\le\\tfrac32$ for $0\\le steadystate\\le\\tfrac12$ we deduce\n$$\n\\frac12-voidcurve^{-1}(finalval)\\ge\\frac23\\left(\\frac12-voidcurveprev^{-1}(finalval)\\right).\n$$\nIntegrating, we arrive at $surprise_{massiveindex}\\ge\\tfrac23\\,surprise_{massiveindex-1}$, completing the induction.\n\n\\medskip\\noindent\\textbf{Second solution.}\\newline\nRetain the preceding notation.  Again $voidcurve(\\tfrac12)=\\tfrac12$, so\n$fixedfield(\\tfrac12)=6\\,fixedfieldprev(\\tfrac12)\\times\\tfrac12\\times\\tfrac12$, whence $fixedfield(\\tfrac12)=(\\tfrac32)^{massiveindex}$ and $fixedfield$ is non-decreasing on $[0,\\tfrac12]$.  Setting\n$$\nfailfunc(knownpoint)=\\begin{cases}(\\tfrac32)^{massiveindex},&0\\le knownpoint\\le\\tfrac12\\,(\\tfrac23)^{massiveindex},\\\\[4pt]0,&\\text{otherwise},\\end{cases}\n$$\none checks (via a rearrangement-inequality argument) that\n$surprise_{massiveindex}\\ge2\\int_0^{1/2}knownpoint\\,failfunc(knownpoint)\\,dknownpoint=\n\\tfrac14\\,(\\tfrac23)^{massiveindex}$, proving the claim.\n\n\\medskip\\noindent\\textbf{Remark.}  For comparison, if one instead takes the median of $2gigacount+1$ independent $\\mathrm U(0,1)$ variables, the density is $improbablepdf(knownpoint)=\\dfrac{(2gigacount+1)!}{gigacount!\\,gigacount!}\\,knownpoint^{gigacount}(1-knownpoint)^{gigacount}$ and the mean absolute deviation from $\\frac12$ equals $2^{-2gigacount-2}\\binom{2gigacount+1}{gigacount}$, which for $gigacount=3^{2021}$ is strictly smaller than the lower bound obtained above."
    },
    "garbled_string": {
      "map": {
        "x": "hqkdmvcz",
        "y": "ptzrsnbe",
        "t": "vmscljya",
        "N": "qbrxlepd",
        "k": "zwfntoyg",
        "n": "lhuvqrje",
        "X": "sdkyimra",
        "X_k": "cplfzwxh",
        "f_k": "djrqpeos",
        "f_k-1": "gobxtram",
        "F_k": "ujnyshcv",
        "F_k-1": "xltpmeqa",
        "F_1": "bkrvsoid",
        "g_k": "ymnadzwe",
        "P_2n+1": "aovfrkji",
        "\\mu": "rsebigwa"
      },
      "question": "Given an ordered list of $3qbrxlepd$ real numbers, we can \\emph{trim} it to form a list of $qbrxlepd$ numbers as follows: We divide the list into $qbrxlepd$ groups of $3$ consecutive numbers, and within each group, discard the highest and lowest numbers, keeping only the median.\n\nConsider generating a random number $sdkyimra$ by the following procedure: Start with a list of $3^{2021}$ numbers, drawn independently and uniformly at random between 0 and 1. Then trim this list as defined above, leaving a list of $3^{2020}$ numbers. Then trim again repeatedly until just one number remains; let $sdkyimra$ be this number. Let $rsebigwa$ be the expected value of $|sdkyimra - \\frac{1}{2}|$. Show that\n\\[\nrsebigwa \\geq \\frac{1}{4} \\left( \\frac{2}{3} \\right)^{2021}.\n\\]",
