{
  "index": "1958-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "COMB",
    "NT"
  ],
  "difficulty": "",
  "question": "3. Real numbers are chosen at random from the interval \\( (0 \\leq x \\leq 1) \\). If after choosing the \\( n \\)th number the sum of the numbers so chosen first exceeds 1 , show that the expected or average value for \\( n \\) is \\( e \\).",
  "solution": "First Solution. We assume that the phrase \"real numbers are chosen at random\" means that the \\( x \\) 's are independent and each has the uniform distribution on \\( [0,1] \\). Then the probability that \\( \\left(x_{1}, x_{2}, \\ldots, x_{n}\\right) \\) falls in a region \\( S \\) of the cube \\( [0,1]^{n} \\) is the \\( n \\)-dimensional content of \\( S \\).\n\nLet \\( p_{n} \\) be the probability that \\( x_{1}+x_{2}+\\cdots+x_{n} \\leq 1 \\). The probability that \\( x_{1}+x_{2}+\\cdots+x_{n}>1 \\) but \\( x_{1}+x_{2}+\\cdots+x_{n-1} \\leq 1 \\) is then\n\\[\nq_{n}=p_{n 1}-p_{n}\n\\]\n\nIt is proved below that \\( p_{n}=1 / n! \\) Hence the expected number of choices required to make the sum exceed one is\n\\[\n\\begin{aligned}\nE=\\sum_{n=1}^{\\infty} n q_{n} & =\\sum_{n=1}^{\\infty} n\\left(\\frac{1}{(n-1)!}-\\frac{1}{n!}\\right)=\\sum_{n=1}^{\\infty} \\frac{1}{(n-1)!}(n-1) \\\\\n& =\\sum_{n=2}^{\\infty} \\frac{1}{(n-2)!}=e\n\\end{aligned}\n\\]\n\nLemma. The \\( n \\)-dimensional content \\( V_{n}(a) \\) of the region in \\( R^{\\prime \\prime} \\) determined by the inequalities \\( x_{i} \\geq 0, i=1,2, \\ldots, n \\) and \\( x_{1}+x_{2}+\\cdots+x_{n} \\leq a \\) is \\( a^{\\prime \\prime} / n \\) !\n\nProof. This is evidently true for \\( n=1 \\). Suppose it is true for \\( n=k \\). Then \\( V_{k+1}(a) \\) can be formed by \"slicing\" perpendicular to the \\( x_{k+1} \\)-axis. The slice for \\( x_{k+1}=b \\) is the \\( k \\)-dimensional region determined by the inequalities \\( x_{i} \\geq 0, i=1,2, \\ldots, k \\) and \\( x_{1}+x_{2}+\\ldots+x_{k} \\leq a-b \\) for \\( 0 \\leq b \\leq a \\). By the inductive hypothesis its \\( k \\)-dimensional content is \\( V_{k}(a-b)=(a-b)^{k} / k! \\). Hence\n\\[\n\\begin{aligned}\nV_{k+1}(a) & =\\int_{0}^{1} V_{k}\\left(a-x_{k+1}\\right) d x_{k+1}=\\int_{0}^{1 \"} \\frac{\\left(a-x_{k+1}\\right)^{k}}{k!} d x_{k+1} \\\\\n& =\\frac{a^{k+1}}{(k+1)!}\n\\end{aligned}\n\\]\n\nThus the formula \\( V_{n}(a)=a^{n} / n! \\) is established for all positive integers \\( n \\).\nEvidently \\( p_{n}=V_{n}(1)=1 / n! \\).\nSecond Solution. Let the expected number of trials required to obtain a score of \\( c \\) or more be \\( E(c) \\). Suppose \\( 0<c \\leq 1 \\). If the first draw is in the interval \\( [x, x+\\Delta x] \\) where \\( x<c \\), then the expected number of draws will be about \\( 1+E(c-x) \\). Hence for \\( 0<c \\leq 1 \\),\n\\[\nE(c)=1+\\int_{0}^{c} E(c-x) d x=1+\\int_{0}^{c} E(u) d u\n\\]\n(The integral exists since \\( E \\) is increasing.) We see from (1) that \\( E \\) is continuous, so by the fundamental theorem of calculus,\n\\[\nE^{\\prime}(c)=E(c)\n\\]\n\nTherefore\n\\[\nE(c)=\\lambda e^{c},\n\\]\nfor some \\( \\lambda \\). In order to satisfy (1), we must have \\( \\lambda=1 \\). Hence \\( E(1)=e \\).\nRemark. D. J. Newman and M. S. Klamkin [American Mathematical Monthly, vol. 66 (1959), pages 50-51] have found the expected value of the least \\( n \\) for which \\( x_{1}{ }^{k}+\\cdots+x_{n}{ }^{k}>1 \\). Klamkin and J. H. van Lint (Statistica Nederlandica, vol. 26 (1972), pages 191-196) have extended the result to more general functions than powers.",
  "vars": [
    "x",
    "x_i",
    "x_1",
    "x_2",
    "x_n",
    "x_k",
    "x_k+1",
    "u"
  ],
  "params": [
    "n",
    "k",
    "i",
    "a",
    "b",
    "c",
    "p_n",
    "q_n",
    "V_n",
    "V_k",
    "V_k+1",
    "E",
    "\\\\lambda"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "randomx",
        "x_i": "randomxi",
        "x_1": "randomxone",
        "x_2": "randomxtwo",
        "x_n": "randomxn",
        "x_k": "randomxk",
        "x_k+1": "randomxkplusone",
        "u": "auxiliaryu",
        "n": "drawcount",
        "k": "slicecount",
        "i": "indexvar",
        "a": "upperlimit",
        "b": "slicevalue",
        "c": "targetsum",
        "p_n": "problessthanone",
        "q_n": "probexceedsone",
        "V_n": "volumeofn",
        "V_k": "volumeofk",
        "V_k+1": "volumeofkplusone",
        "E": "expectationval",
        "\\lambda": "scalingconst"
      },
      "question": "3. Real numbers are chosen at random from the interval \\( (0 \\leq randomx \\leq 1) \\). If after choosing the \\( drawcount \\)th number the sum of the numbers so chosen first exceeds 1 , show that the expected or average value for \\( drawcount \\) is \\( e \\).",
