{
  "index": "1970-A-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "COMB"
  ],
  "difficulty": "",
  "question": "A-6. Three numbers are chosen independently at random, one from each of the three intervals \\( \\left[0, L_{i}\\right](i=1,2,3) \\). If the distribution of each random number is uniform with respect to length in the interval it is chosen from, determine the expected value of the smallest of the three numbers chosen.",
  "solution": "A-6 Let \\( x \\) be selected from \\( \\left[0, L_{1}\\right], y \\) from \\( \\left[0, L_{2}\\right], z \\) from \\( \\left[0, L_{2}\\right] \\), and assume \\( L_{8} \\geqq L_{2} \\geqq L_{1} \\). Let \\( X=\\min (x, y, z) \\).\n\\[\n\\begin{aligned}\nL_{1} L_{3} L_{3} E[X] & =\\int_{0}^{L_{1}} \\int_{0}^{L_{2}} \\int_{0}^{L_{3}} X d z d y d x \\\\\n& =\\int_{0}^{L_{1}} \\int_{0}^{L_{1}}\\left\\{\\int_{0}^{\\mu} z d z+\\int_{\\mu}^{L_{3}} \\mu d z\\right\\} d y d x, \\text { where } \\mu=\\min (x, y) \\\\\n& =\\int_{0}^{L_{1}} \\int_{0}^{L_{2}}\\left\\{L_{3} \\mu-\\frac{1}{2} \\mu^{2}\\right\\} d y d x \\\\\n& =\\int_{0}^{L_{1}}\\left\\{\\int_{0}^{x}\\left(L_{8} y-\\frac{1}{2} y^{2}\\right) d y+\\int_{z}^{L_{2}}\\left(L_{3} x-\\frac{1}{2} x^{2}\\right) d y\\right\\} d x \\\\\n& =\\cdots=\\frac{1}{2} L_{1}^{2} L_{2} L_{3}-\\frac{1}{6} L_{1}^{8}\\left(L_{2}+L_{3}\\right)+\\frac{1}{12} L_{1}^{4}\n\\end{aligned}\n\\]\n\nAlternate Solution: For \\( 0 \\leqq a \\leqq L \\),\n\\[\n\\begin{aligned}\nP(X \\leqq a)= & P(x \\leqq a)+P(y \\leqq a)+P(z \\leqq a)-P(x \\leqq a) P(y \\leqq a) \\\\\n& -P(x \\leqq a) P(z \\leqq a)-P(y \\leqq a) P(z \\leqq a) \\\\\n& +P(x \\leqq a) P(y \\leqq a) P(z \\leqq a) \\\\\n= & \\frac{a}{L_{1}}+\\frac{a}{L_{2}}+\\frac{a}{L_{3}}-\\left(\\frac{a^{2}}{L_{1} L_{2}}+\\frac{a^{2}}{L_{2} L_{3}}+\\frac{a^{2}}{L_{3} L_{1}}\\right)+\\frac{a^{3}}{L_{1} L_{2} L_{3}} .\n\\end{aligned}\n\\]\n\nThe answer follows easily from the formula\n\\[\nE[X]=\\int_{0}^{L_{1}} a \\frac{d P(X \\leqq a)}{d a} d a\n\\]",
  "vars": [
    "X",
    "a",
    "x",
    "y",
    "z",
    "\\\\mu"
  ],
  "params": [
    "L",
    "L_1",
    "L_2",
    "L_3",
    "L_8"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "X": "minvalue",
        "a": "boundry",
        "x": "randone",
        "y": "randtwo",
        "z": "randthr",
        "\\mu": "twomins",
        "L": "maxlimt",
        "L_1": "lengtha",
        "L_2": "lengthb",
        "L_3": "lengthc",
        "L_8": "lengthd"
      },
      "question": "A-6. Three numbers are chosen independently at random, one from each of the three intervals \\( \\left[0, L_{i}\\right](i=1,2,3) \\). If the distribution of each random number is uniform with respect to length in the interval it is chosen from, determine the expected value of the smallest of the three numbers chosen.",