      "solution": "\\noindent\n\\textbf{First solution.}\n(based on a suggestion of Noam Elkies)\nLet $djrqpeos(hqkdmvcz)$ be the probability distribution of $cplfzwxh$, the last number remaining when one repeatedly trims a list of $3^{zwfntoyg}$ random variables chosen with respect to the uniform distribution on $[0,1]$; note that $f_0(hqkdmvcz) = 1$ for $hqkdmvcz \\in [0,1]$.\nLet $ujnyshcv(hqkdmvcz)=\\int_0^{hqkdmvcz} djrqpeos(vmscljya)\\,dvmscljya$ be the cumulative distribution function; by symmetry,\n$ujnyshcv(\\frac{1}{2}) = \\frac{1}{2}$.\nLet $rsebigwa_{zwfntoyg}$ be the expected value of $cplfzwxh - \\frac{1}{2}$; then $rsebigwa_{0} = \\frac{1}{4}$, so it will suffice to prove that $rsebigwa_{zwfntoyg} \\geq \\frac{2}{3} \\, rsebigwa_{zwfntoyg-1}$ for $zwfntoyg > 0$.\n\nBy integration by parts and symmetry, we have\n\\[\nrsebigwa_{zwfntoyg} = 2 \\int_0^{1/2} \\left( \\frac{1}{2} - hqkdmvcz \\right) djrqpeos(hqkdmvcz)\\,dhqkdmvcz  = 2 \\int_0^{1/2} ujnyshcv(hqkdmvcz)\\,dhqkdmvcz;\n\\]\nthat is, $rsebigwa_{zwfntoyg}$ computes twice the area under the curve $ptzrsnbe = ujnyshcv(hqkdmvcz)$ for $0 \\leq hqkdmvcz \\leq\\frac{1}{2}$. Since $ujnyshcv$ is a monotone function from $[0, \\frac{1}{2}]$ \nwith $ujnyshcv(0) = 0$ and $ujnyshcv(\\frac{1}{2}) = \\frac{1}{2}$, we may transpose the axes to obtain\n\\begin{equation} \\label{eq:2021B6 eq4}\nrsebigwa_{zwfntoyg} = 2 \\int_0^{1/2} \\left( \\frac{1}{2} - ujnyshcv^{-1}(ptzrsnbe) \\right)\\,dptzrsnbe.\n\\end{equation}\n\nSince $djrqpeos(hqkdmvcz)$ is the probability distribution of the median of three random variables chosen with respect to the distribution $gobxtram(hqkdmvcz)$,\n\\begin{equation} \\label{eq:2021B6 eq1}\ndjrqpeos(hqkdmvcz) = 6 \\, gobxtram(hqkdmvcz) \\, xltpmeqa(hqkdmvcz) \\, \\bigl( 1- xltpmeqa(hqkdmvcz) \\bigr)\n\\end{equation}\nor equivalently\n\\begin{equation} \\label{eq:2021B6 eq2}\nujnyshcv(hqkdmvcz) = 3 \\, xltpmeqa(hqkdmvcz)^2 - 2 \\, xltpmeqa(hqkdmvcz)^3.