      "solution": "First Solution. We assume that the phrase \"real numbers are chosen at random\" means that the \\( randomx \\) 's are independent and each has the uniform distribution on \\( [0,1] \\). Then the probability that \\( \\left(randomxone, randomxtwo, \\ldots, randomxn\\right) \\) falls in a region \\( S \\) of the cube \\( [0,1]^{drawcount} \\) is the \\( drawcount \\)-dimensional content of \\( S \\).\n\nLet \\( problessthanone \\) be the probability that \\( randomxone+randomxtwo+\\cdots+randomxn \\leq 1 \\). The probability that \\( randomxone+randomxtwo+\\cdots+randomxn>1 \\) but \\( randomxone+randomxtwo+\\cdots+randomx_{drawcount-1} \\leq 1 \\) is then\n\\[\nprobexceedsone=p_{drawcount 1}-problessthanone\n\\]\n\nIt is proved below that \\( problessthanone=1 / drawcount! \\). Hence the expected number of choices required to make the sum exceed one is\n\\[\n\\begin{aligned}\nexpectationval &= \\sum_{drawcount=1}^{\\infty} drawcount\\,probexceedsone \\\\\n&= \\sum_{drawcount=1}^{\\infty} drawcount\\left(\\frac{1}{(drawcount-1)!}-\\frac{1}{drawcount!}\\right) \\\\\n&= \\sum_{drawcount=1}^{\\infty} \\frac{1}{(drawcount-1)!}(drawcount-1) \\\\\n&= \\sum_{drawcount=2}^{\\infty} \\frac{1}{(drawcount-2)!}=e\n\\end{aligned}\n\\]\n\nLemma. The \\( drawcount \\)-dimensional content \\( volumeofn(upperlimit) \\) of the region in \\( R^{\\prime\\prime} \\) determined by the inequalities \\( randomxi \\geq 0,\\; indexvar=1,2, \\ldots, drawcount \\) and \\( randomxone+randomxtwo+\\cdots+randomxn \\leq upperlimit \\) is \\( upperlimit^{\\prime\\prime}/drawcount! \\).\n\nProof. This is evidently true for \\( drawcount=1 \\). Suppose it is true for \\( drawcount=slicecount \\). Then \\( volumeofkplusone(upperlimit) \\) can be formed by \"slicing\" perpendicular to the \\( randomxkplusone \\)-axis. The slice for \\( randomxkplusone=slicevalue \\) is the \\( slicecount \\)-dimensional region determined by the inequalities \\( randomxi \\geq 0,\\; indexvar=1,2, \\ldots, slicecount \\) and \\( randomxone+randomxtwo+\\ldots+randomxk \\leq upperlimit-slicevalue \\) for \\( 0 \\leq slicevalue \\leq upperlimit \\). By the inductive hypothesis its \\( slicecount \\)-dimensional content is \\( volumeofk(upperlimit-slicevalue)=(upperlimit-slicevalue)^{slicecount}/slicecount! \\). Hence\n\\[\n\\begin{aligned}\nvolumeofkplusone(upperlimit) &= \\int_{0}^{1} volumeofk\\left(upperlimit-randomxkplusone\\right) \\, d randomxkplusone \\\\\n&= \\int_{0}^{1} \\frac{\\left(upperlimit-randomxkplusone\\right)^{slicecount}}{slicecount!} \\, d randomxkplusone \\\\\n&= \\frac{upperlimit^{slicecount+1}}{(slicecount+1)!}\n\\end{aligned}\n\\]\n\nThus the formula \\( volumeofn(upperlimit)=upperlimit^{drawcount}/drawcount! \\) is established for all positive integers \\( drawcount \\). Evidently \\( problessthanone=volumeofn(1)=1/drawcount! \\).\n\nSecond Solution. Let the expected number of trials required to obtain a score of \\( targetsum \\) or more be \\( expectationval(targetsum) \\). Suppose \\( 0<targetsum \\leq 1 \\). If the first draw is in the interval \\( [randomx, randomx+\\Delta randomx] \\) where \\( randomx<targetsum \\), then the expected number of draws will be about \\( 1+expectationval(targetsum-randomx) \\). Hence for \\( 0<targetsum \\leq 1 \\),\n\\[\nexpectationval(targetsum)=1+\\int_{0}^{targetsum} expectationval(targetsum-randomx) \\, d randomx=1+\\int_{0}^{targetsum} expectationval(auxiliaryu) \\, d auxiliaryu\n\\]\n(The integral exists since \\( expectationval \\) is increasing.) We see from (1) that \\( expectationval \\) is continuous, so by the fundamental theorem of calculus,\n\\[\nexpectationval^{\\prime}(targetsum)=expectationval(targetsum)\n\\]\nTherefore\n\\[\nexpectationval(targetsum)=scalingconst e^{targetsum},\n\\]\nfor some \\( scalingconst \\). In order to satisfy (1), we must have \\( scalingconst=1 \\). Hence \\( expectationval(1)=e \\).\n\nRemark. D. J. Newman and M. S. Klamkin [American Mathematical Monthly, vol. 66 (1959), pages 50–51] have found the expected value of the least \\( drawcount \\) for which \\( randomxone^{slicecount}+\\cdots+randomxn^{slicecount}>1 \\). Klamkin and J. H. van Lint (Statistica Nederlandica, vol. 26 (1972), pages 191–196) have extended the result to more general functions than powers."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "windsurfer",
        "x_i": "gearshift",
        "x_1": "snowflake",
        "x_2": "toothpick",
        "x_n": "backpacker",
        "x_k": "moonlight",
        "x_k+1": "riverbank",
        "u": "boardgame",
        "n": "lemonade",
        "k": "afterglow",
        "i": "timestamp",
        "a": "sandstone",
        "b": "pomegranate",
        "c": "marshmallow",
        "p_n": "hummingbird",
        "q_n": "thunderbolt",
        "V_n": "rainshower",
        "V_k": "wildfire",
        "V_k+1": "dreamscape",
        "E": "blueberries",
        "\\lambda": "colander"
      },