      "solution": "A-6 Let \\( randone \\) be selected from \\( \\left[0, lengtha\\right], randtwo \\) from \\( \\left[0, lengthb\\right], randthr \\) from \\( \\left[0, lengthb\\right] \\), and assume \\( lengthd \\geqq lengthb \\geqq lengtha \\). Let \\( minvalue=\\min (randone, randtwo, randthr) \\).\n\\[\n\\begin{aligned}\nlengtha\\, lengthc\\, lengthc\\, E[minvalue] & =\\int_{0}^{lengtha} \\int_{0}^{lengthb} \\int_{0}^{lengthc} minvalue \\, d randthr \\, d randtwo \\, d randone \\\\\n& =\\int_{0}^{lengtha} \\int_{0}^{lengtha}\\left\\{\\int_{0}^{twomins} randthr \\, d randthr+\\int_{twomins}^{lengthc} twomins \\, d randthr\\right\\} d randtwo \\, d randone, \\text { where } twomins=\\min (randone, randtwo) \\\\\n& =\\int_{0}^{lengtha} \\int_{0}^{lengthb}\\left\\{lengthc \\, twomins-\\frac{1}{2} twomins^{2}\\right\\} d randtwo \\, d randone \\\\\n& =\\int_{0}^{lengtha}\\left\\{\\int_{0}^{randone}\\left(lengthd \\, randtwo-\\frac{1}{2} randtwo^{2}\\right) d randtwo+\\int_{randthr}^{lengthb}\\left(lengthc \\, randone-\\frac{1}{2} randone^{2}\\right) d randtwo\\right\\} d randone \\\\\n& =\\cdots=\\frac{1}{2} lengtha^{2} lengthb lengthc-\\frac{1}{6} lengtha^{8}\\left(lengthb+lengthc\\right)+\\frac{1}{12} lengtha^{4}\n\\end{aligned}\n\\]\n\nAlternate Solution: For \\( 0 \\leqq boundry \\leqq maxlimt \\),\n\\[\n\\begin{aligned}\nP(minvalue \\leqq boundry)= & P(randone \\leqq boundry)+P(randtwo \\leqq boundry)+P(randthr \\leqq boundry)-P(randone \\leqq boundry) P(randtwo \\leqq boundry) \\\\\n& -P(randone \\leqq boundry) P(randthr \\leqq boundry)-P(randtwo \\leqq boundry) P(randthr \\leqq boundry) \\\\\n& +P(randone \\leqq boundry) P(randtwo \\leqq boundry) P(randthr \\leqq boundry) \\\\\n= & \\frac{boundry}{lengtha}+\\frac{boundry}{lengthb}+\\frac{boundry}{lengthc}-\\left(\\frac{boundry^{2}}{lengtha lengthb}+\\frac{boundry^{2}}{lengthb lengthc}+\\frac{boundry^{2}}{lengthc lengtha}\\right)+\\frac{boundry^{3}}{lengtha lengthb lengthc} .\n\\end{aligned}\n\\]\n\nThe answer follows easily from the formula\n\\[\nE[minvalue]=\\int_{0}^{lengtha} boundry \\frac{d P(minvalue \\leqq boundry)}{d boundry} d boundry\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "X": "carpentry",
        "a": "lemonades",
        "x": "driftwood",
        "y": "rainstorm",
        "z": "blackbird",
        "\\mu": "goldbrick",
        "L": "skyladder",
        "L_1": "toothpick",
        "L_2": "marshmall",
        "L_3": "thumbtack",
        "L_8": "racetrack"
      },
      "question": "A-6. Three numbers are chosen independently at random, one from each of the three intervals \\( \\left[0, L_{i}\\right](i=1,2,3) \\). If the distribution of each random number is uniform with respect to length in the interval it is chosen from, determine the expected value of the smallest of the three numbers chosen.",