\n\\end{equation}\nBy induction, $ujnyshcv$ is the $zwfntoyg$-th iterate of $bkrvsoid(hqkdmvcz) = 3 hqkdmvcz^2 -2 hqkdmvcz^3$, so\n\\begin{equation} \\label{eq:2021B6 eq5}\nujnyshcv(hqkdmvcz) = xltpmeqa\\bigl(bkrvsoid(hqkdmvcz)\\bigr).\n\\end{equation}\nSince $djrqpeos(vmscljya) = 6 vmscljya(1-vmscljya) \\leq \\frac{3}{2}$ for $vmscljya \\in [0,\\frac{1}{2}]$,\n\\[\n\\frac{1}{2} - bkrvsoid(hqkdmvcz) = \\int_{hqkdmvcz}^{1/2} 6 vmscljya(1-vmscljya)\\,dvmscljya \\leq \\frac{3}{2}\\left(\\frac{1}{2}-hqkdmvcz\\right);\n\\]\nfor $ptzrsnbe \\in [0, \\frac{1}{2}]$, we may take $hqkdmvcz = ujnyshcv^{-1}(ptzrsnbe)$ to obtain\n\\begin{equation} \\label{eq:2021B6 eq3}\n\\frac{1}{2} - ujnyshcv^{-1}(ptzrsnbe) \\geq \\frac{2}{3} \\left( \\frac{1}{2} - xltpmeqa^{-1}(ptzrsnbe) \\right).\n\\end{equation}\nUsing \\eqref{eq:2021B6 eq5} and \\eqref{eq:2021B6 eq3}, we obtain\n\\begin{align*}\nrsebigwa_{zwfntoyg} &= 2 \\int_0^{1/2} \\left( \\frac{1}{2} - ujnyshcv^{-1}(ptzrsnbe) \\right) \\,dptzrsnbe \\\\\n&\\geq \\frac{4}{3} \\int_0^{1/2} \\left( \\frac{1}{2} - xltpmeqa^{-1}(ptzrsnbe) \\right) \\,dptzrsnbe = \\frac{2}{3}\\, rsebigwa_{zwfntoyg-1}\n\\end{align*}\nas desired.\n\n\\noindent\n\\textbf{Second solution.}\nRetain notation as in the first solution. Again $ujnyshcv(\\frac{1}{2}) = \\frac{1}{2}$, so \\eqref{eq:2021B6 eq1} implies\n\\[\ndjrqpeos\\left( \\frac{1}{2} \\right) = 6 \\, gobxtram \\left( \\frac{1}{2} \\right) \\times \\frac{1}{2} \\times \\frac{1}{2}.\n\\]\nBy induction on $zwfntoyg$, we deduce that %$djrqpeos(hqkdmvcz)$ is a polynomial in $hqkdmvcz$,\n$djrqpeos(\\frac{1}{2}) = \\left(\\frac{3}{2}\\right)^{zwfntoyg}$\nand $djrqpeos(hqkdmvcz)$ is nondecreasing on $[0,\\frac{1}{2}]$.\n(More precisely, besides \\eqref{eq:2021B6 eq1}, the second assertion uses that $xltpmeqa(hqkdmvcz)$ increases from $0$ to $1/2$\nand $ptzrsnbe \\mapsto ptzrsnbe - ptzrsnbe^2$ is nondecreasing on $[0, 1/2]$.)\n\nThe expected value of $|cplfzwxh-\\frac{1}{2}|$ equals\n\\begin{align*}\nrsebigwa_{zwfntoyg} &= 2 \\int_0^{1/2} \\left( \\frac{1}{2} - hqkdmvcz \\right) djrqpeos(hqkdmvcz)\\,dhqkdmvcz \\\\\n&= 2 \\int_0^{1/2} hqkdmvcz \\, djrqpeos\\left( \\frac{1}{2} - hqkdmvcz \\right)\\,dhqkdmvcz.% \\\\\n%&= \\int_0^{1/2} \\left( \\frac{1}{2} - ujnyshcv\\left( \\frac{1}{2} - hqkdmvcz \\right)\\right)\\,dhqkdmvcz \\\\\n\\end{align*}\n%where the last step is integration by parts. Define the function\nDefine the function\n\\[\nymnadzwe(hqkdmvcz) = \\begin{cases} \\left( \\frac{3}{2} \\right)^{zwfntoyg} & hqkdmvcz \\in \\left[ 0, \\frac{1}{2} \\left( \\frac{2}{3} \\right)^{zwfntoyg} \\right] \\\\ 0 & \\mbox{otherwise}.