      "question": "Real numbers are chosen at random from the interval \\( (0 \\leq windsurfer \\leq 1) \\). If after choosing the \\( lemonade \\)th number the sum of the numbers so chosen first exceeds 1 , show that the expected or average value for \\( lemonade \\) is \\( e \\).",
      "solution": "First Solution. We assume that the phrase \"real numbers are chosen at random\" means that the \\( windsurfer \\) 's are independent and each has the uniform distribution on \\( [0,1] \\). Then the probability that \\( \\left(snowflake, toothpick, \\ldots, backpacker\\right) \\) falls in a region \\( S \\) of the cube \\( [0,1]^{lemonade} \\) is the \\( lemonade \\)-dimensional content of \\( S \\).\n\nLet \\( hummingbird \\) be the probability that \\( snowflake+toothpick+\\cdots+backpacker \\leq 1 \\). The probability that \\( snowflake+toothpick+\\cdots+backpacker>1 \\) but \\( snowflake+toothpick+\\cdots+x_{n-1} \\leq 1 \\) is then\n\\[\nthunderbolt = p_{n 1}-hummingbird\n\\]\n\nIt is proved below that \\( hummingbird = 1 / lemonade! \\). Hence the expected number of choices required to make the sum exceed one is\n\\[\n\\begin{aligned}\nblueberries &= \\sum_{lemonade=1}^{\\infty} lemonade\\; thunderbolt = \\sum_{lemonade=1}^{\\infty} lemonade\\left(\\frac{1}{(lemonade-1)!}-\\frac{1}{lemonade!}\\right)=\\sum_{lemonade=1}^{\\infty} \\frac{1}{(lemonade-1)!}(lemonade-1) \\\\\n& =\\sum_{lemonade=2}^{\\infty} \\frac{1}{(lemonade-2)!}=e\n\\end{aligned}\n\\]\n\nLemma. The \\( lemonade \\)-dimensional content \\( rainshower(sandstone) \\) of the region in \\( R^{\\prime \\prime} \\) determined by the inequalities \\( gearshift \\geq 0, timestamp=1,2, \\ldots, lemonade \\) and \\( snowflake+toothpick+\\cdots+backpacker \\leq sandstone \\) is \\( sandstone^{\\prime \\prime} / lemonade !\\).\n\nProof. This is evidently true for \\( lemonade=1 \\). Suppose it is true for \\( lemonade=afterglow \\). Then \\( dreamscape(sandstone) \\) can be formed by \"slicing\" perpendicular to the \\( riverbank \\)-axis. The slice for \\( riverbank=pomegranate \\) is the \\( afterglow \\)-dimensional region determined by the inequalities \\( gearshift \\geq 0, timestamp=1,2, \\ldots, afterglow \\) and \\( snowflake+toothpick+\\ldots+moonlight \\leq sandstone-pomegranate \\) for \\( 0 \\leq pomegranate \\leq sandstone \\). By the inductive hypothesis its \\( afterglow \\)-dimensional content is \\( wildfire(sandstone-pomegranate)=(sandstone-pomegranate)^{afterglow} / afterglow! \\). Hence\n\\[\n\\begin{aligned}\ndreamscape(sandstone) &= \\int_{0}^{1} wildfire\\left(sandstone-riverbank\\right) d\\, riverbank = \\int_{0}^{1} \\frac{\\left(sandstone-riverbank\\right)^{afterglow}}{afterglow!} d\\, riverbank \\\\\n& = \\frac{sandstone^{afterglow+1}}{(afterglow+1)!}\n\\end{aligned}\n\\]\n\nThus the formula \\( rainshower(sandstone)=sandstone^{lemonade} / lemonade! \\) is established for all positive integers \\( lemonade \\).\nEvidently \\( hummingbird = rainshower(1)=1 / lemonade! \\).\n\nSecond Solution. Let the expected number of trials required to obtain a score of \\( marshmallow \\) or more be \\( blueberries(marshmallow) \\). Suppose \\( 0<marshmallow \\leq 1 \\). If the first draw is in the interval \\[windsurfer, windsurfer+\\Delta windsurfer] where \\( windsurfer<marshmallow \\), then the expected number of draws will be about \\( 1+blueberries(marshmallow-windsurfer) \\). Hence for \\( 0<marshmallow \\leq 1 \\),\n\\[\nblueberries(marshmallow)=1+\\int_{0}^{marshmallow} blueberries(marshmallow-windsurfer) d\\, windsurfer = 1+\\int_{0}^{marshmallow} blueberries(boardgame) d\\, boardgame\n\\]\n(The integral exists since \\( blueberries \\) is increasing.) We see from (1) that \\( blueberries \\) is continuous, so by the fundamental theorem of calculus,\n\\[\nblueberries^{\\prime}(marshmallow)=blueberries(marshmallow)\n\\]\n\nTherefore\n\\[\nblueberries(marshmallow)=colander e^{marshmallow},\n\\]\nfor some \\( colander \\). In order to satisfy (1), we must have \\( colander=1 \\). Hence \\( blueberries(1)=e \\).\n\nRemark. D. J. Newman and M. S. Klamkin [American Mathematical Monthly, vol. 66 (1959), pages 50-51] have found the expected value of the least \\( lemonade \\) for which \\( snowflake{ }^{afterglow}+\\cdots+backpacker{ }^{afterglow}>1 \\). Klamkin and J. H. van Lint (Statistica Nederlandica, vol. 26 (1972), pages 191-196) have extended the result to more general functions than powers."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "x_i": "fixedvalueindex",
        "x_1": "fixedvaluefirst",
        "x_2": "fixedvaluesecond",
        "x_n": "fixedvaluefinal",
        "x_k": "fixedvaluekappa",
        "x_k+1": "fixedvaluekplusone",
        "u": "constantfactor",
        "n": "staticcount",
        "k": "staticstep",
        "i": "staticiterator",
        "a": "bottomlimit",
        "b": "toplimit",
        "c": "zerothreshold",
        "p_n": "certaintyseries",
        "q_n": "uncertaintyseries",
        "V_n": "surfacecollection",
        "V_k": "surfacekappa",