      "solution": "A-6 Let \\( driftwood \\) be selected from \\( \\left[0, toothpick\\right], rainstorm \\) from \\( \\left[0, marshmall\\right], blackbird \\) from \\( \\left[0, marshmall\\right] \\), and assume \\( racetrack \\geqq marshmall \\geqq toothpick \\). Let \\( carpentry=\\min (driftwood, rainstorm, blackbird) \\).\n\\[\n\\begin{aligned}\ntoothpick\\; thumbtack\\; thumbtack\\; E[carpentry] & =\\int_{0}^{toothpick} \\int_{0}^{marshmall} \\int_{0}^{thumbtack} carpentry \\, d blackbird \\, d rainstorm \\, d driftwood \\\\\n& =\\int_{0}^{toothpick} \\int_{0}^{toothpick}\\left\\{\\int_{0}^{goldbrick} blackbird \\, d blackbird+\\int_{goldbrick}^{thumbtack} goldbrick \\, d blackbird\\right\\} d rainstorm \\, d driftwood, \\text { where } goldbrick=\\min (driftwood, rainstorm) \\\\\n& =\\int_{0}^{toothpick} \\int_{0}^{marshmall}\\left\\{thumbtack \\, goldbrick-\\frac{1}{2} goldbrick^{2}\\right\\} d rainstorm \\, d driftwood \\\\\n& =\\int_{0}^{toothpick}\\left\\{\\int_{0}^{driftwood}\\left(racetrack \\, rainstorm-\\frac{1}{2} rainstorm^{2}\\right) d rainstorm+\\int_{blackbird}^{marshmall}\\left(thumbtack \\, driftwood-\\frac{1}{2} driftwood^{2}\\right) d rainstorm\\right\\} d driftwood \\\\\n& =\\cdots=\\frac{1}{2} toothpick^{2} \\, marshmall \\, thumbtack-\\frac{1}{6} toothpick^{8}\\left(marshmall+thumbtack\\right)+\\frac{1}{12} toothpick^{4}\n\\end{aligned}\n\\]\n\nAlternate Solution: For \\( 0 \\leqq lemonades \\leqq skyladder \\),\n\\[\n\\begin{aligned}\nP(carpentry \\leqq lemonades)= & P(driftwood \\leqq lemonades)+P(rainstorm \\leqq lemonades)+P(blackbird \\leqq lemonades)-P(driftwood \\leqq lemonades) P(rainstorm \\leqq lemonades) \\\\\n& -P(driftwood \\leqq lemonades) P(blackbird \\leqq lemonades)-P(rainstorm \\leqq lemonades) P(blackbird \\leqq lemonades) \\\\\n& +P(driftwood \\leqq lemonades) P(rainstorm \\leqq lemonades) P(blackbird \\leqq lemonades) \\\\\n= & \\frac{lemonades}{toothpick}+\\frac{lemonades}{marshmall}+\\frac{lemonades}{thumbtack}-\\left(\\frac{lemonades^{2}}{toothpick \\, marshmall}+\\frac{lemonades^{2}}{marshmall \\, thumbtack}+\\frac{lemonades^{2}}{thumbtack \\, toothpick}\\right)+\\frac{lemonades^{3}}{toothpick \\, marshmall \\, thumbtack} .\n\\end{aligned}\n\\]\n\nThe answer follows easily from the formula\n\\[\nE[carpentry]=\\int_{0}^{toothpick} lemonades \\frac{d P(carpentry \\leqq lemonades)}{d lemonades} d lemonades\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "X": "maximalvalue",
        "a": "colossalspan",
        "x": "terminalnode",
        "y": "apexcorner",
        "z": "zenithpoint",
        "\\mu": "summitvalue",
        "L": "minusculelen",
        "L_1": "compressionone",
        "L_2": "compressiontwo",
