{
  "index": "1989-B-6",
  "type": "COMB",
  "tag": [
    "COMB",
    "ANA"
  ],
  "difficulty": "",
  "question": "Let $(x_1,\\,x_2,\\,\\ldots\\,x_n)$ be a point chosen at random from the\n$n$-dimensional region defined by\n$0<x_1<x_2<\\cdots < x_n<1.$  Let $f$ be a continuous function on\n$[0,1]$ with $f(1)=0$.\nSet $x_0=0$ and $x_{n+1}=1$. Show that the expected value of the\nRiemann sum\n\\[\n\\sum_{i=0}^n (x_{i+1}-x_i) f(x_{i+1})\n\\]\nis $\\int_0^1 f(t)P(t)\\, dt$, where $P$ is a polynomial of degree $n$,\nindependent of $f$, with $0\\le P(t)\\le 1$ for $0\\le t \\le 1$.\n\n\\end{itemize}\n\\end{document}",
  "solution": "Solution 1. We may drop the \\( i=n \\) term in the Riemann sum, since \\( f\\left(x_{n+1}\\right)= \\) \\( f(1)=0 \\). The volume of the region \\( 0<x_{1}<\\cdots<x_{n}<1 \\) in \\( \\mathbb{R}^{n} \\) is\n\\[\n\\int_{0}^{1} \\int_{0}^{x_{n}} \\cdots \\int_{0}^{x_{2}} d x_{1} d x_{2} \\cdots d x_{n}=1 / n!\n\\]\n\nTherefore the expected value of the Riemann sum is \\( E=\\left(\\sum_{i=0}^{n-1} M_{i}\\right) /(1 / n!) \\), where\n\\[\n\\begin{aligned}\nM_{i} & =\\int_{0}^{1} \\int_{0}^{x_{n}} \\cdots \\int_{0}^{x_{2}}\\left(x_{i+1}-x_{i}\\right) f\\left(x_{i+1}\\right) d x_{1} \\cdots d x_{n} \\\\\n& =\\int_{0}^{1} \\int_{0}^{x_{n}} \\cdots \\int_{0}^{x_{i+1}}\\left(x_{i+1}-x_{i}\\right) \\frac{x_{i}^{i-1}}{(i-1)!} f\\left(x_{i+1}\\right) d x_{i} d x_{i+1} \\cdots d x_{n} \\\\\n& =\\int_{0}^{1} \\int_{0}^{x_{n}} \\cdots \\int_{0}^{x_{i+2}}\\left(x_{i+1} \\frac{x_{i+1}^{i}}{i!}-\\frac{x_{i+1}^{i+1}}{(i+1) \\cdot(i-1)!}\\right) f\\left(x_{i+1}\\right) d x_{i+1} \\cdots d x_{n} \\\\\n& =\\int_{0}^{1} \\int_{0}^{x_{n}} \\cdots \\int_{0}^{x_{i+2}} \\frac{x_{i+1}^{i+1}}{(i+1)!} f\\left(x_{i+1}\\right) d x_{i+1} \\cdots d x_{n} \\\\\n& =\\int_{0}^{1} \\int_{x_{i+1}}^{1} \\int_{x_{i+1}}^{x_{n}} \\int_{x_{i+1}}^{x_{i+3}} \\frac{x_{i+1}^{i+1}}{(i+1)!} f\\left(x_{i+1}\\right) d x_{i+2} \\cdots d x_{n} d x_{i+1} \\\\\n& =\\int_{0}^{1} \\frac{\\left(1-x_{i+1}\\right)^{n-(i+1)}}{(n-(i+1))!} \\frac{x_{i+1}^{i+1}}{(i+1)!} f\\left(x_{i+1}\\right) d x_{i+1} \\\\\n& =\\frac{1}{n!} \\int_{0}^{1}\\binom{n}{i+1}\\left(1-x_{i+1}\\right)^{n-(i+1)} x_{i+1}^{i+1} f\\left(x_{i+1}\\right) d x_{i+1} .\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\begin{aligned}\nE & =\\int_{0}^{1}\\left(\\sum_{i=0}^{n-1}\\binom{n}{i+1}\\left(1-x_{i+1}\\right)^{n-(i+1)} x_{i+1}^{i+1}\\right) f\\left(x_{i+1}\\right) d x_{i+1} \\\\\n& =\\int_{0}^{1}\\left(\\sum_{j=1}^{n}\\binom{n}{j}(1-t)^{n-j} t^{j}\\right) f(t) d t \\quad\\left(\\text { we set } j=i+1 \\text { and } t=x_{i+1}\\right) \\\\\n& =\\int_{0}^{1}\\left(1-(1-t)^{n}\\right) f(t) d t\n\\end{aligned}\n\\]\nsince the sum is the binomial expansion of \\( ((1-t)+t)^{n} \\) except that the \\( j=0 \\) term, which is \\( (1-t)^{n} \\), is missing. Hence \\( E=\\int_{0}^{1} f(t) P(t) d t \\) where \\( P(t)=1-(1-t)^{n} \\). Clearly \\( 0 \\leq P(t) \\leq 1 \\) for \\( 0 \\leq t \\leq 1 \\).\n\nSolution 2. Fix \\( n \\geq 0 \\) and let \\( \\left(y_{1}, \\ldots, y_{n}\\right) \\) be chosen uniformly from \\( [0,1]^{n} \\). Fix \\( t \\in[0,1] \\), and let \\( E_{n}(t) \\) denote the expected value of\n\\[\nR_{n}=t-\\max \\left(\\{0\\} \\cup\\left(\\left\\{y_{1}, \\ldots, y_{n}\\right\\} \\cap[0, t]\\right)\\right) .\n\\]\n\nLet \\( \\left(x_{1}, \\ldots, x_{n}\\right) \\) be the permutation of \\( \\left(y_{1}, \\ldots, y_{n}\\right) \\) such that \\( x_{1} \\leq \\cdots \\leq x_{n} \\). The distribution of \\( \\left(x_{1}, \\ldots, x_{n}\\right) \\) equals the distribution in the problem. Define\n\\[\nS=\\sum_{i=0}^{n-1}\\left(y_{i+1}-z_{i+1}\\right) f\\left(y_{i+1}\\right)\n\\]\nwhere \\( z_{i+1} \\) is the largest \\( y_{j} \\) less than \\( y_{i+1} \\), or 0 if no such \\( y_{j} \\) exists. The terms in \\( S \\) are a permutation of those in the original Riemann sum (ignoring the final term in the Riemann sum, which is zero), so the expected values of the Riemann sum and of \\( S \\) are equal. Each term in \\( S \\) has expected value \\( \\int_{0}^{1} E_{n-1}(t) f(t) d t \\), since the expected value of \\( y_{i+1}-z_{i+1} \\) conditioned on the event \\( y_{i+1}=t \\) for some \\( t \\in[0,1] \\) is the definition of \\( E_{n-1}(t) \\) using \\( \\left(y_{j}\\right)_{j \\neq i+1} \\in[0,1]^{n-1} \\). Therefore the expected value of the Riemann sum is \\( n \\int_{0}^{1} E_{n-1}(t) f(t) d t \\).\n\nIt remains to determine \\( E_{n}(t) \\) for \\( n \\geq 0 \\). Let \\( y_{n+1} \\) be chosen uniformly in \\( [0,1] \\), independently of the \\( y_{1}, \\ldots, y_{n} \\) in the definition of \\( E_{n}(t) \\). Then \\( E_{n}(t) \\) equals the probability that \\( y_{n+1} \\) is in \\( [0, t] \\) and is closer to \\( t \\) than any other \\( y_{j} \\), since this probability conditioned on a choice of \\( y_{1}, \\ldots, y_{n} \\) equals \\( R_{n} \\). On the other hand, this probability equals \\( \\left(1-(1-t)^{n+1}\\right) /(n+1) \\), since the probability that at least one of \\( y_{1}, \\ldots, y_{n+1} \\) lies in \\( [0, t] \\) equals \\( 1-(1-t)^{n+1} \\), and conditioned on this, the probability that the \\( y_{j} \\) in \\( [0, t] \\) closest to \\( t \\) is \\( y_{n+1} \\) is \\( 1 /(n+1) \\), since all possible indices for this closest \\( y_{j} \\) are equally likely. Thus \\( E_{n}(t)=\\left(1-(1-t)^{n+1}\\right) /(n+1) \\), and we may take \\( P(t)=n E_{n-1}(t)=1-(1-t)^{n} \\).\n\nLiterature note. For more on Riemann sums, see [Spv, Ch. 13, Appendix 1].",
  "vars": [
    "x_0",
    "x_1",
    "x_2",
    "x_n",
    "x_i",
    "x_i+1",
    "x_i+2",
    "x_i+3",
    "x_n+1",
    "t",
    "i",
    "j",
    "y_1",
    "y_n",
    "y_i+1",
    "y_n+1",
    "y_j",
    "z_i+1",
    "R_n",
    "E_n",
    "E_n-1",
    "M_i",
    "S"
  ],
  "params": [
    "n",
    "f",
    "P",
    "E"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x_0": "initialpoint",
        "x_1": "firstpoint",
        "x_2": "secondpoint",
        "x_n": "nthpoint",
        "x_i": "indexpoint",
        "x_{i+1}": "nextpoint",
        "x_{i+2}": "secondnextpoint",
        "x_{i+3}": "thirdnextpoint",
        "x_{n+1}": "finalpoint",