        "V_k+1": "surfacekplusone",
        "E": "variance",
        "\\\\lambda": "emptiness"
      },
      "question": "3. Real numbers are chosen at random from the interval \\( (0 \\leq fixedvalue \\leq 1) \\). If after choosing the \\( staticcount \\)th number the sum of the numbers so chosen first exceeds 1 , show that the expected or average value for \\( staticcount \\) is \\( e \\).",
      "solution": "First Solution. We assume that the phrase \"real numbers are chosen at random\" means that the \\( fixedvalue \\) 's are independent and each has the uniform distribution on \\( [0,1] \\). Then the probability that \\( \\left(fixedvaluefirst, fixedvaluesecond, \\ldots, fixedvaluefinal\\right) \\) falls in a region \\( S \\) of the cube \\( [0,1]^{staticcount} \\) is the \\( staticcount \\)-dimensional content of \\( S \\).\n\nLet \\( certaintyseries \\) be the probability that \\( fixedvaluefirst+fixedvaluesecond+\\cdots+fixedvaluefinal \\leq 1 \\). The probability that \\( fixedvaluefirst+fixedvaluesecond+\\cdots+fixedvaluefinal>1 \\) but \\( fixedvaluefirst+fixedvaluesecond+\\cdots+x_{staticcount-1} \\leq 1 \\) is then\n\\[\nuncertaintyseries=certaintyseries_{staticcount-1}-certaintyseries\n\\]\n\nIt is proved below that \\( certaintyseries=1 / staticcount! \\) Hence the expected number of choices required to make the sum exceed one is\n\\[\n\\begin{aligned}\nvariance=\\sum_{staticcount=1}^{\\infty} staticcount\\,uncertaintyseries & =\\sum_{staticcount=1}^{\\infty} staticcount\\left(\\frac{1}{(staticcount-1)!}-\\frac{1}{staticcount!}\\right)=\\sum_{staticcount=1}^{\\infty} \\frac{1}{(staticcount-1)!}(staticcount-1) \\\\ & =\\sum_{staticcount=2}^{\\infty} \\frac{1}{(staticcount-2)!}=e\n\\end{aligned}\n\\]\n\nLemma. The \\( staticcount \\)-dimensional content \\( surfacecollection(bottomlimit) \\) of the region in \\( R^{\\prime \\prime} \\) determined by the inequalities \\( fixedvalueindex \\geq 0,\\; staticiterator=1,2, \\ldots, staticcount \\) and \\( fixedvaluefirst+fixedvaluesecond+\\cdots+fixedvaluefinal \\leq bottomlimit \\) is \\( bottomlimit^{\\prime \\prime} / staticcount \\) !\n\nProof. This is evidently true for \\( staticcount=1 \\). Suppose it is true for \\( staticcount=staticstep \\). Then \\( surfacekplusone(bottomlimit) \\) can be formed by \"slicing\" perpendicular to the \\( fixedvaluekplusone \\)-axis. The slice for \\( fixedvaluekplusone=toplimit \\) is the \\( staticstep \\)-dimensional region determined by the inequalities \\( fixedvalueindex \\geq 0,\\; staticiterator=1,2, \\ldots, staticstep \\) and \\( fixedvaluefirst+fixedvaluesecond+\\ldots+fixedvaluekappa \\leq bottomlimit-toplimit \\) for \\( 0 \\leq toplimit \\leq bottomlimit \\). By the inductive hypothesis its \\( staticstep \\)-dimensional content is \\( surfacekappa(bottomlimit-toplimit)=(bottomlimit-toplimit)^{staticstep} / staticstep! \\). Hence\n\\[\n\\begin{aligned}\nsurfacekplusone(bottomlimit) & =\\int_{0}^{1} surfacekappa\\left(bottomlimit-fixedvaluekplusone\\right) d fixedvaluekplusone=\\int_{0}^{1 } \\frac{\\left(bottomlimit-fixedvaluekplusone\\right)^{staticstep}}{staticstep!} d fixedvaluekplusone \\\\ & =\\frac{bottomlimit^{staticstep+1}}{(staticstep+1)!}\n\\end{aligned}\n\\]\n\nThus the formula \\( surfacecollection(bottomlimit)=bottomlimit^{staticcount} / staticcount! \\) is established for all positive integers \\( staticcount \\).\nEvidently \\( certaintyseries=surfacecollection(1)=1 / staticcount! \\).\n\nSecond Solution. Let the expected number of trials required to obtain a score of \\( zerothreshold \\) or more be \\( variance(zerothreshold) \\). Suppose \\( 0<zerothreshold \\leq 1 \\). If the first draw is in the interval \\[fixedvalue, fixedvalue+\\Delta fixedvalue] where \\( fixedvalue<zerothreshold \\), then the expected number of draws will be about \\( 1+variance(zerothreshold-fixedvalue) \\). Hence for \\( 0<zerothreshold \\leq 1 \\),\n\\[\nvariance(zerothreshold)=1+\\int_{0}^{zerothreshold} variance(zerothreshold-fixedvalue) d fixedvalue=1+\\int_{0}^{zerothreshold} variance(constantfactor) d constantfactor\n\\]\n(The integral exists since \\( variance \\) is increasing.) We see from (1) that \\( variance \\) is continuous, so by the fundamental theorem of calculus,\n\\[\nvariance^{\\prime}(zerothreshold)=variance(zerothreshold)\n\\]\n\nTherefore\n\\[\nvariance(zerothreshold)=emptiness e^{zerothreshold},\n\\]\nfor some \\( emptiness \\). In order to satisfy (1), we must have \\( emptiness=1 \\). Hence \\( variance(1)=e \\).\n\nRemark. D. J. Newman and M. S. Klamkin [American Mathematical Monthly, vol. 66 (1959), pages 50-51] have found the expected value of the least \\( staticcount \\) for which \\( fixedvaluefirst^{staticstep}+\\cdots+fixedvaluefinal^{staticstep}>1 \\). Klamkin and J. H. van Lint (Statistica Nederlandica, vol. 26 (1972), pages 191-196) have extended the result to more general functions than powers."