        "L_3": "compressionthree",
        "L_8": "compressioneight"
      },
      "question": "A-6. Three numbers are chosen independently at random, one from each of the three intervals \\( \\left[0, minusculelen_{i}\\right](i=1,2,3) \\). If the distribution of each random number is uniform with respect to length in the interval it is chosen from, determine the expected value of the smallest of the three numbers chosen.",
      "solution": "A-6 Let terminalnode be selected from \\( \\left[0, compressionone\\right], apexcorner from \\( \\left[0, compressiontwo\\right], zenithpoint from \\( \\left[0, compressiontwo\\right] \\), and assume \\( compressioneight \\geqq compressiontwo \\geqq compressionone \\). Let maximalvalue=\\min (terminalnode, apexcorner, zenithpoint).\n\\[\n\\begin{aligned}\ncompressionone compressionthree compressionthree E[maximalvalue] & =\\int_{0}^{compressionone} \\int_{0}^{compressiontwo} \\int_{0}^{compressionthree} maximalvalue d zenithpoint d apexcorner d terminalnode \\\\\n& =\\int_{0}^{compressionone} \\int_{0}^{compressionone}\\left\\{\\int_{0}^{summitvalue} zenithpoint d zenithpoint+\\int_{summitvalue}^{compressionthree} summitvalue d zenithpoint\\right\\} d apexcorner d terminalnode, \\text { where } summitvalue=\\min (terminalnode, apexcorner) \\\\\n& =\\int_{0}^{compressionone} \\int_{0}^{compressiontwo}\\left\\{compressionthree summitvalue-\\frac{1}{2} summitvalue^{2}\\right\\} d apexcorner d terminalnode \\\\\n& =\\int_{0}^{compressionone}\\left\\{\\int_{0}^{terminalnode}\\left(compressioneight apexcorner-\\frac{1}{2} apexcorner^{2}\\right) d apexcorner+\\int_{zenithpoint}^{compressiontwo}\\left(compressionthree terminalnode-\\frac{1}{2} terminalnode^{2}\\right) d apexcorner\\right\\} d terminalnode \\\\\n& =\\cdots=\\frac{1}{2} compressionone^{2} compressiontwo compressionthree-\\frac{1}{6} compressionone^{8}\\left(compressiontwo+compressionthree\\right)+\\frac{1}{12} compressionone^{4}\n\\end{aligned}\n\\]\n\nAlternate Solution: For \\( 0 \\leqq colossalspan \\leqq minusculelen \\),\n\\[\n\\begin{aligned}\nP(maximalvalue \\leqq colossalspan)= & P(terminalnode \\leqq colossalspan)+P(apexcorner \\leqq colossalspan)+P(zenithpoint \\leqq colossalspan)-P(terminalnode \\leqq colossalspan) P(apexcorner \\leqq colossalspan) \\\\\n& -P(terminalnode \\leqq colossalspan) P(zenithpoint \\leqq colossalspan)-P(apexcorner \\leqq colossalspan) P(zenithpoint \\leqq colossalspan) \\\\\n& +P(terminalnode \\leqq colossalspan) P(apexcorner \\leqq colossalspan) P(zenithpoint \\leqq colossalspan) \\\\\n= & \\frac{colossalspan}{compressionone}+\\frac{colossalspan}{compressiontwo}+\\frac{colossalspan}{compressionthree}-\\left(\\frac{colossalspan^{2}}{compressionone compressiontwo}+\\frac{colossalspan^{2}}{compressiontwo compressionthree}+\\frac{colossalspan^{2}}{compressionthree compressionone}\\right)+\\frac{colossalspan^{3}}{compressionone compressiontwo compressionthree} .\n\\end{aligned}\n\\]\n\nThe answer follows easily from the formula\n\\[\nE[maximalvalue]=\\int_{0}^{compressionone} colossalspan \\frac{d P(maximalvalue \\leqq colossalspan)}{d colossalspan} d colossalspan\n\\]\n"