\\end{cases}\n\\]\nNote that for $hqkdmvcz \\in [0, 1/2]$ we have\n\\[\n\\int_0^{hqkdmvcz} (ymnadzwe(vmscljya) - djrqpeos(1/2-vmscljya))\\,dvmscljya \\geq 0\n\\]\nwith equality at $hqkdmvcz=0$ or $hqkdmvcz=1/2$. (On the interval $\\left[0, (1/2)(2/3)^{zwfntoyg}\\right]$ the integrand is nonnegative, so the function increases from 0; on the interval $\\left[(1/2)(2/3)^{zwfntoyg}, 1/2\\right]$ the integrand is nonpositive, so the function decreases to 0.)\nHence by integration by parts,\n\\begin{align*}\n&rsebigwa_{zwfntoyg} - 2 \\int_0^{1/2} hqkdmvcz \\, ymnadzwe(hqkdmvcz) \\,dhqkdmvcz \\\\\n&\\quad = \\int_0^{1/2} 2 hqkdmvcz \\bigl(djrqpeos\\left( \\frac{1}{2} - hqkdmvcz \\right) - ymnadzwe(hqkdmvcz)\\bigr) \\,dhqkdmvcz \\\\\n&\\quad = \\int_0^{1/2} hqkdmvcz^2 \\left( \\int_0^{hqkdmvcz} ymnadzwe(vmscljya) - \\int_0^{hqkdmvcz} djrqpeos\\left( \\frac{1}{2} - vmscljya \\right)\\,dvmscljya \\right)\\,dhqkdmvcz \\geq 0. \n\\end{align*}\n(This can also be interpreted as an instance of the \\emph{rearrangement inequality}.)\n\nWe now see that\n\\begin{align*}\nrsebigwa_{zwfntoyg} &\\geq 2\\int_0^{1/2} hqkdmvcz \\, ymnadzwe(hqkdmvcz)\\,dhqkdmvcz \\\\\n&\\quad \\geq 2 \\left( \\frac{3}{2} \\right)^{zwfntoyg} \\int_0^{(1/2)(2/3)^{zwfntoyg}} hqkdmvcz\\,dhqkdmvcz\\\\\n&\\quad = 2 \\left( \\frac{3}{2} \\right)^{zwfntoyg} \\left. \\frac{1}{2} hqkdmvcz^2 \\right|_0^{(1/2)(2/3)^{zwfntoyg}} \\\\\n&\\quad = 2 \\left( \\frac{3}{2} \\right)^{zwfntoyg} \\frac{1}{8} \\left( \\frac{2}{3} \\right)^{2zwfntoyg}  = \\frac{1}{4} \\left( \\frac{2}{3} \\right)^{zwfntoyg}\n\\end{align*}\nas desired.\n\n\n\n\\noindent\n\\textbf{Remark.}\nFor comparison, if we instead take the median of a list of $lhuvqrje$ numbers, the probability distribution is given by\n\\[\naovfrkji(hqkdmvcz) = \\frac{(2lhuvqrje+1)!}{lhuvqrje!lhuvqrje!} hqkdmvcz^{lhuvqrje} (1-hqkdmvcz)^{lhuvqrje}.\n\\]\nThe expected value of the absolute difference between $1/2$ and the median is \n\\[\n2 \\int_0^{1/2} \\left(\\frac{1}{2} - hqkdmvcz\\right) aovfrkji(hqkdmvcz) \\, dhqkdmvcz = 2^{-2lhuvqrje-2}{{2lhuvqrje+1}\\choose lhuvqrje}.\n\\]\nFor $lhuvqrje = 3^{2021}$, using Stirling's approximation this can be estimated as\n$1.13 (0.577)^{2021} < 0.25 (0.667)^{2021}$. This shows that the trimming procedure produces a quantity that is on average further away from 1/2 than the median."