    },
    "garbled_string": {
      "map": {
        "x": "ghtyplms",
        "x_i": "wervbklx",
        "x_1": "zxvmlkqp",
        "x_2": "ljhgfdsq",
        "x_n": "juytrewa",
        "x_k": "pqowieur",
        "x_k+1": "mzxcbvna",
        "u": "asdfghjk",
        "n": "plokijuh",
        "k": "mnbvcxzy",
        "i": "qwertyui",
        "a": "qazwsxed",
        "b": "rfvtgbyh",
        "c": "yhnujmko",
        "p_n": "lkjhgfdp",
        "q_n": "poiuytre",
        "V_n": "cvbnmzas",
        "V_k": "dfghjkqw",
        "V_k+1": "edcrfvgt",
        "E": "xswedcvf",
        "\\lambda": "vfrtgbhu"
      },
      "question": "3. Real numbers are chosen at random from the interval \\( (0 \\leq ghtyplms \\leq 1) \\). If after choosing the \\( plokijuh \\)th number the sum of the numbers so chosen first exceeds 1 , show that the expected or average value for \\( plokijuh \\) is \\( e \\).",
      "solution": "First Solution. We assume that the phrase \"real numbers are chosen at random\" means that the \\( ghtyplms \\) 's are independent and each has the uniform distribution on \\( [0,1] \\). Then the probability that \\( \\left(zxvmlkqp, ljhgfdsq, \\ldots, juytrewa\\right) \\) falls in a region \\( S \\) of the cube \\( [0,1]^{plokijuh} \\) is the \\( plokijuh \\)-dimensional content of \\( S \\).\n\nLet \\( lkjhgfdp \\) be the probability that \\( zxvmlkqp+ljhgfdsq+\\cdots+juytrewa \\leq 1 \\). The probability that \\( zxvmlkqp+ljhgfdsq+\\cdots+juytrewa>1 \\) but \\( zxvmlkqp+ljhgfdsq+\\cdots+ghtyplms_{plokijuh-1} \\leq 1 \\) is then\n\\[\npoiuytre=lkjhgfdp 1-lkjhgfdp\n\\]\n\nIt is proved below that \\( lkjhgfdp=1 / plokijuh! \\) Hence the expected number of choices required to make the sum exceed one is\n\\[\n\\begin{aligned}\nxswedcvf=\\sum_{plokijuh=1}^{\\infty} plokijuh\\, poiuytre & =\\sum_{plokijuh=1}^{\\infty} plokijuh\\left(\\frac{1}{(plokijuh-1)!}-\\frac{1}{plokijuh!}\\right)=\\sum_{plokijuh=1}^{\\infty} \\frac{1}{(plokijuh-1)!}(plokijuh-1) \\\\\n& =\\sum_{plokijuh=2}^{\\infty} \\frac{1}{(plokijuh-2)!}=e\n\\end{aligned}\n\\]\n\nLemma. The \\( plokijuh \\)-dimensional content \\( cvbnmzas(qazwsxed) \\) of the region in \\( R^{\\prime \\prime} \\) determined by the inequalities \\( wervbklx \\geq 0, qwertyui=1,2, \\ldots, plokijuh \\) and \\( zxvmlkqp+ljhgfdsq+\\cdots+juytrewa \\leq qazwsxed \\) is \\( qazwsxed^{\\prime \\prime} / plokijuh \\) !\n\nProof. This is evidently true for \\( plokijuh=1 \\). Suppose it is true for \\( plokijuh=mnbvcxzy \\). Then \\( edcrfvgt(qazwsxed) \\) can be formed by \"slicing\" perpendicular to the \\( mzxcbvna \\)-axis. The slice for \\( mzxcbvna=rfvtgbyh \\) is the \\( mnbvcxzy \\)-dimensional region determined by the inequalities \\( wervbklx \\geq 0, qwertyui=1,2, \\ldots, mnbvcxzy \\) and \\( zxvmlkqp+ljhgfdsq+\\ldots+pqowieur \\leq qazwsxed-rfvtgbyh \\) for \\( 0 \\leq rfvtgbyh \\leq qazwsxed \\). By the inductive hypothesis its \\( mnbvcxzy \\)-dimensional content is \\( dfghjkqw(qazwsxed-rfvtgbyh)=(qazwsxed-rfvtgbyh)^{mnbvcxzy} / mnbvcxzy! \\). Hence\n\\[\n\\begin{aligned}\nedcrfvgt(qazwsxed) & =\\int_{0}^{1} dfghjkqw\\left(qazwsxed-mzxcbvna\\right) d mzxcbvna=\\int_{0}^{1 \"} \\frac{\\left(qazwsxed-mzxcbvna\\right)^{mnbvcxzy}}{mnbvcxzy!} d mzxcbvna \\\\\n& =\\frac{qazwsxed^{mnbvcxzy+1}}{(mnbvcxzy+1)!}\n\\end{aligned}\n\\]\n\nThus the formula \\( cvbnmzas(qazwsxed)=qazwsxed^{plokijuh} / plokijuh! \\) is established for all positive integers \\( plokijuh \\).\nEvidently \\( lkjhgfdp=cvbnmzas(1)=1 / plokijuh! \\).\n\nSecond Solution. Let the expected number of trials required to obtain a score of \\( yhnujmko \\) or more be \\( xswedcvf(yhnujmko) \\). Suppose \\( 0<yhnujmko \\leq 1 \\). If the first draw is in the interval \\( [ghtyplms, ghtyplms+\\Delta ghtyplms] \\) where \\( ghtyplms<yhnujmko \\), then the expected number of draws will be about \\( 1+xswedcvf(yhnujmko-ghtyplms) \\). Hence for \\( 0<yhnujmko \\leq 1 \\),\n\\[\nxswedcvf(yhnujmko)=1+\\int_{0}^{yhnujmko} xswedcvf(yhnujmko-ghtyplms) d ghtyplms=1+\\int_{0}^{yhnujmko} xswedcvf(asdfghjk) d asdfghjk\n\\]\n(The integral exists since \\( xswedcvf \\) is increasing.) We see from (1) that \\( xswedcvf \\) is continuous, so by the fundamental theorem of calculus,\n\\[\nxswedcvf^{\\prime}(yhnujmko)=xswedcvf(yhnujmko)\n\\]\n\nTherefore\n\\[\nxswedcvf(yhnujmko)=vfrtgbhu e^{yhnujmko},\n\\]\nfor some \\( vfrtgbhu \\). In order to satisfy (1), we must have \\( vfrtgbhu=1 \\). Hence \\( xswedcvf(1)=e \\).\n\nRemark. D. J. Newman and M. S. Klamkin [American Mathematical Monthly, vol. 66 (1959), pages 50-51] have found the expected value of the least \\( plokijuh \\) for which \\( zxvmlkqp^{mnbvcxzy}+\\cdots+juytrewa^{mnbvcxzy}>1 \\). Klamkin and J. H. van Lint (Statistica Nederlandica, vol. 26 (1972), pages 191-196) have extended the result to more general functions than powers."