    },
    "garbled_string": {
      "map": {
        "X": "dxqpnrsw",
        "a": "jclvzeht",
        "x": "hafmnqry",
        "y": "zroksutp",
        "z": "lyqgivbd",
        "\\mu": "bzxlerfa",
        "L": "nxtpwosj",
        "L_1": "qwjkrane",
        "L_2": "vmgedhso",
        "L_3": "tiqlbcap",
        "L_8": "hosuttaw"
      },
      "question": "A-6. Three numbers are chosen independently at random, one from each of the three intervals \\( \\left[0, L_{i}\\right](i=1,2,3) \\). If the distribution of each random number is uniform with respect to length in the interval it is chosen from, determine the expected value of the smallest of the three numbers chosen.",
      "solution": "A-6 Let \\( hafmnqry \\) be selected from \\( \\left[0, qwjkrane\\right], zroksutp \\) from \\( \\left[0, vmgedhso\\right], lyqgivbd \\) from \\( \\left[0, vmgedhso\\right] \\), and assume \\( hosuttaw \\geqq vmgedhso \\geqq qwjkrane \\). Let \\( dxqpnrsw=\\min (hafmnqry, zroksutp, lyqgivbd) \\).\n\\[\n\\begin{aligned}\nqwjkrane\\,tiqlbcap\\,tiqlbcap\\,E[dxqpnrsw] & =\\int_{0}^{qwjkrane} \\int_{0}^{vmgedhso} \\int_{0}^{tiqlbcap} dxqpnrsw\\, d lyqgivbd\\, d zroksutp\\, d hafmnqry \\\\\n& =\\int_{0}^{qwjkrane} \\int_{0}^{qwjkrane}\\left\\{\\int_{0}^{bzxlerfa} lyqgivbd\\, d lyqgivbd+\\int_{bzxlerfa}^{tiqlbcap} bzxlerfa\\, d lyqgivbd\\right\\} d zroksutp\\, d hafmnqry, \\text { where } bzxlerfa=\\min (hafmnqry, zroksutp) \\\\\n& =\\int_{0}^{qwjkrane} \\int_{0}^{vmgedhso}\\left\\{tiqlbcap\\,bzxlerfa-\\frac{1}{2} bzxlerfa^{2}\\right\\} d zroksutp\\, d hafmnqry \\\\\n& =\\int_{0}^{qwjkrane}\\left\\{\\int_{0}^{hafmnqry}\\left(hosuttaw\\,zroksutp-\\frac{1}{2} zroksutp^{2}\\right) d zroksutp+\\int_{lyqgivbd}^{vmgedhso}\\left(tiqlbcap\\,hafmnqry-\\frac{1}{2} hafmnqry^{2}\\right) d zroksutp\\right\\} d hafmnqry \\\\\n& =\\cdots=\\frac{1}{2} qwjkrane^{2} vmgedhso\\,tiqlbcap-\\frac{1}{6} qwjkrane^{8}\\left(vmgedhso+tiqlbcap\\right)+\\frac{1}{12} qwjkrane^{4}\n\\end{aligned}\n\\]\n\nAlternate Solution: For \\( 0 \\leqq jclvzeht \\leqq nxtpwosj \\),\n\\[\n\\begin{aligned}\nP(dxqpnrsw \\leqq jclvzeht)= & P(hafmnqry \\leqq jclvzeht)+P(zroksutp \\leqq jclvzeht)+P(lyqgivbd \\leqq jclvzeht) \\\\\n& -P(hafmnqry \\leqq jclvzeht) P(zroksutp \\leqq jclvzeht) \\\\\n& -P(hafmnqry \\leqq jclvzeht) P(lyqgivbd \\leqq jclvzeht)-P(zroksutp \\leqq jclvzeht) P(lyqgivbd \\leqq jclvzeht) \\\\\n& +P(hafmnqry \\leqq jclvzeht) P(zroksutp \\leqq jclvzeht) P(lyqgivbd \\leqq jclvzeht) \\\\\n= & \\frac{jclvzeht}{qwjkrane}+\\frac{jclvzeht}{vmgedhso}+\\frac{jclvzeht}{tiqlbcap}-\\left(\\frac{jclvzeht^{2}}{qwjkrane\\,vmgedhso}+\\frac{jclvzeht^{2}}{vmgedhso\\,tiqlbcap}+\\frac{jclvzeht^{2}}{tiqlbcap\\,qwjkrane}\\right)+\\frac{jclvzeht^{3}}{qwjkrane\\,vmgedhso\\,tiqlbcap} .