    },
    "kernel_variant": {
      "question": "Let k = 2023.  Start with a list of 5^{k} independent real numbers, each chosen uniformly at random from the interval [0,\\tfrac12].  Divide the list into consecutive blocks of five and, within every block, discard the two largest and the two smallest entries, keeping only the median (the third–smallest).  The selected medians form a new list of 5^{k-1} numbers.  Repeat the same “trim-by-five’’ operation on this new list and continue for a total of k trimming stages, until a single number X remains.\n\nSet\n\\[\\mu\\;=\\;\\mathbb E\\bigl\\lvert X-\\tfrac14\\bigr\\rvert .\\]\nProve that\n\\[\\boxed{\\displaystyle \\mu\\;\\ge\\;\\frac18\\Bigl(\\frac4{15}\\Bigr)^{2023}}.\\]",
      "solution": "Throughout we write k∈{0,1,2,…} for an arbitrary number of trimming steps and substitute k = 2023 only in the last line.\n\n1.  Centring the variables.\n   After j trimming steps (0≤j≤k) denote the surviving list by (X_{j,i})_{1≤i≤5^{k-j}} and write simply X_j for an arbitrary element of this list.  Put\n   \\[\n      Y_j := X_j-\\frac14, \\qquad j=0,1,\\dots ,k.\n   \\]\n   Hence Y_0 is uniform on [−\\tfrac14,\\tfrac14]; for j≥1, Y_j equals the median of five independent copies of Y_{j-1}.  Let f_j and F_j be respectively the density and cdf of Y_j, and set\n   \\[\n      \\mu_j := \\mathbb E|Y_j| = 2\\int_0^{\\infty}\\bigl(1-F_j(t)\\bigr)\\,dt \\qquad (\\mu_0 = \\tfrac18).\n   \\]\n   We shall show\n   \\[\n      \\mu_j \\;\\ge\\; \\frac18\\Bigl(\\frac4{15}\\Bigr)^j \\quad (\\forall j\\ge0),\\tag{⋆}\n   \\]\n   which for j = k = 2023 yields the desired bound for \\mu = \\mu_{2023}.\n\n2.  The median–of–five recursion and the density at the origin.\n   For any distribution function H write\n   \\[G(H) = 10H^3(1-H)^2 + 5H^4(1-H) + H^5\\]\n   — the cdf of the median of five i.i.d. variables with cdf H.  Consequently\n   \\[\n      F_j(x) = G\\bigl(F_{j-1}(x)\\bigr), \\qquad\n      f_j(x) = g\\bigl(F_{j-1}(x)\\bigr)\\,f_{j-1}(x), \\qquad\n      g(y):=30y^{2}(1-y)^{2}.\\tag{1}\n   \\]\n   Because every Y_j is symmetric, F_{j-1}(0)=\\tfrac12.  Since f_0(0)=2, an induction using (1) gives\n   \\[\n      f_j(0) = g(\\tfrac12) f_{j-1}(0) = \\frac{15}{8} f_{j-1}(0)\n              = 2\\Bigl(\\frac{15}{8}\\Bigr)^{j}.\\tag{2}\n   \\]\n\n3.  A uniform upper bound for the density.\n   We claim that for every j≥0\n   \\[\n      f_j(x) \\;\\le\\; f_j(0) \\qquad(\\forall x\\in\\mathbb R).\\tag{3}\n   \\]\n   Proof by induction.  For j=0 the density is the constant 2 on [−\\tfrac14,\\tfrac14], so (3) is clear.  Assume (3) for j−1.  Because F_{j-1}(x)≥\\tfrac12 for x≥0 and g is decreasing on [\\tfrac12,1], we obtain g(F_{j-1}(x)) ≤ g(\\tfrac12)=\\tfrac{15}{8}.  Therefore, using (1) and the induction hypothesis,\n   \\[\n      f_j(x)=g(F_{j-1}(x))\\,f_{j-1}(x)\\le\\frac{15}{8} f_{j-1}(0)=f_j(0),\\qquad x\\ge0.\n   \\]\n   Symmetry gives the same bound for x≤0, establishing (3).\n\n4.  A neighbourhood where the survival function is large.\n   Fix j≥0 and put\n   \\[h_j:=f_j(0)=2\\Bigl(\\tfrac{15}{8}\\Bigr)^{j},\\qquad \\delta_j:=\\frac1{4h_j}.