    },
    "kernel_variant": {
      "question": "For a fixed integer \\(d\\ge 2\\) let  \n\\[\n\\mathbf X^{(1)},\\mathbf X^{(2)},\\dots ,\n\\qquad   \n\\mathbf X^{(k)}=(X^{(k)}_1,\\dots ,X^{(k)}_d)\\;(k\\ge 1)\n\\]\nbe an i.i.d. sequence of random vectors whose coordinates are independent and each\ncoordinate is uniformly distributed on the open interval \\((0,1)\\).\nFor every \\(n\\ge 1\\) write the coordinate–wise partial sums  \n\\[\n\\mathbf S_n=\\mathbf X^{(1)}+\\dots+\\mathbf X^{(n)},\\qquad \nS_{n,j}:=\\sum_{k=1}^{n}X^{(k)}_j\\quad(j=1,\\dots ,d).\n\\]\n\nDefine the stopping time  \n\\[\nN=\\min\\Bigl\\{n\\ge 1:\\;S_{n,1}>1,\\;S_{n,2}>1,\\dots ,S_{n,d}>1\\Bigr\\},\n\\]\ni.e. the first round in which every one of the \\(d\\) coordinate–sums exceeds \\(1\\).\n\n(a)  Prove the identity  \n\\[\n\\boxed{\\;\n\\mathbb E[N]=\\sum_{n=0}^{\\infty}\n         \\Bigl[1-\\bigl(1-\\tfrac1{n!}\\bigr)^{d}\\Bigr]\n      =\\sum_{k=1}^{d}(-1)^{k+1}\\binom{d}{k}\\!\n       \\sum_{n=0}^{\\infty}\\frac1{(n!)^{\\,k}}\\;} .\n\\]\n\n(b)  Evaluate \\(\\mathbb E[N]\\) numerically for \\(d=2\\) and \\(d=3\\) with an absolute\nerror not exceeding \\(5\\cdot10^{-7}\\).\n\n(c)  Show that, as \\(d\\to\\infty\\),\n\\[\n\\boxed{\\;\n  \\mathbb E[N]=\\bigl(1+o(1)\\bigr)\\;\n               \\frac{\\log d}{\\log\\log d}}\\!.\n\\]",
      "solution": "Step 1.  Reduction to one–dimensional stopping times  \nFor each coordinate put  \n\\[\nN_j=\\min\\{n\\ge 1:S_{n,j}>1\\}\\qquad(j=1,\\dots ,d).\n\\]\nBecause the coordinates inside every \\(\\mathbf X^{(k)}\\) are independent and the vectors themselves are i.i.d., the variables \\(N_1,\\dots ,N_d\\) are i.i.d.; moreover  \n\\(N=\\max\\{N_1,\\dots ,N_d\\}\\).\n\nStep 2.  Law of a single coordinate  \nFor \\(n\\ge 0\\) let \\(A_n=\\{S_{n,1}\\le 1\\}\\).  The random vector  \n\\((X^{(1)}_1,\\dots ,X^{(n)}_1)\\) is distributed uniformly on the cube \\((0,1)^n\\),\nso\n\\[\n\\mathbb P(A_n)=\\operatorname{Vol}_n\n\\{(x_1,\\dots ,x_n)\\in(0,1)^n:\\;x_1+\\dots +x_n\\le 1\\}.\n\\]\nThe volume of this simplex equals \\(1/n!\\) (standard inductive\nargument or beta–integral), hence\n\\[\n\\mathbb P(S_{n,1}\\le 1)=\\frac1{n!}\\quad(n\\ge 0).\n\\]\nConsequently for \\(n\\ge 1\\)\n\\[\n\\mathbb P(N_1=n)\n  =\\mathbb P(S_{n-1,1}\\le 1)-\\mathbb P(S_{n,1}\\le 1)\n  =\\frac1{(n-1)!}-\\frac1{n!},\n\\]\nand\n\\[\nF(n):=\\mathbb P(N_1\\le n)=1-\\frac1{n!}\\quad(n\\ge 1),\\qquad F(0)=0.\n\\]\n\nStep 3.  Distribution of \\(N=\\max N_j\\)  \nIndependence yields\n\\[\n\\mathbb P(N\\le n)=\\bigl[F(n)\\bigr]^d\n                 =\\Bigl(1-\\frac1{n!}\\Bigr)^{\\!d},\n\\]\n\\[\n\\mathbb P(N>n)=1-\\Bigl(1-\\frac1{n!}\\Bigr)^{\\!d}\n              =\\sum_{k=1}^{d}(-1)^{k+1}\\binom{d}{k}\\frac1{(n!)^{\\,k}}.\n\\]\n\nStep 4.  Expected value (part (a))  \nFor any non–negative integer–valued r.v.\\ \\(X\\) one has the tail representation\n\\(\\mathbb E[X]=\\sum_{n=0}^{\\infty}\\mathbb P(X>n)\\).\nApplying this to \\(N\\) and substituting the expression from Step 3 gives the two boxed series.  \nAbsolute convergence follows because for each fixed \\(k\\ge 1\\)\n\\(\\sum_{n\\ge 0}1/(n!)^{k}\\le e\\).\n\nStep 5.  High–precision evaluation (part (b))  \nWrite \\(S_k=\\sum_{n=0}^{\\infty}\\!1/(n!)^{k}\\).\nUsing, say, 30 terms one gets  \n\\[\nS_1=e=2.718\\,281\\,828\\,459\\ldots,\\qquad\nS_2=2.279\\,585\\,302\\,336\\ldots,\\qquad\nS_3=2.129\\,702\\,553\\,462\\ldots\n\\]\nand hence  \n\\[\n\\mathbb E[N]_{d=2}=2S_1-S_2\n                 =3.156\\,978\\,351\\,71\\ldots\\quad\\Longrightarrow\\;\n                 \\boxed{\\mathbb E[N]_{2}=3.156978},\n\\]\n\\[\n\\mathbb E[N]_{d=3}=3S_1-3S_2+S_3\n                 =3.445\\,792\\,126\\,78\\ldots\\quad\\Longrightarrow\\;\n                 \\boxed{\\mathbb E[N]_{3}=3.445792}.\n\\]\nBoth answers are within \\(5\\cdot10^{-7}\\) of the true values.\n\nStep 6.  Asymptotics (part (c))  \n\nNotation.  Put  \n\\[\na_n:=\\frac1{n!},\\qquad g_d(n):=1-(1-a_n)^{d},\\qquad\nn^{*}=n^{*}(d):=\\min\\{n:\\,a_n\\le 1/d\\}.\n\\]\nThen \\(a_{n^{*}-1}>1/d\\ge a_{n^{*}}\\).\n\n6.1  Stirling approximation  \nStirling’s formula gives  \n\\[\n\\log\\frac1{a_n}=n\\log n-n+\\tfrac12\\log(2\\pi n)+O(1).\n\\]\nSolving \\(\\log(1/a_n)=\\log d\\) yields  \n\\[\nn^{*}=\\frac{\\log d}{\\log\\log d}\\bigl(1+o(1)\\bigr).\n\\]\n\n6.2  Two–regime analysis  \n\n(i)  Lower range \\(n\\le (1-\\varepsilon)n^{*}\\).  \nThen \\(\\log (1/a_n)\\le(1-\\varepsilon)\\log d\\), so \\(d\\,a_n\\ge d^{\\varepsilon}\\to\\infty\\).\nHence\n\\[\ng_d(n)=1-(1-a_n)^{d}\n      =1-\\exp\\!\\bigl(d\\log(1-a_n)\\bigr)\n      \\ge 1-\\exp(-d a_n)=1-o(1).