        "t": "timevalue",
        "j": "altindex",
        "y_1": "ysampleone",
        "y_n": "ysamplen",
        "y_{i+1}": "ysamplenext",
        "y_{n+1}": "ysampleextra",
        "y_{j}": "ysamplej",
        "z_{i+1}": "zprev",
        "R_{n}": "riemannrand",
        "E_{n}": "expectn",
        "E_{n-1}": "expectprev",
        "M_{i}": "midterm",
        "S": "sumall",
        "n": "dimcount",
        "f": "contifunc",
        "P": "polyfunc",
        "E": "expectval"
      },
      "question": "Let $(firstpoint,\\,secondpoint,\\,\\ldots\\,nthpoint)$ be a point chosen at random from the\n$dimcount$-dimensional region defined by\n$0<firstpoint<secondpoint<\\cdots < nthpoint<1.$  Let $contifunc$ be a continuous function on\n$[0,1]$ with $contifunc(1)=0$.\nSet $initialpoint=0$ and $finalpoint=1$. Show that the expected value of the\nRiemann sum\n\\[\n\\sum_{i=0}^{dimcount} (nextpoint-indexpoint)\\,contifunc(nextpoint)\n\\]\nis $\\int_0^1 contifunc(timevalue)\\,polyfunc(timevalue)\\, d timevalue$, where $polyfunc$ is a polynomial of degree $dimcount$,\nindependent of $contifunc$, with $0\\le polyfunc(timevalue)\\le 1$ for $0\\le timevalue \\le 1$.",
      "solution": "Solution 1. We may drop the \\( i=dimcount \\) term in the Riemann sum, since \\( contifunc\\left(finalpoint\\right)=contifunc(1)=0 \\). The volume of the region \\( 0<firstpoint<\\cdots<nthpoint<1 \\) in \\( \\mathbb{R}^{dimcount} \\) is\n\\[\n\\int_{0}^{1} \\int_{0}^{nthpoint} \\cdots \\int_{0}^{secondpoint}\\, d firstpoint\\, d secondpoint \\cdots d nthpoint = 1/ dimcount!.\n\\]\nTherefore the expected value of the Riemann sum is\n\\[\nexpectval = \\frac{\\displaystyle\\sum_{i=0}^{dimcount-1} midterm}{1/ dimcount!},\n\\]\nwhere\n\\[\n\\begin{aligned}\nmidterm &= \\int_{0}^{1} \\int_{0}^{nthpoint} \\cdots \\int_{0}^{secondpoint}\\bigl(nextpoint-indexpoint\\bigr)\\,contifunc\\bigl(nextpoint\\bigr)\\, d firstpoint \\cdots d nthpoint\\\\\n&= \\int_{0}^{1} \\int_{0}^{nthpoint} \\cdots \\int_{0}^{nextpoint}\\bigl(nextpoint-indexpoint\\bigr)\\frac{indexpoint^{i-1}}{(i-1)!}\\,contifunc(nextpoint)\\, d indexpoint\\, d nextpoint \\cdots d nthpoint\\\\\n&= \\int_{0}^{1} \\int_{0}^{nthpoint} \\cdots \\int_{0}^{secondnextpoint}\\Bigl(nextpoint\\frac{nextpoint^{i}}{i!}-\\frac{nextpoint^{i+1}}{(i+1)\\,(i-1)!}\\Bigr)\\,contifunc(nextpoint)\\, d nextpoint \\cdots d nthpoint\\\\\n&= \\int_{0}^{1} \\int_{0}^{nthpoint} \\cdots \\int_{0}^{secondnextpoint} \\frac{nextpoint^{i+1}}{(i+1)!}\\,contifunc(nextpoint)\\, d nextpoint \\cdots d nthpoint\\\\\n&= \\int_{0}^{1} \\int_{nextpoint}^{1} \\int_{nextpoint}^{nthpoint} \\int_{nextpoint}^{thirdnextpoint} \\frac{nextpoint^{i+1}}{(i+1)!}\\,contifunc(nextpoint)\\, d secondnextpoint \\cdots d nthpoint\\, d nextpoint\\\\\n&= \\int_{0}^{1} \\frac{(1-nextpoint)^{dimcount-(i+1)}}{(dimcount-(i+1))!}\\,\\frac{nextpoint^{i+1}}{(i+1)!}\\,contifunc(nextpoint)\\, d nextpoint\\\\\n&= \\frac1{dimcount!}\\int_{0}^{1}\\binom{dimcount}{i+1}(1-nextpoint)^{dimcount-(i+1)} nextpoint^{i+1}\\,contifunc(nextpoint)\\, d nextpoint.\n\\end{aligned}\n\\]\nHence\n\\[\n\\begin{aligned}\nexpectval &= \\int_{0}^{1}\\Bigl(\\sum_{i=0}^{dimcount-1}\\binom{dimcount}{i+1}(1-nextpoint)^{dimcount-(i+1)} nextpoint^{i+1}\\Bigr)\\,contifunc(nextpoint)\\, d nextpoint\\\\\n&= \\int_{0}^{1}\\Bigl(\\sum_{altindex=1}^{dimcount}\\binom{dimcount}{altindex}(1-timevalue)^{dimcount-altindex} timevalue^{altindex}\\Bigr)\\,contifunc(timevalue)\\, d timevalue\\\\\n&= \\int_{0}^{1}\\bigl(1-(1-timevalue)^{dimcount}\\bigr)\\,contifunc(timevalue)\\, d timevalue,\n\\end{aligned}\n\\]\nsince the sum is the binomial expansion of \\(((1-timevalue)+timevalue)^{dimcount}\\) with the \\(altindex=0\\) term \\((1-timevalue)^{dimcount}\\) omitted. Thus we have\n\\[\nexpectval = \\int_{0}^{1} contifunc(timevalue)\\,\\polyfunc(timevalue)\\, d timevalue,\\qquad \\text{where } \\polyfunc(timevalue)=1-(1-timevalue)^{dimcount}.\n\\]\nClearly $0\\le \\polyfunc(timevalue)\\le 1$ for $0\\le timevalue\\le1$.\n\nSolution 2. Fix $dimcount\\ge0$ and let $(ysampleone,\\ldots,ysamplen)$ be chosen uniformly from $[0,1]^{dimcount}$.  Fix $timevalue\\in[0,1]$, and let $expectn(timevalue)$ denote the expected value of\n\\[\nriemannrand = timevalue-\\max\\bigl(\\{0\\}\\cup\\bigl(\\{ysampleone,\\ldots,ysamplen\\}\\cap[0,timevalue]\\bigr)\\bigr).\n\\]\nLet $(firstpoint,\\ldots,nthpoint)$ be the permutation of $(ysampleone,\\ldots,ysamplen)$ with $firstpoint\\le\\cdots\\le nthpoint$.  This has the same distribution as in the problem.  Define\n\\[\nsumall = \\sum_{i=0}^{dimcount-1}(ysamplenext-zprev)\\,contifunc(ysamplenext),\n\\]\nwhere $zprev$ is the largest $y_{j}$ less than $ysamplenext$, or $0$ if none exists.  The terms in $\\sumall$ form a permutation of those in the original Riemann sum (the last term is zero), so the expected values of the two sums coincide.  Each term in $\\sumall$ has expected value $\\int_{0}^{1}expectprev(timevalue)\\,contifunc(timevalue)\\,d timevalue$, because, conditioning on $ysamplenext=timevalue$, the expectation of $ysamplenext-zprev$ is exactly $expectprev(timevalue)$ with the other $dimcount-1$ sample points.  Consequently the expected value of the Riemann sum is\n\\[\n dimcount\\int_{0}^{1}expectprev(timevalue)\\,contifunc(timevalue)\\,d timevalue.\n\\]\nIt remains to compute $expectn(timevalue)$ for $dimcount\\ge0$.  Let $ysampleextra$ be chosen uniformly from $[0,1]$, independently of the previous points.  Then $expectn(timevalue)$ equals the probability that $ysampleextra\\in[0,timevalue]$ and is closer to $timevalue$ than any other $y_{j}$.  This probability is\n\\[\n\\frac{1-(1-timevalue)^{dimcount+1}}{dimcount+1},\n\\]\nfor the numerator is the probability that at least one of the $dimcount+1$ points lies in $[0,timevalue]$, and, conditional on this, each of the $dimcount+1$ indices is equally likely to be the closest.  Hence\n\\[\nexpectn(timevalue)=\\frac{1-(1-timevalue)^{dimcount+1}}{dimcount+1},\\qquad \\text{so } \\polyfunc(timevalue)=dimcount\\,expectprev(timevalue)=1-(1-timevalue)^{dimcount}.\n\\]\nLiterature note.  For more on Riemann sums, see [Spv, Ch.\u0000A0 13, Appendix 1]."