\\]\n   For t≥0, (3) implies\n   \\[F_j(t)-\\tfrac12 = \\int_0^{t}f_j(s)ds \\le h_j t.\\]\n   Hence for 0≤t≤δ_j,\n   \\[\n      F_j(t) \\le \\tfrac12 + h_j\\delta_j = \\tfrac12 + \\tfrac14 = \\tfrac34, \\quad\\text{i.e. } 1-F_j(t)\\ge\\tfrac14.\\tag{4}\n   \\]\n\n5.  Lower–bounding the first moment.\n   Using (4) we obtain\n   \\[\n      \\mu_j = 2\\int_0^{\\infty}(1-F_j(t))\\,dt \\;\\ge\\; 2\\int_0^{\\delta_j}\\frac14\\,dt = \\frac{\\delta_j}{2} = \\frac1{8h_j}.\n   \\]\n   Substituting h_j from (2) yields\n   \\[\n      \\mu_j \\;\\ge\\; \\frac1{16}\\Bigl(\\frac{8}{15}\\Bigr)^{j}.\\tag{5}\n   \\]\n   Now observe that for j≥1 we have 2^{j-1}≥1, and\n   \\[\n      \\frac1{16}\\Bigl(\\frac{8}{15}\\Bigr)^{j} = \\frac18\\,2^{j-1}\\Bigl(\\frac4{15}\\Bigr)^{j} \\ge \\frac18\\Bigl(\\frac4{15}\\Bigr)^{j}.\n   \\]\n   For j=0 inequality (⋆) is immediate because \\mu_0 = \\tfrac18 = \\tfrac18(\\tfrac4{15})^{0}.  Combining the two cases proves (⋆) for all j≥0.\n\n6.  Finally, with j=k=2023 we have\n   \\[\\mu = \\mu_{2023} \\;\\ge\\; \\frac18\\Bigl(\\frac4{15}\\Bigr)^{2023},\\]\n   exactly as claimed.\n\n\\medskip\nRemark.  The only property of the densities used in Step 4 is the pointwise bound (3); no monotonicity of f_j on (0,∞) is required, so the argument avoids the pitfall noted in the original draft.",
      "_meta": {
        "core_steps": [
          "Describe the kth trimming result by its pdf f_k and cdf F_k (with F_0(x)=x for the uniform start).",
          "Use the median-of-three formula to get the recursion  F_k(x)=3F_{k-1}(x)^2-2F_{k-1}(x)^3.",
          "Rewrite the target mean as  μ_k = 2∫_0^{1/2}(1/2-F_k^{-1}(y))dy (area under the inverse–cdf).",
          "Bound f_1 on [0,1/2] by a constant (3/2), giving (1/2-F_k^{-1}) ≥ (2/3)(1/2-F_{k-1}^{-1}).",
          "Integrate to obtain μ_k ≥ (2/3)μ_{k-1}; iterate from μ_0=1/4 to reach μ=μ_{2021} ≥ 1/4·(2/3)^{2021}."
        ],
        "mutable_slots": {
          "slot1": {
            "description": "number of trimming stages (any positive integer k)",
            "original": 2021
          },
          "slot2": {
            "description": "initial list length, namely 3^k so that the list can be trimmed k times",
            "original": "3^{2021}"
          },
          "slot3": {
            "description": "size of each block whose median is kept; determines the polynomial in step 2",
            "original": 3
          },
          "slot4": {
            "description": "end-points of the starting uniform distribution; scaling them rescales every μ_k uniformly",
            "original": "[0,1]"
          },
          "slot5": {
            "description": "symmetry point about which deviation is measured",
            "original": "1/2"
          },
          "slot6": {
            "description": "max-pdf constant on [0,1/2] that yields the contraction; its reciprocal is the factor in step 4",
            "original": "3/2 (hence contraction factor 2/3)"
          },
          "slot7": {
            "description": "initial expected deviation μ_0 coming from the uniform start",
            "original": "1/4"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}