\n\\]\nTherefore\n\\[\n\\sum_{n\\le(1-\\varepsilon)n^{*}}g_d(n)\n=(1-\\varepsilon)n^{*}\\bigl(1+o(1)\\bigr).\n\\]\n\n(ii)  Upper range \\(n\\ge(1+\\varepsilon)n^{*}\\).  \nNow \\(\\log (1/a_n)\\ge(1+\\varepsilon)\\log d\\), whence \\(d\\,a_n\\le d^{-\\varepsilon}\\to0\\).  \nUsing \\(1-e^{-x}\\le x\\) for \\(x\\ge0\\):\n\\[\ng_d(n)\\le d\\,a_n=\\frac{d}{n!}.\n\\]\nBecause \\(a_{n+1}=a_n/(n+1)\\), the tail satisfies\n\\[\n\\sum_{n\\ge (1+\\varepsilon)n^{*}}\\!g_d(n)\n\\le d\\,a_{(1+\\varepsilon)n^{*}}\n     \\sum_{m\\ge 0}\\frac1{\\prod_{k=1}^{m}((1+\\varepsilon)n^{*}+k)}\n =o(n^{*}).\n\\]\n\n6.3  Conclusion  \nCombining (i) and (ii) we have, for every fixed\n\\(0<\\varepsilon<1/2\\),\n\\[\n(1-\\varepsilon)n^{*}(1+o(1))\n\\le \\mathbb E[N]\n\\le\\bigl((1+\\varepsilon)n^{*}+o(n^{*})\\bigr).\n\\]\nSince \\(\\varepsilon>0\\) is arbitrary,\n\\[\n\\boxed{\\;\n\\mathbb E[N]=n^{*}(1+o(1))\n            =\\frac{\\log d}{\\log\\log d}\\bigl(1+o(1)\\bigr)\\,}.\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.504167",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: each “draw’’ is now a \\(d\\)-vector; the stopping time is the\nmaximum of \\(d\\) interacting partial-sum processes.\n\n• Additional constraints: all \\(d\\) coordinate sums must exceed\nthe threshold **simultaneously**; this forces an analysis of the joint\ndistribution of several dependent stopping times.\n\n• Sophisticated structures: the solution hinges on representing the\nmultivariate stopping time as the maximum of i.i.d.\\ one-dimensional\nstopping times, then combining tail distributions via inclusion–exclusion\nto obtain a convergent **alternating series** with combinatorial\ncoefficients.\n\n• Deeper theory: asymptotics require Stirling’s formula, analysis of\nextreme–value behavior for discrete distributions, and careful error\nestimation (leading to the \\(\\tfrac{\\log d}{\\log\\log d}\\) law).\n\n• Multiple interacting concepts: probability of regions in high-dimensional\nhypercubes, order statistics (maxima), generating-function style tail\nsums, combinatorial identities, and classical analytic asymptotics.\nAll of these go far beyond the one-dimensional, single-threshold setting of\nboth the original problem and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "For a fixed integer \\(d\\ge 2\\) let  \n\\[\n\\mathbf X^{(1)},\\mathbf X^{(2)},\\dots ,\n\\qquad   \n\\mathbf X^{(k)}=(X^{(k)}_1,\\dots ,X^{(k)}_d)\\;(k\\ge 1)\n\\]\nbe an i.i.d. sequence of random vectors whose coordinates are independent and each\ncoordinate is uniformly distributed on the open interval \\((0,1)\\).\nFor every \\(n\\ge 1\\) write the coordinate–wise partial sums  \n\\[\n\\mathbf S_n=\\mathbf X^{(1)}+\\dots+\\mathbf X^{(n)},\\qquad \nS_{n,j}:=\\sum_{k=1}^{n}X^{(k)}_j\\quad(j=1,\\dots ,d).\n\\]\n\nDefine the stopping time  \n\\[\nN=\\min\\Bigl\\{n\\ge 1:\\;S_{n,1}>1,\\;S_{n,2}>1,\\dots ,S_{n,d}>1\\Bigr\\},\n\\]\ni.e. the first round in which every one of the \\(d\\) coordinate–sums exceeds \\(1\\).\n\n(a)  Prove the identity  \n\\[\n\\boxed{\\;\n\\mathbb E[N]=\\sum_{n=0}^{\\infty}\n         \\Bigl[1-\\bigl(1-\\tfrac1{n!}\\bigr)^{d}\\Bigr]\n      =\\sum_{k=1}^{d}(-1)^{k+1}\\binom{d}{k}\\!\n       \\sum_{n=0}^{\\infty}\\frac1{(n!)^{\\,k}}\\;} .\n\\]\n\n(b)  Evaluate \\(\\mathbb E[N]\\) numerically for \\(d=2\\) and \\(d=3\\) with an absolute\nerror not exceeding \\(5\\cdot10^{-7}\\).\n\n(c)  Show that, as \\(d\\to\\infty\\),\n\\[\n\\boxed{\\;\n  \\mathbb E[N]=\\bigl(1+o(1)\\bigr)\\;\n               \\frac{\\log d}{\\log\\log d}}\\!.\n\\]",
      "solution": "Step 1.  Reduction to one–dimensional stopping times  \nFor each coordinate put  \n\\[\nN_j=\\min\\{n\\ge 1:S_{n,j}>1\\}\\qquad(j=1,\\dots ,d).\n\\]\nBecause the coordinates inside every \\(\\mathbf X^{(k)}\\) are independent and the vectors themselves are i.i.d., the variables \\(N_1,\\dots ,N_d\\) are i.i.d.; moreover  \n\\(N=\\max\\{N_1,\\dots ,N_d\\}\\).\n\nStep 2.  Law of a single coordinate  \nFor \\(n\\ge 0\\) let \\(A_n=\\{S_{n,1}\\le 1\\}\\).  The random vector  \n\\((X^{(1)}_1,\\dots ,X^{(n)}_1)\\) is distributed uniformly on the cube \\((0,1)^n\\),\nso\n\\[\n\\mathbb P(A_n)=\\operatorname{Vol}_n\n\\{(x_1,\\dots ,x_n)\\in(0,1)^n:\\;x_1+\\dots +x_n\\le 1\\}.\n\\]\nThe volume of this simplex equals \\(1/n!\\) (standard inductive\nargument or beta–integral), hence\n\\[\n\\mathbb P(S_{n,1}\\le 1)=\\frac1{n!}\\quad(n\\ge 0).\n\\]\nConsequently for \\(n\\ge 1\\)\n\\[\n\\mathbb P(N_1=n)\n  =\\mathbb P(S_{n-1,1}\\le 1)-\\mathbb P(S_{n,1}\\le 1)\n  =\\frac1{(n-1)!