    },
    "descriptive_long_confusing": {
      "map": {
        "x_0": "lighthouse",
        "x_1": "butterfly",
        "x_2": "watermelon",
        "x_n": "orangutang",
        "x_i": "chandelier",
        "x_i+1": "toothpick",
        "x_i+2": "campfire",
        "x_i+3": "snowflake",
        "x_n+1": "storyboard",
        "t": "peppermint",
        "i": "compassrose",
        "j": "goldfinch",
        "y_1": "rainstorm",
        "y_n": "tumbleweed",
        "y_i+1": "nightingale",
        "y_n+1": "starflower",
        "y_j": "honeycomb",
        "z_i+1": "marshmallow",
        "R_n": "arrowhead",
        "E_n": "seashell",
        "E_n-1": "sunflower",
        "M_i": "playhouse",
        "S": "moonstone",
        "n": "pinecone",
        "f": "riverbank",
        "P": "blacksmith",
        "E": "whirlwind"
      },
      "question": "Let $(butterfly,\\,watermelon,\\,\\ldots\\,orangutang)$ be a point chosen at random from the\n$pinecone$-dimensional region defined by\n$0<butterfly<watermelon<\\cdots < orangutang<1.$  Let $riverbank$ be a continuous function on\n$[0,1]$ with $riverbank(1)=0$.\nSet $lighthouse=0$ and $storyboard=1$. Show that the expected value of the\nRiemann sum\n\\[\n\\sum_{compassrose=0}^{pinecone} (toothpick-chandelier) riverbank(toothpick)\n\\]\nis $\\int_0^1 riverbank(peppermint)blacksmith(peppermint)\\, dpeppermint$, where $blacksmith$ is a polynomial of degree $pinecone$,\nindependent of $riverbank$, with $0\\le blacksmith(peppermint)\\le 1$ for $0\\le peppermint \\le 1$.",
      "solution": "Solution 1. We may drop the \\( compassrose=pinecone \\) term in the Riemann sum, since \\( riverbank\\left(storyboard\\right)= \\) \\( riverbank(1)=0 \\). The volume of the region \\( 0<butterfly<\\cdots<orangutang<1 \\) in \\( \\mathbb{R}^{pinecone} \\) is\n\\[\n\\int_{0}^{1} \\int_{0}^{orangutang} \\cdots \\int_{0}^{watermelon} d butterfly d watermelon \\cdots d orangutang=1 / pinecone!\n\\]\n\nTherefore the expected value of the Riemann sum is \\( whirlwind=\\left(\\sum_{compassrose=0}^{pinecone-1} playhouse\\right) /(1 / pinecone!) \\), where\n\\[\n\\begin{aligned}\nplayhouse & =\\int_{0}^{1} \\int_{0}^{orangutang} \\cdots \\int_{0}^{watermelon}\\left(toothpick-chandelier\\right) riverbank\\left(toothpick\\right) d butterfly \\cdots d orangutang \\\\\n& =\\int_{0}^{1} \\int_{0}^{orangutang} \\cdots \\int_{0}^{toothpick}\\left(toothpick-chandelier\\right) \\frac{chandelier^{compassrose-1}}{(compassrose-1)!} riverbank\\left(toothpick\\right) d chandelier d toothpick \\cdots d orangutang \\\\\n& =\\int_{0}^{1} \\int_{0}^{orangutang} \\cdots \\int_{0}^{campfire}\\left(toothpick \\frac{toothpick^{compassrose}}{compassrose!}-\\frac{toothpick^{compassrose+1}}{(compassrose+1) \\cdot(compassrose-1)!}\\right) riverbank\\left(toothpick\\right) d toothpick \\cdots d orangutang \\\\\n& =\\int_{0}^{1} \\int_{0}^{orangutang} \\cdots \\int_{0}^{campfire} \\frac{toothpick^{compassrose+1}}{(compassrose+1)!} riverbank\\left(toothpick\\right) d toothpick \\cdots d orangutang \\\\\n& =\\int_{0}^{1} \\int_{toothpick}^{1} \\int_{toothpick}^{orangutang} \\int_{toothpick}^{snowflake} \\frac{toothpick^{compassrose+1}}{(compassrose+1)!} riverbank\\left(toothpick\\right) d campfire \\cdots d orangutang d toothpick \\\\\n& =\\int_{0}^{1} \\frac{\\left(1-toothpick\\right)^{pinecone-(compassrose+1)}}{(pinecone-(compassrose+1))!} \\frac{toothpick^{compassrose+1}}{(compassrose+1)!} riverbank\\left(toothpick\\right) d toothpick \\\\\n& =\\frac{1}{pinecone!} \\int_{0}^{1}\\\\binom{pinecone}{compassrose+1}\\left(1-toothpick\\right)^{pinecone-(compassrose+1)} toothpick^{compassrose+1} riverbank\\left(toothpick\\right) d toothpick .\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\begin{aligned}\nwhirlwind & =\\int_{0}^{1}\\left(\\sum_{compassrose=0}^{pinecone-1}\\\\binom{pinecone}{compassrose+1}\\left(1-toothpick\\right)^{pinecone-(compassrose+1)} toothpick^{compassrose+1}\\right) riverbank\\left(toothpick\\right) d toothpick \\\\\n& =\\int_{0}^{1}\\left(\\sum_{goldfinch=1}^{pinecone}\\\\binom{pinecone}{goldfinch}(1-peppermint)^{pinecone-goldfinch} peppermint^{goldfinch}\\right) riverbank(peppermint) d peppermint \\quad\\left(\\text { we set } goldfinch=compassrose+1 \\text { and } peppermint=toothpick\\right) \\\\\n& =\\int_{0}^{1}\\left(1-(1-peppermint)^{pinecone}\\right) riverbank(peppermint) d peppermint\n\\end{aligned}\n\\]\nsince the sum is the binomial expansion of \\( ((1-peppermint)+peppermint)^{pinecone} \\) except that the \\( goldfinch=0 \\) term, which is \\( (1-peppermint)^{pinecone} \\), is missing. Hence \\( whirlwind=\\int_{0}^{1} riverbank(peppermint) blacksmith(peppermint) d peppermint \\) where \\( blacksmith(peppermint)=1-(1-peppermint)^{pinecone} \\). Clearly \\( 0 \\leq blacksmith(peppermint) \\leq 1 \\) for \\( 0 \\leq peppermint \\leq 1 \\).\n\nSolution 2. Fix \\( pinecone \\geq 0 \\) and let \\( \\left(rainstorm, \\ldots, tumbleweed\\right) \\) be chosen uniformly from \\( [0,1]^{pinecone} \\). Fix \\( peppermint \\in[0,1] \\), and let \\( seashell(peppermint) \\) denote the expected value of\n\\[\narrowhead=peppermint-\\max \\left(\\{0\\} \\cup\\left(\\left\\{rainstorm, \\ldots, tumbleweed\\right\\} \\cap[0, peppermint]\\right)\\right) .\n\\]\n\nLet \\( \\left(butterfly, \\ldots, orangutang\\right) \\) be the permutation of \\( \\left(rainstorm, \\ldots, tumbleweed\\right) \\) such that \\( butterfly \\leq \\cdots \\leq orangutang \\). The distribution of \\( \\left(butterfly, \\ldots, orangutang\\right) \\) equals the distribution in the problem. Define\n\\[\nmoonstone=\\sum_{compassrose=0}^{pinecone-1}\\left(nightingale-marshmallow\\right) riverbank\\left(nightingale\\right)\n\\]\nwhere \\( marshmallow \\) is the largest \\( honeycomb \\) less than \\( nightingale \\), or 0 if no such \\( honeycomb \\) exists. The terms in \\( moonstone \\) are a permutation of those in the original Riemann sum (ignoring the final term in the Riemann sum, which is zero), so the expected values of the Riemann sum and of \\( moonstone \\) are equal. Each term in \\( moonstone \\) has expected value \\( \\int_{0}^{1} sunflower(peppermint) riverbank(peppermint) d peppermint \\), since the expected value of \\( nightingale-marshmallow \\) conditioned on the event \\( nightingale=peppermint \\) for some \\( peppermint \\in[0,1] \\) is the definition of \\( sunflower(peppermint) \\) using \\( \\left(honeycomb\\right)_{goldfinch \\neq compassrose+1} \\in[0,1]^{pinecone-1} \\). Therefore the expected value of the Riemann sum is \\( pinecone \\int_{0}^{1} sunflower(peppermint) riverbank(peppermint) d peppermint \\).\n\nIt remains to determine \\( seashell(peppermint) \\) for \\( pinecone \\geq 0 \\). Let \\( starflower \\) be chosen uniformly in \\( [0,1] \\), independently of the \\( rainstorm, \\ldots, tumbleweed \\) in the definition of \\( seashell(peppermint) \\). Then \\( seashell(peppermint) \\) equals the probability that \\( starflower \\) is in \\( [0, peppermint] \\) and is closer to \\( peppermint \\) than any other \\( honeycomb \\), since this probability conditioned on a choice of \\( rainstorm, \\ldots, tumbleweed \\) equals \\( arrowhead \\). On the other hand, this probability equals \\( \\left(1-(1-peppermint)^{pinecone+1}\\right) /(pinecone+1) \\), since the probability that at least one of \\( rainstorm, \\ldots, starflower \\) lies in \\( [0, peppermint] \\) equals \\( 1-(1-peppermint)^{pinecone+1} \\), and conditioned on this, the probability that the \\( honeycomb \\) in \\( [0, peppermint] \\) closest to \\( peppermint \\) is \\( starflower \\) is \\( 1 /(pinecone+1) \\), since all possible indices for this closest \\( honeycomb \\) are equally likely. Thus \\( seashell(peppermint)=\\left(1-(1-peppermint)^{pinecone+1}\\right) /(pinecone+1) \\), and we may take \\( blacksmith(peppermint)=pinecone sunflower(peppermint)=1-(1-peppermint)^{pinecone} \\).\n\nLiterature note. For more on Riemann sums, see [Spv, Ch. 13, Appendix 1]."
    },
    "descriptive_long_misleading": {
      "map": {
        "x_0": "finalpoint",
        "x_1": "lastcoordinate",
        "x_2": "middlecoordinate",
        "x_n": "firstcoordinate",
        "x_i": "fixedcoordinate",
        "x_i+1": "previouscoordinate",
        "x_i+2": "earliercoordinate",
        "x_i+3": "ancientcoordinate",
        "x_n+1": "zerocoordinate",
        "t": "staticvalue",
        "i": "fixedindex",
        "j": "stableindex",
        "y_1": "outsidepoint",
        "y_n": "boundarypoint",
        "y_i+1": "precedingpoint",
        "y_n+1": "exteriorpoint",
        "y_j": "genericpoint",
        "z_i+1": "summitpoint",
        "R_n": "steadysegment",
        "E_n": "randomnessvalue",
        "E_n-1": "unpredictvalue",
        "M_i": "macrovalue",
        "S": "voidness",
        "n": "singulardim",
        "f": "discretefunct",
        "P": "irrationalcurve",
        "E": "observedvalue"
      },
      "question": "Let $(lastcoordinate,\\,middlecoordinate,\\,\\ldots\\,firstcoordinate)$ be a point chosen at random from the\nsingulardim-dimensional region defined by\n$0<lastcoordinate<middlecoordinate<\\cdots < firstcoordinate<1.$  Let $discretefunct$ be a continuous function on\n$[0,1]$ with $discretefunct(1)=0$.\nSet $finalpoint=0$ and $zerocoordinate=1$. Show that the expected value of the\nRiemann sum\n\\[\n\\sum_{fixedindex=0}^{singulardim} (previouscoordinate-fixedcoordinate) \\,discretefunct(previouscoordinate)\n\\]\nis $\\int_0^1 discretefunct(staticvalue)\\,irrationalcurve(staticvalue)\\, dstaticvalue$, where $irrationalcurve$ is a polynomial of degree $singulardim$,\nindependent of $discretefunct$, with $0\\le irrationalcurve(staticvalue)\\le 1$ for $0\\le staticvalue \\le 1$.",
      "solution": "Solution 1. We may drop the \\( fixedindex=singulardim \\) term in the Riemann sum, since \\( discretefunct\\!\\left(zerocoordinate\\right)=discretefunct(1)=0 \\). The volume of the region \\( 0<lastcoordinate<\\cdots<firstcoordinate<1 \\) in \\( \\mathbb{R}^{singulardim} \\) is\n\\[\n\\int_{0}^{1} \\int_{0}^{firstcoordinate} \\cdots \\int_{0}^{middlecoordinate} d lastcoordinate \\, d middlecoordinate \\cdots d firstcoordinate =\\frac{1}{singulardim!}.\n\\]\n\nTherefore the expected value of the Riemann sum is\n\\[\nobservedvalue=\\left(\\sum_{fixedindex=0}^{singulardim-1} \\macrovalue_{fixedindex}\\right)\\Big/\\left(\\frac1{singulardim!}\\right),\n\\]\nwhere\n\\[\n\\begin{aligned}\n\\macrovalue_{fixedindex} & =\\int_{0}^{1} \\int_{0}^{firstcoordinate} \\cdots \\int_{0}^{middlecoordinate}\\left(previouscoordinate-fixedcoordinate\\right) \\,discretefunct\\!\\left(previouscoordinate\\right) \\,d lastcoordinate \\cdots d firstcoordinate \\\\\n& =\\int_{0}^{1} \\int_{0}^{firstcoordinate} \\cdots \\int_{0}^{previouscoordinate}\\left(previouscoordinate-fixedcoordinate\\right) \\frac{fixedcoordinate^{fixedindex-1}}{(fixedindex-1)!}\\,discretefunct\\!\\left(previouscoordinate\\right) \\,d fixedcoordinate \\,d previouscoordinate \\cdots d firstcoordinate \\\\\n& =\\int_{0}^{1} \\int_{0}^{firstcoordinate} \\cdots \\int_{0}^{earliercoordinate}\\left(previouscoordinate \\frac{previouscoordinate^{fixedindex}}{fixedindex!}-\\frac{previouscoordinate^{fixedindex+1}}{(fixedindex+1)\\,(fixedindex-1)!}\\right) \\,discretefunct\\!\\left(previouscoordinate\\right) \\,d previouscoordinate \\cdots d firstcoordinate \\\\\n& =\\int_{0}^{1} \\int_{0}^{firstcoordinate} \\cdots \\int_{0}^{earliercoordinate} \\frac{previouscoordinate^{fixedindex+1}}{(fixedindex+1)!}\\,discretefunct\\!\\left(previouscoordinate\\right) \\,d previouscoordinate \\cdots d firstcoordinate \\\\\n& =\\int_{0}^{1} \\int_{previouscoordinate}^{1} \\int_{previouscoordinate}^{firstcoordinate} \\int_{previouscoordinate}^{ancientcoordinate} \\frac{previouscoordinate^{fixedindex+1}}{(fixedindex+1)!}\\,discretefunct\\!\\left(previouscoordinate\\right) \\,d earliercoordinate \\cdots d firstcoordinate \\,d previouscoordinate \\\\\n& =\\int_{0}^{1} \\frac{\\left(1-previouscoordinate\\right)^{singulardim-(fixedindex+1)}}{(singulardim-(fixedindex+1))!} \\frac{previouscoordinate^{fixedindex+1}}{(fixedindex+1)!}\\,discretefunct\\!\\left(previouscoordinate\\right) \\,d previouscoordinate \\\\\n& =\\frac{1}{singulardim!} \\int_{0}^{1}\\binom{singulardim}{fixedindex+1}\\left(1-previouscoordinate\\right)^{singulardim-(fixedindex+1)} previouscoordinate^{fixedindex+1}\\,discretefunct\\!\\left(previouscoordinate\\right) \\,d previouscoordinate .\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\begin{aligned}\nobservedvalue & =\\int_{0}^{1}\\left(\\sum_{fixedindex=0}^{singulardim-1}\\binom{singulardim}{fixedindex+1}\\left(1-previouscoordinate\\right)^{singulardim-(fixedindex+1)} previouscoordinate^{fixedindex+1}\\right)\\,discretefunct\\!\\left(previouscoordinate\\right) \\,d previouscoordinate \\\\\n& =\\int_{0}^{1}\\left(\\sum_{stableindex=1}^{singulardim}\\binom{singulardim}{stableindex}(1-staticvalue)^{singulardim-stableindex} staticvalue^{stableindex}\\right)\\,discretefunct(staticvalue) \\,d staticvalue \\quad\\left(\\text { we set } stableindex=fixedindex+1 \\text { and } staticvalue=previouscoordinate\\right) \\\\\n& =\\int_{0}^{1}\\left(1-(1-staticvalue)^{singulardim}\\right)\\,discretefunct(staticvalue) \\,d staticvalue ,\n\\end{aligned}\n\\]\nsince the sum is the binomial expansion of \\( ((1-staticvalue)+staticvalue)^{singulardim} \\) except that the \\( stableindex=0 \\) term, which is \\( (1-staticvalue)^{singulardim} \\), is missing. Hence \\( observedvalue=\\int_{0}^{1} \\discretefunct(staticvalue)\\, \\irrationalcurve(staticvalue)\\, d staticvalue \\) where \\( \\irrationalcurve(staticvalue)=1-(1-staticvalue)^{singulardim} \\). Clearly \\( 0 \\leq \\irrationalcurve(staticvalue) \\leq 1 \\) for \\( 0 \\leq staticvalue \\leq 1 \\).\n\nSolution 2. Fix \\( singulardim \\geq 0 \\) and let \\( \\left(outsidepoint, \\ldots, boundarypoint\\right) \\) be chosen uniformly from \\( [0,1]^{singulardim} \\). Fix \\( staticvalue \\in[0,1] \\), and let \\( randomnessvalue_{singulardim}(staticvalue) \\) denote the expected value of\n\\[\nsteadysegment_{singulardim}=staticvalue-\\max \\left(\\{0\\} \\cup\\left(\\left\\{outsidepoint, \\ldots, boundarypoint\\right\\} \\cap[0, staticvalue]\\right)\\right) .\n\\]\n\nLet \\( \\left(lastcoordinate, \\ldots, firstcoordinate\\right) \\) be the permutation of \\( \\left(outsidepoint, \\ldots, boundarypoint\\right) \\) such that \\( lastcoordinate \\leq \\cdots \\leq firstcoordinate \\). The distribution of \\( \\left(lastcoordinate, \\ldots, firstcoordinate\\right) \\) equals the distribution in the problem. Define\n\\[\nvoidness=\\sum_{fixedindex=0}^{singulardim-1}\\left(precedingpoint-summitpoint\\right)\\, \\discretefunct\\!\\left(precedingpoint\\right)\n\\]\nwhere \\( summitpoint \\) is the largest \\( genericpoint \\) less than \\( precedingpoint \\), or 0 if no such \\( genericpoint \\) exists. The terms in \\( voidness \\) are a permutation of those in the original Riemann sum (ignoring the final term in the Riemann sum, which is zero), so the expected values of the Riemann sum and of \\( voidness \\) are equal. Each term in \\( voidness \\) has expected value \\( \\int_{0}^{1} \\unpredictvalue(staticvalue) \\,\\discretefunct(staticvalue) \\,d staticvalue \\), since the expected value of \\( precedingpoint-summitpoint \\) conditioned on the event \\( precedingpoint=staticvalue \\) for some \\( staticvalue \\in[0,1] \\) is the definition of \\( \\unpredictvalue(staticvalue) \\) using \\( \\left(genericpoint\\right)_{genericpoint \\neq precedingpoint} \\in[0,1]^{singulardim-1} \\). Therefore the expected value of the Riemann sum is \\( singulardim \\int_{0}^{1} \\unpredictvalue(staticvalue)\\, \\discretefunct(staticvalue) \\,d staticvalue \\).\n\nIt remains to determine \\( randomnessvalue_{singulardim}(staticvalue) \\) for \\( singulardim \\geq 0 \\). Let \\( exteriorpoint \\) be chosen uniformly in \\( [0,1] \\), independently of the \\( outsidepoint, \\ldots, boundarypoint \\) in the definition of \\( randomnessvalue_{singulardim}(staticvalue) \\). Then \\( randomnessvalue_{singulardim}(staticvalue) \\) equals the probability that \\( exteriorpoint \\) is in \\( [0, staticvalue] \\) and is closer to \\( staticvalue \\) than any other \\( genericpoint \\), since this probability conditioned on a choice of \\( outsidepoint, \\ldots, boundarypoint \\) equals \\( steadysegment_{singulardim} \\). On the other hand, this probability equals \\( \\left(1-(1-staticvalue)^{singulardim+1}\\right)/(singulardim+1) \\), since the probability that at least one of \\( outsidepoint, \\ldots, exteriorpoint \\) lies in \\( [0, staticvalue] \\) equals \\( 1-(1-staticvalue)^{singulardim+1} \\), and conditioned on this, the probability that the \\( genericpoint \\) in \\( [0, staticvalue] \\) closest to \\( staticvalue \\) is \\( exteriorpoint \\) is \\( 1 /(singulardim+1) \\), since all possible indices for this closest \\( genericpoint \\) are equally likely. Thus\n\\[\nrandomnessvalue_{singulardim}(staticvalue)=\\frac{1-(1-staticvalue)^{singulardim+1}}{singulardim+1},\n\\]\nand we may take \\( \\irrationalcurve(staticvalue)=singulardim \\,\\unpredictvalue(staticvalue)=1-(1-staticvalue)^{singulardim}. \\)\n\nLiterature note. For more on Riemann sums, see [Spv, Ch. 13, Appendix 1]."