}-\\frac1{n!},\n\\]\nand\n\\[\nF(n):=\\mathbb P(N_1\\le n)=1-\\frac1{n!}\\quad(n\\ge 1),\\qquad F(0)=0.\n\\]\n\nStep 3.  Distribution of \\(N=\\max N_j\\)  \nIndependence yields\n\\[\n\\mathbb P(N\\le n)=\\bigl[F(n)\\bigr]^d\n                 =\\Bigl(1-\\frac1{n!}\\Bigr)^{\\!d},\n\\]\n\\[\n\\mathbb P(N>n)=1-\\Bigl(1-\\frac1{n!}\\Bigr)^{\\!d}\n              =\\sum_{k=1}^{d}(-1)^{k+1}\\binom{d}{k}\\frac1{(n!)^{\\,k}}.\n\\]\n\nStep 4.  Expected value (part (a))  \nFor any non–negative integer–valued r.v.\\ \\(X\\) one has the tail representation\n\\(\\mathbb E[X]=\\sum_{n=0}^{\\infty}\\mathbb P(X>n)\\).\nApplying this to \\(N\\) and substituting the expression from Step 3 gives the two boxed series.  \nAbsolute convergence follows because for each fixed \\(k\\ge 1\\)\n\\(\\sum_{n\\ge 0}1/(n!)^{k}\\le e\\).\n\nStep 5.  High–precision evaluation (part (b))  \nWrite \\(S_k=\\sum_{n=0}^{\\infty}\\!1/(n!)^{k}\\).\nUsing, say, 30 terms one gets  \n\\[\nS_1=e=2.718\\,281\\,828\\,459\\ldots,\\qquad\nS_2=2.279\\,585\\,302\\,336\\ldots,\\qquad\nS_3=2.129\\,702\\,553\\,462\\ldots\n\\]\nand hence  \n\\[\n\\mathbb E[N]_{d=2}=2S_1-S_2\n                 =3.156\\,978\\,351\\,71\\ldots\\quad\\Longrightarrow\\;\n                 \\boxed{\\mathbb E[N]_{2}=3.156978},\n\\]\n\\[\n\\mathbb E[N]_{d=3}=3S_1-3S_2+S_3\n                 =3.445\\,792\\,126\\,78\\ldots\\quad\\Longrightarrow\\;\n                 \\boxed{\\mathbb E[N]_{3}=3.445792}.\n\\]\nBoth answers are within \\(5\\cdot10^{-7}\\) of the true values.\n\nStep 6.  Asymptotics (part (c))  \n\nNotation.  Put  \n\\[\na_n:=\\frac1{n!},\\qquad g_d(n):=1-(1-a_n)^{d},\\qquad\nn^{*}=n^{*}(d):=\\min\\{n:\\,a_n\\le 1/d\\}.\n\\]\nThen \\(a_{n^{*}-1}>1/d\\ge a_{n^{*}}\\).\n\n6.1  Stirling approximation  \nStirling’s formula gives  \n\\[\n\\log\\frac1{a_n}=n\\log n-n+\\tfrac12\\log(2\\pi n)+O(1).\n\\]\nSolving \\(\\log(1/a_n)=\\log d\\) yields  \n\\[\nn^{*}=\\frac{\\log d}{\\log\\log d}\\bigl(1+o(1)\\bigr).\n\\]\n\n6.2  Two–regime analysis  \n\n(i)  Lower range \\(n\\le (1-\\varepsilon)n^{*}\\).  \nThen \\(\\log (1/a_n)\\le(1-\\varepsilon)\\log d\\), so \\(d\\,a_n\\ge d^{\\varepsilon}\\to\\infty\\).\nHence\n\\[\ng_d(n)=1-(1-a_n)^{d}\n      =1-\\exp\\!\\bigl(d\\log(1-a_n)\\bigr)\n      \\ge 1-\\exp(-d a_n)=1-o(1).\n\\]\nTherefore\n\\[\n\\sum_{n\\le(1-\\varepsilon)n^{*}}g_d(n)\n=(1-\\varepsilon)n^{*}\\bigl(1+o(1)\\bigr).\n\\]\n\n(ii)  Upper range \\(n\\ge(1+\\varepsilon)n^{*}\\).  \nNow \\(\\log (1/a_n)\\ge(1+\\varepsilon)\\log d\\), whence \\(d\\,a_n\\le d^{-\\varepsilon}\\to0\\).  \nUsing \\(1-e^{-x}\\le x\\) for \\(x\\ge0\\):\n\\[\ng_d(n)\\le d\\,a_n=\\frac{d}{n!}.\n\\]\nBecause \\(a_{n+1}=a_n/(n+1)\\), the tail satisfies\n\\[\n\\sum_{n\\ge (1+\\varepsilon)n^{*}}\\!g_d(n)\n\\le d\\,a_{(1+\\varepsilon)n^{*}}\n     \\sum_{m\\ge 0}\\frac1{\\prod_{k=1}^{m}((1+\\varepsilon)n^{*}+k)}\n =o(n^{*}).\n\\]\n\n6.3  Conclusion  \nCombining (i) and (ii) we have, for every fixed\n\\(0<\\varepsilon<1/2\\),\n\\[\n(1-\\varepsilon)n^{*}(1+o(1))\n\\le \\mathbb E[N]\n\\le\\bigl((1+\\varepsilon)n^{*}+o(n^{*})\\bigr).\n\\]\nSince \\(\\varepsilon>0\\) is arbitrary,\n\\[\n\\boxed{\\;\n\\mathbb E[N]=n^{*}(1+o(1))\n            =\\frac{\\log d}{\\log\\log d}\\bigl(1+o(1)\\bigr)\\,}.\n\\]",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.421158",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension: each “draw’’ is now a \\(d\\)-vector; the stopping time is the\nmaximum of \\(d\\) interacting partial-sum processes.\n\n• Additional constraints: all \\(d\\) coordinate sums must exceed\nthe threshold **simultaneously**; this forces an analysis of the joint\ndistribution of several dependent stopping times.\n\n• Sophisticated structures: the solution hinges on representing the\nmultivariate stopping time as the maximum of i.i.d.\\ one-dimensional\nstopping times, then combining tail distributions via inclusion–exclusion\nto obtain a convergent **alternating series** with combinatorial\ncoefficients.\n\n• Deeper theory: asymptotics require Stirling’s formula, analysis of\nextreme–value behavior for discrete distributions, and careful error\nestimation (leading to the \\(\\tfrac{\\log d}{\\log\\log d}\\) law).\n\n• Multiple interacting concepts: probability of regions in high-dimensional\nhypercubes, order statistics (maxima), generating-function style tail\nsums, combinatorial identities, and classical analytic asymptotics.\nAll of these go far beyond the one-dimensional, single-threshold setting of\nboth the original problem and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}