    },
    "garbled_string": {
      "map": {
        "x_0": "hjgrksla",
        "x_1": "qvnsdryk",
        "x_2": "mzkplfau",
        "x_n": "wjdqopcz",
        "x_i": "brghtnke",
        "x_{i+1}": "syctamuv",
        "x_{i+2}": "lbhqveop",
        "x_{i+3}": "zrxwcldi",
        "x_{n+1}": "pfjadkmu",
        "t": "gloxrime",
        "i": "dlmqyrnt",
        "j": "skvdohuf",
        "y_1": "aclwznot",
        "y_n": "qerspfhc",
        "y_{i+1}": "udnhqlej",
        "y_{n+1}": "xpjvoitr",
        "y_j": "hybkzuma",
        "z_{i+1}": "nwsagxre",
        "R_n": "kejmpvla",
        "E_n": "wvthuibc",
        "E_{n-1}": "rgkamcye",
        "M_i": "otpwlzfk",
        "S": "pshargwe",
        "n": "vbanmcot",
        "f": "sirdmafe",
        "P": "ltqirhvm",
        "E": "kxpvozlh"
      },
      "question": "Let $(qvnsdryk,\\,mzkplfau,\\,\\ldots\\,wjdqopcz)$ be a point chosen at random from the\n$vbanmcot$-dimensional region defined by\n$0<qvnsdryk<mzkplfau<\\cdots < wjdqopcz<1.$  Let $sirdmafe$ be a continuous function on\n$[0,1]$ with $sirdmafe(1)=0$.\nSet $hjgrksla=0$ and $pfjadkmu=1$. Show that the expected value of the\nRiemann sum\n\\[\n\\sum_{dlmqyrnt=0}^{vbanmcot} (syctamuv-brghtnke) sirdmafe(syctamuv)\n\\]\nis $\\int_0^1 sirdmafe(gloxrime)ltqirhvm(gloxrime)\\, dgloxrime$, where $ltqirhvm$ is a polynomial of degree $vbanmcot$,\nindependent of $sirdmafe$, with $0\\le ltqirhvm(gloxrime)\\le 1$ for $0\\le gloxrime \\le 1$.\n\n\\end{itemize}\n\\end{document}",
      "solution": "Solution 1. We may drop the $(dlmqyrnt=vbanmcot)$ term in the Riemann sum, since $sirdmafe\\left(pfjadkmu\\right)=sirdmafe(1)=0$. The volume of the region $0<qvnsdryk<\\cdots<wjdqopcz<1$ in $\\mathbb{R}^{vbanmcot}$ is\n\\[\n\\int_{0}^{1} \\int_{0}^{wjdqopcz} \\cdots \\int_{0}^{mzkplfau} d qvnsdryk d mzkplfau \\cdots d wjdqopcz=1 / vbanmcot!\n\\]\n\nTherefore the expected value of the Riemann sum is $kxpvozlh=\\left(\\sum_{dlmqyrnt=0}^{vbanmcot-1} otpwlzfk\\right) /(1 / vbanmcot!)$, where\n\\[\n\\begin{aligned}\notpwlzfk & =\\int_{0}^{1} \\int_{0}^{wjdqopcz} \\cdots \\int_{0}^{mzkplfau}\\left(syctamuv-brghtnke\\right) sirdmafe\\left(syctamuv\\right) d qvnsdryk \\cdots d wjdqopcz \\\\\n& =\\int_{0}^{1} \\int_{0}^{wjdqopcz} \\cdots \\int_{0}^{syctamuv}\\left(syctamuv-brghtnke\\right) \\frac{brghtnke^{dlmqyrnt-1}}{(dlmqyrnt-1)!} sirdmafe\\left(syctamuv\\right) d brghtnke d syctamuv \\cdots d wjdqopcz \\\\\n& =\\int_{0}^{1} \\int_{0}^{wjdqopcz} \\cdots \\int_{0}^{lbhqveop}\\left(syctamuv \\frac{syctamuv^{dlmqyrnt}}{dlmqyrnt!}-\\frac{syctamuv^{dlmqyrnt+1}}{(dlmqyrnt+1) \\cdot(dlmqyrnt-1)!}\\right) sirdmafe\\left(syctamuv\\right) d syctamuv \\cdots d wjdqopcz \\\\\n& =\\int_{0}^{1} \\int_{0}^{wjdqopcz} \\cdots \\int_{0}^{lbhqveop} \\frac{syctamuv^{dlmqyrnt+1}}{(dlmqyrnt+1)!} sirdmafe\\left(syctamuv\\right) d syctamuv \\cdots d wjdqopcz \\\\\n& =\\int_{0}^{1} \\int_{syctamuv}^{1} \\int_{syctamuv}^{wjdqopcz} \\int_{syctamuv}^{zrxwcldi} \\frac{syctamuv^{dlmqyrnt+1}}{(dlmqyrnt+1)!} sirdmafe\\left(syctamuv\\right) d lbhqveop \\cdots d wjdqopcz d syctamuv \\\\\n& =\\int_{0}^{1} \\frac{\\left(1-syctamuv\\right)^{vbanmcot-(dlmqyrnt+1)}}{(vbanmcot-(dlmqyrnt+1))!} \\frac{syctamuv^{dlmqyrnt+1}}{(dlmqyrnt+1)!} sirdmafe\\left(syctamuv\\right) d syctamuv \\\\\n& =\\frac{1}{vbanmcot!} \\int_{0}^{1}\\binom{vbanmcot}{dlmqyrnt+1}\\left(1-syctamuv\\right)^{vbanmcot-(dlmqyrnt+1)} syctamuv^{dlmqyrnt+1} sirdmafe\\left(syctamuv\\right) d syctamuv .\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\begin{aligned}\nkxpvozlh & =\\int_{0}^{1}\\left(\\sum_{dlmqyrnt=0}^{vbanmcot-1}\\binom{vbanmcot}{dlmqyrnt+1}\\left(1-syctamuv\\right)^{vbanmcot-(dlmqyrnt+1)} syctamuv^{dlmqyrnt+1}\\right) sirdmafe\\left(syctamuv\\right) d syctamuv \\\\\n& =\\int_{0}^{1}\\left(\\sum_{skvdohuf=1}^{vbanmcot}\\binom{vbanmcot}{skvdohuf}(1-gloxrime)^{vbanmcot-skvdohuf} gloxrime^{skvdohuf}\\right) sirdmafe(gloxrime) d gloxrime \\quad\\left(\\text { we set } skvdohuf=dlmqyrnt+1 \\text { and } gloxrime=syctamuv\\right) \\\\\n& =\\int_{0}^{1}\\left(1-(1-gloxrime)^{vbanmcot}\\right) sirdmafe(gloxrime) d gloxrime\n\\end{aligned}\n\\]\nsince the sum is the binomial expansion of $((1-gloxrime)+gloxrime)^{vbanmcot}$ except that the $skvdohuf=0$ term, which is $(1-gloxrime)^{vbanmcot}$, is missing. Hence $kxpvozlh=\\int_{0}^{1} sirdmafe(gloxrime) ltqirhvm(gloxrime) d gloxrime$ where $ltqirhvm(gloxrime)=1-(1-gloxrime)^{vbanmcot}$. Clearly $0 \\leq ltqirhvm(gloxrime) \\leq 1$ for $0 \\leq gloxrime \\leq 1$.\n\nSolution 2. Fix $vbanmcot \\geq 0$ and let $\\left(aclwznot, \\ldots, qerspfhc\\right)$ be chosen uniformly from $[0,1]^{vbanmcot}$. Fix $gloxrime \\in[0,1]$, and let $wvthuibc(gloxrime)$ denote the expected value of\n\\[\nkejmpvla=gloxrime-\\max \\left(\\{0\\} \\cup\\left(\\left\\{aclwznot, \\ldots, qerspfhc\\right\\} \\cap[0, gloxrime]\\right)\\right) .\n\\]\n\nLet $\\left(qvnsdryk, \\ldots, wjdqopcz\\right)$ be the permutation of $\\left(aclwznot, \\ldots, qerspfhc\\right)$ such that $qvnsdryk \\leq \\cdots \\leq wjdqopcz$. The distribution of $\\left(qvnsdryk, \\ldots, wjdqopcz\\right)$ equals the distribution in the problem. Define\n\\[\npshargwe=\\sum_{dlmqyrnt=0}^{vbanmcot-1}\\left(udnhqlej-nwsagxre\\right) sirdmafe\\left(udnhqlej\\right)\n\\]\nwhere $nwsagxre$ is the largest $hybkzuma$ less than $udnhqlej$, or 0 if no such $hybkzuma$ exists. The terms in $pshargwe$ are a permutation of those in the original Riemann sum (ignoring the final term in the Riemann sum, which is zero), so the expected values of the Riemann sum and of $pshargwe$ are equal. Each term in $pshargwe$ has expected value $\\int_{0}^{1} rgkamcye(gloxrime) sirdmafe(gloxrime) d gloxrime$, since the expected value of $udnhqlej-nwsagxre$ conditioned on the event $udnhqlej=gloxrime$ for some $gloxrime \\in[0,1]$ is the definition of $rgkamcye(gloxrime)$ using $\\left(hybkzuma\\right)_{skvdohuf \\neq dlmqyrnt+1} \\in[0,1]^{vbanmcot-1}$. Therefore the expected value of the Riemann sum is $vbanmcot \\int_{0}^{1} rgkamcye(gloxrime) sirdmafe(gloxrime) d gloxrime$.\n\nIt remains to determine $wvthuibc(gloxrime)$ for $vbanmcot \\geq 0$. Let $xpjvoitr$ be chosen uniformly in $[0,1]$, independently of the $aclwznot, \\ldots, qerspfhc$ in the definition of $wvthuibc(gloxrime)$. Then $wvthuibc(gloxrime)$ equals the probability that $xpjvoitr$ is in $[0, gloxrime]$ and is closer to $gloxrime$ than any other $hybkzuma$, since this probability conditioned on a choice of $aclwznot, \\ldots, qerspfhc$ equals $kejmpvla$. On the other hand, this probability equals $\\left(1-(1-gloxrime)^{vbanmcot+1}\\right) /(vbanmcot+1)$, since the probability that at least one of $aclwznot, \\ldots, xpjvoitr$ lies in $[0, gloxrime]$ equals $1-(1-gloxrime)^{vbanmcot+1}$, and conditioned on this, the probability that the $hybkzuma$ in $[0, gloxrime]$ closest to $gloxrime$ is $xpjvoitr$ is $1 /(vbanmcot+1)$, since all possible indices for this closest $hybkzuma$ are equally likely. Thus $wvthuibc(gloxrime)=\\left(1-(1-gloxrime)^{vbanmcot+1}\\right) /(vbanmcot+1)$, and we may take $ltqirhvm(gloxrime)=vbanmcot\\, rgkamcye(gloxrime)=1-(1-gloxrime)^{vbanmcot}$.\n\nLiterature note. For more on Riemann sums, see [Spv, Ch. 13, Appendix 1]."
    },
    "kernel_variant": {
      "question": "Let d∈ℕ, d≥1, and let n₁,…,n_d be fixed positive integers.  \nFor every r∈{1,…,d} choose a random point  \n\n   (x_{r,1},…,x_{r,n_r})        (∗)\n\nuniformly (with respect to Lebesgue volume) from the simplex  \n\n   0 < x_{r,1} < ⋯ < x_{r,n_r} < 2 .\n\nThe d choices are made independently and we set  \n\n   x_{r,0}=0 , x_{r,n_r+1}=2  (r=1,…,d).\n\nFix a binary vector ε=(ε₁,…,ε_d)∈{0,1}^d and define the two boundary values  \n\n   a_r := 0 if ε_r=0 ,   a_r := 2 if ε_r=1 .   (1)\n\nBoundary condition.  \nThroughout we work with integrable functions  \n\n   f:[0,2]^d→ℝ such that f(t)=0\n   whenever t_r = a_r for at least one r.   (2)\n\n(Thus f vanishes on the union of the 2d coordinate hyper-planes on which the “chosen’’ evaluation component touches the corresponding end-point.  \nThis hypothesis will be indispensable for the one–dimensional computation below.)\n\nDefine the ε–“mixed end-point’’ Riemann sum\n\n   S_{ε} :=  Σ_{i₁=0}^{n₁} ⋯ Σ_{i_d=0}^{n_d}\n        (x_{1,i₁+1}-x_{1,i₁}) ⋯ (x_{d,i_d+1}-x_{d,i_d})\n        f( x_{1,i₁+ε₁}, …, x_{d,i_d+ε_d} ).   (3)\n\n(a)  Prove that there exists a polynomial P_{ε}:[0,2]^d→[0,1],\n     independent of f, such that  \n\n     𝔼[S_{ε}] = ∫_{[0,2]^d} f(t) P_{ε}(t) dt.     (4)\n\n(b)  Determine this polynomial explicitly, prove 0≤P_{ε}≤1 on [0,2]^d,\n     and show the global Lipschitz estimate  \n\n     |P_{ε}(t)−P_{ε}(s)| ≤ ½(n₁+⋯+n_d) ‖t−s‖₁  (t,s∈[0,2]^d). (5)\n\n(c)  For p∈[1,∞] define the linear functional  \n\n     T_{ε,n₁,…,n_d}: L^{p}([0,2]^d) → ℝ,\n      T_{ε,n₁,…,n_d}(f) := 𝔼[S_{ε}]  (f subject to (2)).    (6)\n\n     Prove that T_{ε,n₁,…,n_d} is bounded on every L^{p} space and that  \n\n     ‖T_{ε,n₁,…,n_d}‖ = ‖P_{ε}‖_{L^{p'}([0,2]^d)},  p' := p/(p−1)  \n     (with the usual convention p'=∞ when p=1 and p'=1 when p=∞). (7)",
      "solution": "Throughout we write  \n\n  N:=(n₁,…,n_d),  ϕ_{R,n}(t):=1−(1−t/2)^{n},  ϕ_{L,n}(t):=(1−t/2)^{n}.  \n\n(All polynomial weights are evaluated for t∈[0,2].)\n\nStep 0.  A one-dimensional computation with the correct boundary hypothesis  \n\nFix n≥1 and draw (x₁,…,x_n) uniformly from 0<x₁<⋯<x_n<2.  \nPut x₀:=0, x_{n+1}:=2.  For an integrable function g:[0,2]→ℝ define the two Riemann sums  \n\n R_L(g) := Σ_{i=0}^{n} (x_{i+1}−x_i) g(x_i),     (8a)  \n R_R(g) := Σ_{i=0}^{n} (x_{i+1}−x_i) g(x_{i+1}).  (8b)\n\nImportant convention.  \nWhen the left-endpoint version is needed we assume g(0)=0;  \nfor the right-endpoint version we assume g(2)=0.  \nThat hypothesis will eliminate the last term in (8a)/(8b) and, as we shall see, justifies the forthcoming integral identities – they are false without it.\n\nDerivation of the right-endpoint weight ϕ_{R,n}.  \nAssume g(2)=0.  Rescale  \n\n  y_i:=x_i/2  and  h(t):=g(2t).  \n\nThen (8b) becomes  \n\n  R_R(g)=2 Σ_{i=0}^{n} (y_{i+1}-y_i) h(y_{i+1})  with 0<y₁<⋯<y_n<1.\n\nLet Δ_n:= {0<y₁<⋯<y_n<1}.  Its Lebesgue volume is 1/n!.  Therefore\n\n 𝔼[R_R(g)] = 2·n! ∫_{Δ_n} Σ_{i=0}^{n−1} (y_{i+1}-y_i) h(y_{i+1}) dy. (9a')\n\n(The summation stops at i=n−1 because for i=n, x_{n+1}=2 and the term\nvanishes by g(2)=0.)  \nProceed exactly as in the original derivation: fix i ≤ n−1, apply\nFubini, obtain\n\n ∫_{Δ_n} (y_{i+1}-y_i) h(y_{i+1}) dy\n  = 1/[n!(i+1)] ∫_{0}^{1} h(t) t^{\\,i+1}(1−t)^{\\,n−(i+1)} dt. (9b')\n\nSummation over i=0,…,n−1 gives\n\n Σ_{i=0}^{n−1} t^{\\,i+1}(1−t)^{\\,n−(i+1)} = 1−(1−t)^{n}. (9c')\n\nPutting the pieces together and reverting to t=2y yields\n\n 𝔼[R_R(g)] = ∫_{0}^{2} g(t) [1−(1−t/2)^{n}] dt\n       = ∫_{0}^{2} g(t) ϕ_{R,n}(t) dt. (10)\n\nDerivation of the left-endpoint weight ϕ_{L,n}.  \nNow assume g(0)=0 and apply the previous result to the reflected function ǧ(t):=g(2−t).  After the change of variables u:=2−t one obtains  \n\n 𝔼[R_L(g)] = ∫_{0}^{2} g(t) (1−t/2)^{n} dt\n       = ∫_{0}^{2} g(t) ϕ_{L,n}(t) dt. (11)\n\nElementary calculus gives, for both symbols •=L,R,  \n\n 0≤ϕ_{•,n}(t)≤1,   |ϕ'_{•,n}(t)| = n/2·|(1−t/2)^{\\,n−1}| ≤ n/2 (t∈[0,2]). (12)\n\nStep 1.  Iterated expectation – proof of part (a)  \n\nFix all random coordinates belonging to r=2,…,d and denote their realisation by x^{(2:d)}.  \nDefine the auxiliary function depending on t₁∈[0,2]  \n\n g_1(t₁) := Σ_{i₂=0}^{n₂}⋯Σ_{i_d=0}^{n_d}\n      (x_{2,i₂+1}-x_{2,i₂})⋯(x_{d,i_d+1}-x_{d,i_d})\n      f(t₁, x_{2,i₂+ε₂},…,x_{d,i_d+ε_d}). (13)\n\nBecause of the boundary condition (2) we have g_1(a₁)=0, where a₁ is the corresponding end-point (0 or 2).  Hence the hypotheses of Step 0 are satisfied and we may use (10) if ε₁=1 and (11) if ε₁=0.  Conditioning on x^{(2:d)} gives  \n\n 𝔼[S_{ε} | x^{(2:d)} ] = ∫_{0}^{2} g_1(t₁) ϕ_{ε₁,n₁}(t₁) dt₁. (14)\n\nRepeat the argument successively for r=2,3,…,d; each time Fubini and (10)/(11) insert the factor ϕ_{ε_r,n_r}.  After d steps we arrive at\n\n 𝔼[S_{ε}] = ∫_{[0,2]^d} f(t)  Π_{r=1}^{d} ϕ_{ε_r,n_r}(t_r) dt. (15)\n\nThus statement (4) holds with  \n\n P_{ε}(t) := Π_{r=1}^{d} ϕ_{ε_r,n_r}(t_r). (16)\n\nStep 2.  Properties of P_{ε} – proof of part (b)  \n\nSince every factor of (16) lies in [0,1], we immediately have 0≤P_{ε}≤1.  \nFor its Lipschitz constant compute\n\n ∂_{t_r}P_{ε}(u) = ϕ'_{ε_r,n_r}(u_r)⋅Π_{q≠r}ϕ_{ε_q,n_q}(u_q),\n\nhence, by (12),\n\n |∂_{t_r}P_{ε}(u)| ≤ n_r/2 (u∈[0,2]^d).  \n\nThe mean-value theorem yields\n\n |P_{ε}(t)−P_{ε}(s)| ≤ Σ_{r=1}^{d} (n_r/2)|t_r−s_r|\n          = ½(n₁+⋯+n_d)‖t−s‖₁  (t,s∈[0,2]^d), which is (5).\n\nStep 3.  The operator norm – proof of part (c)  \n\nUpper bound.  \nFor every f∈L^{p} subject to (2), Hölder’s inequality gives\n\n |T_{ε,N}(f)| = |∫ fP_{ε}| ≤ ‖f‖_{p}‖P_{ε}‖_{p'}, (17)\n\nso ‖T_{ε,N}‖ ≤ ‖P_{ε}‖_{p'}.  \n\nLower bound – three cases, taking care of the boundary hyper-planes.  \nWhenever we construct a test function that does not vanish on the\nboundary, we modify it on that measure-zero set so that (2) holds\nwithout changing its L^p norm or the value of the integral.\n\n(i) 1<p<∞.  \nDefine f₀(t):=sign P_{ε}(t)·|P_{ε}(t)|^{p'−1}.  \nAfter the above modification f₀ satisfies (2) and ‖f₀‖_{p}=‖P_{ε}‖_{p'}^{p'/p}=‖P_{ε}‖_{p'}^{p'-1}.  Equality holds in (17), hence ‖T‖≥‖P_{ε}‖_{p'}.\n\n(ii) p=∞ (so p'=1).  \nTake f(t)=sign P_{ε}(t) modified on the boundary.  Then ‖f‖_{∞}=1 and  \nT(f)=∫|P_{ε}| = ‖P_{ε}‖_{1}, establishing equality.\n\n(iii) p=1 (so p'=∞).  \nPut M:=‖P_{ε}‖_{∞} and, for k∈ℕ, let  \n\n A_k := {t∈[0,2]^d : P_{ε}(t) ≥ M−1/k}.\n\nEach A_k has positive measure, and χ_{A_k} can be altered on the boundary so that (2) holds.  Set  \n\n f_k(t):=sign P_{ε}(t)·χ_{A_k}(t).\n\nThen ‖f_k‖_{1}=|A_k| and\n\n T(f_k)=∫_{A_k} |P_{ε}| ≥ (M−1/k)|A_k| = (M−1/k)‖f_k‖_{1}.\n\nHence ‖T‖ ≥ M−1/k for every k, so ‖T‖ ≥ M = ‖P_{ε}‖_{∞}.  \n\nCombining with the upper bound we have equality in all cases p∈[1,∞],\nestablishing (7).\n\nThe solution is now fully rigorous.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.711022",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimensional interaction:  The problem replaces a single random\n    partition with d independent partitions, forcing the solver to pass\n    from one–dimensional Dirichlet/Beta reasoning to genuine\n    multi-variable considerations and to keep careful track of indices.\n\n2.  Mixed endpoints:  Allowing an arbitrary ε∈{0,1}^d means that both left-\n    and right–endpoint Riemann sums occur simultaneously, so two distinct\n    polynomial kernels have to be derived and then combined.\n\n3.  Additional quantitative requirement:  Besides computing P_{ε},\n    the solver must establish the global Lipschitz bound (3) and determine\n    the exact operator norm (5).  These estimates are absent from both the\n    original problem and the existing kernel variant and require a\n    delicate derivative analysis and a functional-analytic argument.\n\n4.  Depth of techniques:  The solution demands\n       •  repeated use of the Beta–function computation from the classical\n          one-dimensional case;\n       •  factorisation of expectations over independent random partitions;\n       •  multivariate calculus for the Lipschitz constant;\n       •  Hölder duality to compute the operator norm.\n    No step can be disposed of by simple pattern matching; each needs\n    explicit, coordinated reasoning about several interacting concepts.\n\nHence the enhanced variant is substantially more intricate, both\ncombinatorially and analytically, than either the original problem or the\ncurrent kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let d∈ℕ, d≥1, and let n₁,…,n_d be fixed positive integers.  \nFor every r∈{1,…,d} choose a random point  \n\n   (x_{r,1},…,x_{r,n_r})        (∗)\n\nuniformly (with respect to Lebesgue volume) from the simplex  \n\n   0 < x_{r,1} < ⋯ < x_{r,n_r} < 2 .\n\nThe d choices are made independently and we set  \n\n   x_{r,0}=0 , x_{r,n_r+1}=2  (r=1,…,d).\n\nFix a binary vector ε=(ε₁,…,ε_d)∈{0,1}^d and define the two boundary values  \n\n   a_r := 0 if ε_r=0 ,   a_r := 2 if ε_r=1 .   (1)\n\nBoundary condition.  \nThroughout we work with integrable functions  \n\n   f:[0,2]^d→ℝ such that f(t)=0\n   whenever t_r = a_r for at least one r.   (2)\n\n(Thus f vanishes on the union of the 2d coordinate hyper-planes on which the “chosen’’ evaluation component touches the corresponding end-point.  \nThis hypothesis will be indispensable for the one–dimensional computation below.)\n\nDefine the ε–“mixed end-point’’ Riemann sum\n\n   S_{ε} :=  Σ_{i₁=0}^{n₁} ⋯ Σ_{i_d=0}^{n_d}\n        (x_{1,i₁+1}-x_{1,i₁}) ⋯ (x_{d,i_d+1}-x_{d,i_d})\n        f( x_{1,i₁+ε₁}, …, x_{d,i_d+ε_d} ).   (3)\n\n(a)  Prove that there exists a polynomial P_{ε}:[0,2]^d→[0,1],\n     independent of f, such that  \n\n     𝔼[S_{ε}] = ∫_{[0,2]^d} f(t) P_{ε}(t) dt.     (4)\n\n(b)  Determine this polynomial explicitly, prove 0≤P_{ε}≤1 on [0,2]^d,\n     and show the global Lipschitz estimate  \n\n     |P_{ε}(t)−P_{ε}(s)| ≤ ½(n₁+⋯+n_d) ‖t−s‖₁  (t,s∈[0,2]^d). (5)\n\n(c)  For p∈[1,∞] define the linear functional  \n\n     T_{ε,n₁,…,n_d}: L^{p}([0,2]^d) → ℝ,\n      T_{ε,n₁,…,n_d}(f) := 𝔼[S_{ε}]  (f subject to (2)).    (6)\n\n     Prove that T_{ε,n₁,…,n_d} is bounded on every L^{p} space and that  \n\n     ‖T_{ε,n₁,…,n_d}‖ = ‖P_{ε}‖_{L^{p'}([0,2]^d)},  p' := p/(p−1)  \n     (with the usual convention p'=∞ when p=1 and p'=1 when p=∞). (7)",
      "solution": "Throughout we write  \n\n  N:=(n₁,…,n_d),  ϕ_{R,n}(t):=1−(1−t/2)^{n},  ϕ_{L,n}(t):=(1−t/2)^{n}.  \n\n(All polynomial weights are evaluated for t∈[0,2].)\n\nStep 0.  A one-dimensional computation with the correct boundary hypothesis  \n\nFix n≥1 and draw (x₁,…,x_n) uniformly from 0<x₁<⋯<x_n<2.  \nPut x₀:=0, x_{n+1}:=2.  For an integrable function g:[0,2]→ℝ define the two Riemann sums  \n\n R_L(g) := Σ_{i=0}^{n} (x_{i+1}−x_i) g(x_i),     (8a)  \n R_R(g) := Σ_{i=0}^{n} (x_{i+1}−x_i) g(x_{i+1}).  (8b)\n\nImportant convention.  \nWhen the left-endpoint version is needed we assume g(0)=0;  \nfor the right-endpoint version we assume g(2)=0.  \nThat hypothesis will eliminate the last term in (8a)/(8b) and, as we shall see, justifies the forthcoming integral identities – they are false without it.\n\nDerivation of the right-endpoint weight ϕ_{R,n}.  \nAssume g(2)=0.  Rescale  \n\n  y_i:=x_i/2  and  h(t):=g(2t).  \n\nThen (8b) becomes  \n\n  R_R(g)=2 Σ_{i=0}^{n} (y_{i+1}-y_i) h(y_{i+1})  with 0<y₁<⋯<y_n<1.\n\nLet Δ_n:= {0<y₁<⋯<y_n<1}.  Its Lebesgue volume is 1/n!.  Therefore\n\n 𝔼[R_R(g)] = 2·n! ∫_{Δ_n} Σ_{i=0}^{n−1} (y_{i+1}-y_i) h(y_{i+1}) dy. (9a')\n\n(The summation stops at i=n−1 because for i=n, x_{n+1}=2 and the term\nvanishes by g(2)=0.)  \nProceed exactly as in the original derivation: fix i ≤ n−1, apply\nFubini, obtain\n\n ∫_{Δ_n} (y_{i+1}-y_i) h(y_{i+1}) dy\n  = 1/[n!(i+1)] ∫_{0}^{1} h(t) t^{\\,i+1}(1−t)^{\\,n−(i+1)} dt. (9b')\n\nSummation over i=0,…,n−1 gives\n\n Σ_{i=0}^{n−1} t^{\\,i+1}(1−t)^{\\,n−(i+1)} = 1−(1−t)^{n}. (9c')\n\nPutting the pieces together and reverting to t=2y yields\n\n 𝔼[R_R(g)] = ∫_{0}^{2} g(t) [1−(1−t/2)^{n}] dt\n       = ∫_{0}^{2} g(t) ϕ_{R,n}(t) dt. (10)\n\nDerivation of the left-endpoint weight ϕ_{L,n}.  \nNow assume g(0)=0 and apply the previous result to the reflected function ǧ(t):=g(2−t).  After the change of variables u:=2−t one obtains  \n\n 𝔼[R_L(g)] = ∫_{0}^{2} g(t) (1−t/2)^{n} dt\n       = ∫_{0}^{2} g(t) ϕ_{L,n}(t) dt. (11)\n\nElementary calculus gives, for both symbols •=L,R,  \n\n 0≤ϕ_{•,n}(t)≤1,   |ϕ'_{•,n}(t)| = n/2·|(1−t/2)^{\\,n−1}| ≤ n/2 (t∈[0,2]). (12)\n\nStep 1.  Iterated expectation – proof of part (a)  \n\nFix all random coordinates belonging to r=2,…,d and denote their realisation by x^{(2:d)}.  \nDefine the auxiliary function depending on t₁∈[0,2]  \n\n g_1(t₁) := Σ_{i₂=0}^{n₂}⋯Σ_{i_d=0}^{n_d}\n      (x_{2,i₂+1}-x_{2,i₂})⋯(x_{d,i_d+1}-x_{d,i_d})\n      f(t₁, x_{2,i₂+ε₂},…,x_{d,i_d+ε_d}). (13)\n\nBecause of the boundary condition (2) we have g_1(a₁)=0, where a₁ is the corresponding end-point (0 or 2).  Hence the hypotheses of Step 0 are satisfied and we may use (10) if ε₁=1 and (11) if ε₁=0.  Conditioning on x^{(2:d)} gives  \n\n 𝔼[S_{ε} | x^{(2:d)} ] = ∫_{0}^{2} g_1(t₁) ϕ_{ε₁,n₁}(t₁) dt₁. (14)\n\nRepeat the argument successively for r=2,3,…,d; each time Fubini and (10)/(11) insert the factor ϕ_{ε_r,n_r}.  After d steps we arrive at\n\n 𝔼[S_{ε}] = ∫_{[0,2]^d} f(t)  Π_{r=1}^{d} ϕ_{ε_r,n_r}(t_r) dt. (15)\n\nThus statement (4) holds with  \n\n P_{ε}(t) := Π_{r=1}^{d} ϕ_{ε_r,n_r}(t_r). (16)\n\nStep 2.  Properties of P_{ε} – proof of part (b)  \n\nSince every factor of (16) lies in [0,1], we immediately have 0≤P_{ε}≤1.  \nFor its Lipschitz constant compute\n\n ∂_{t_r}P_{ε}(u) = ϕ'_{ε_r,n_r}(u_r)⋅Π_{q≠r}ϕ_{ε_q,n_q}(u_q),\n\nhence, by (12),\n\n |∂_{t_r}P_{ε}(u)| ≤ n_r/2 (u∈[0,2]^d).  \n\nThe mean-value theorem yields\n\n |P_{ε}(t)−P_{ε}(s)| ≤ Σ_{r=1}^{d} (n_r/2)|t_r−s_r|\n          = ½(n₁+⋯+n_d)‖t−s‖₁  (t,s∈[0,2]^d), which is (5).\n\nStep 3.  The operator norm – proof of part (c)  \n\nUpper bound.  \nFor every f∈L^{p} subject to (2), Hölder’s inequality gives\n\n |T_{ε,N}(f)| = |∫ fP_{ε}| ≤ ‖f‖_{p}‖P_{ε}‖_{p'}, (17)\n\nso ‖T_{ε,N}‖ ≤ ‖P_{ε}‖_{p'}.  \n\nLower bound – three cases, taking care of the boundary hyper-planes.  \nWhenever we construct a test function that does not vanish on the\nboundary, we modify it on that measure-zero set so that (2) holds\nwithout changing its L^p norm or the value of the integral.\n\n(i) 1<p<∞.  \nDefine f₀(t):=sign P_{ε}(t)·|P_{ε}(t)|^{p'−1}.  \nAfter the above modification f₀ satisfies (2) and ‖f₀‖_{p}=‖P_{ε}‖_{p'}^{p'/p}=‖P_{ε}‖_{p'}^{p'-1}.  Equality holds in (17), hence ‖T‖≥‖P_{ε}‖_{p'}.\n\n(ii) p=∞ (so p'=1).  \nTake f(t)=sign P_{ε}(t) modified on the boundary.  Then ‖f‖_{∞}=1 and  \nT(f)=∫|P_{ε}| = ‖P_{ε}‖_{1}, establishing equality.\n\n(iii) p=1 (so p'=∞).  \nPut M:=‖P_{ε}‖_{∞} and, for k∈ℕ, let  \n\n A_k := {t∈[0,2]^d : P_{ε}(t) ≥ M−1/k}.\n\nEach A_k has positive measure, and χ_{A_k} can be altered on the boundary so that (2) holds.  Set  \n\n f_k(t):=sign P_{ε}(t)·χ_{A_k}(t).\n\nThen ‖f_k‖_{1}=|A_k| and\n\n T(f_k)=∫_{A_k} |P_{ε}| ≥ (M−1/k)|A_k| = (M−1/k)‖f_k‖_{1}.\n\nHence ‖T‖ ≥ M−1/k for every k, so ‖T‖ ≥ M = ‖P_{ε}‖_{∞}.  \n\nCombining with the upper bound we have equality in all cases p∈[1,∞],\nestablishing (7).\n\nThe solution is now fully rigorous.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.554441",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimensional interaction:  The problem replaces a single random\n    partition with d independent partitions, forcing the solver to pass\n    from one–dimensional Dirichlet/Beta reasoning to genuine\n    multi-variable considerations and to keep careful track of indices.\n\n2.  Mixed endpoints:  Allowing an arbitrary ε∈{0,1}^d means that both left-\n    and right–endpoint Riemann sums occur simultaneously, so two distinct\n    polynomial kernels have to be derived and then combined.\n\n3.  Additional quantitative requirement:  Besides computing P_{ε},\n    the solver must establish the global Lipschitz bound (3) and determine\n    the exact operator norm (5).  These estimates are absent from both the\n    original problem and the existing kernel variant and require a\n    delicate derivative analysis and a functional-analytic argument.\n\n4.  Depth of techniques:  The solution demands\n       •  repeated use of the Beta–function computation from the classical\n          one-dimensional case;\n       •  factorisation of expectations over independent random partitions;\n       •  multivariate calculus for the Lipschitz constant;\n       •  Hölder duality to compute the operator norm.\n    No step can be disposed of by simple pattern matching; each needs\n    explicit, coordinated reasoning about several interacting concepts.\n\nHence the enhanced variant is substantially more intricate, both\ncombinatorially and analytically, than either the original problem or the\ncurrent kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}