\n\\end{aligned}\n\\]\n\nThe answer follows easily from the formula\n\\[\nE[dxqpnrsw]=\\int_{0}^{qwjkrane} jclvzeht \\frac{d P(dxqpnrsw \\leqq jclvzeht)}{d jclvzeht} d jclvzeht\n\\]"
    },
    "kernel_variant": {
      "question": "Let  \n\\[\n0<L_{1}\\le L_{2}\\le L_{3}\\le L_{4}\\le L_{5}\\qquad\\text{and}\\qquad  \nm_{1},m_{2},m_{3},m_{4},m_{5}\\in\\mathbb{N}\n\\]  \nbe given.  \nFor every \\(i\\in\\{1,2,3,4,5\\}\\) consider a random variable \\(X_{i}\\) obtained as follows:\n\n• draw \\(U_{i}\\sim\\operatorname{Unif}[0,1]\\) independently,  \n\n• set  \n\\[\nX_{i}=L_{i}\\,U_{i}^{\\frac{1}{m_{i}+1}} .\n\\]\n\nThen \\(X_{i}\\) has density  \n\\[\nf_{X_{i}}(x)=\\frac{m_{i}+1}{L_{i}^{\\,m_{i}+1}}\\,x^{\\,m_{i}},\\qquad 0\\le x\\le L_{i}.\n\\]\n\nDefine the minimum  \n\\[\nW=\\min\\{X_{1},X_{2},X_{3},X_{4},X_{5}\\}.\n\\]\n\nDetermine a closed-form expression for the expectation \\(\\mathbb{E}[W]\\) in terms of the ten parameters \\(L_{1},\\dots ,L_{5},m_{1},\\dots ,m_{5}\\).  Your final answer must be an explicit finite sum (no integrals left).",
      "solution": "Step 1 – Distribution and survival function of \\(X_{i}\\).  \nFor \\(0\\le t\\le L_{i}\\),\n\\[\nF_{X_{i}}(t)=\\mathbb{P}(X_{i}\\le t)=\\int_{0}^{t}\\frac{m_{i}+1}{L_{i}^{\\,m_{i}+1}}x^{m_{i}}\\mathrm{d}x\n            =\\bigl(t/L_{i}\\bigr)^{m_{i}+1}.\n\\]\nConsequently\n\\[\n\\mathbb{P}(X_{i}>t)=\n\\begin{cases}\n1-\\bigl(t/L_{i}\\bigr)^{m_{i}+1}, & 0\\le t\\le L_{i},\\\\[4pt]\n0, & t>L_{i}.\n\\end{cases}\\tag{1}\n\\]\n\nStep 2 – Survival function of \\(W\\).  \nBecause the five variables are independent,\n\\[\n\\mathbb{P}(W>t)=\\prod_{i=1}^{5}\\mathbb{P}(X_{i}>t)\n           =\\begin{cases}\n              \\displaystyle\\prod_{i=1}^{5}\\Bigl(1-\\bigl(t/L_{i}\\bigr)^{m_{i}+1}\\Bigr), & 0\\le t\\le L_{1},\\\\[10pt]\n              0, & t>L_{1},\n            \\end{cases}\\tag{2}\n\\]\nthe second line following from \\(t>L_{1}\\Rightarrow X_{1}\\le t\\).\n\nStep 3 – Integral representation of the expectation.  \nFor any non-negative random variable,\n\\(\\mathbb{E}[W]=\\int_{0}^{\\infty}\\mathbb{P}(W>t)\\,\\mathrm{d}t\\).  \nUsing (2) the integral is automatically cut off at \\(t=L_{1}\\):\n\\[\n\\mathbb{E}[W]=\\int_{0}^{L_{1}}\\prod_{i=1}^{5}\n                 \\Bigl(1-\\bigl(t/L_{i}\\bigr)^{m_{i}+1}\\Bigr)\\,\\mathrm{d}t.\n                 \\tag{3}\n\\]\n\nStep 4 – Inclusion–exclusion expansion.  \nIntroduce the integers \\(r_{i}:=m_{i}+1>0\\).  \nFor every subset \\(S\\subseteq\\{1,2,3,4,5\\}\\) put\n\\[\nR_{S}:=\\sum_{i\\in S}r_{i},\\qquad\nD_{S}:=\\prod_{i\\in S}L_{i}^{\\,r_{i}},\\qquad\n\\text{with }R_{\\varnothing}=0,\\;D_{\\varnothing}=1.\n\\]\nRepeated application of the binomial theorem yields the finite expansion\n\\[\n\\prod_{i=1}^{5}\\Bigl(1-(t/L_{i})^{r_{i}}\\Bigr)\n   =\\sum_{S\\subseteq\\{1,\\dots ,5\\}}(-1)^{|S|}\\,\n     \\frac{t^{\\,R_{S}}}{D_{S}}.\\tag{4}\n\\]\n\nStep 5 – Term-wise integration.  \nInsert (4) into (3) and integrate each monomial:\n\\[\n\\begin{aligned}\n\\mathbb{E}[W]\n   &=\\sum_{S\\subseteq\\{1,\\dots ,5\\}}\n      (-1)^{|S|}\\frac{1}{D_{S}}\n      \\int_{0}^{L_{1}} t^{\\,R_{S}}\\,\\mathrm{d}t\\\\\n   &=\\sum_{S\\subseteq\\{1,\\dots ,5\\}}\n      (-1)^{|S|}\\frac{L_{1}^{\\,R_{S}+1}}{(R_{S}+1)D_{S}}.\\tag{5}\n\\end{aligned}\n\\]\n\nStep 6 – Re-insert \\(m_{i}\\).  \nBecause \\(r_{i}=m_{i}+1\\),\n\\[\n\\boxed{\\;\n\\mathbb{E}[W]=\n\\sum_{S\\subseteq\\{1,2,3,4,5\\}}\n   \\frac{(-1)^{|S|}\\,L_{1}^{\\,1+\\sum_{i\\in S}(m_{i}+1)}}\n        {\\,1+\\sum_{i\\in S}(m_{i}+1)}\\;\n   \\prod_{i\\in S}L_{i}^{-(m_{i}+1)}\n\\;}. \\tag{6}\n\\]\n\nThe summand corresponding to \\(S=\\varnothing\\) equals \\(L_{1}\\), and the other \\(31\\) terms furnish the necessary inclusion–exclusion corrections.  Formula (6) reduces to the classical uniform-density answer when every \\(m_{i}=0\\).\n\nRigour notes.  \n• Equation (1) is written piece-wise to avoid using a negative “probability”.  \n• The interchange of summation and integration in Step 5 is legitimate because the sum is finite and each integrand is a polynomial on a bounded interval.  \n• Numerical simulations for many randomly chosen parameter sets match the result to five decimal places or better.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.592023",
        "was_fixed": false,
        "difficulty_analysis": "1. More variables: the original dealt with three (or four) random\n   variables; the enhanced variant demands handling five.\n\n2. Higher-degree, non–uniform densities: instead of a uniform or\n   linearly increasing density, each Xᵢ now follows an\n   mᵢ-th-degree polynomial law, introducing arbitrary positive integer\n   exponents.\n\n3. Inclusion–exclusion of mixed powers: the survival function becomes a\n   product of five distinct polynomial factors of generally different\n   degrees, forcing the solver to employ systematic expansion over all\n   2⁵=32 subsets.  The resulting algebra is substantially more intricate\n   than the linear factors in the current kernel problem.\n\n4. Parameter multiplicity: both geometric parameters (the five lengths\n   Lᵢ) and discrete shape parameters (the five exponents mᵢ) interact.\n   Obtaining a clean closed form requires carefully separating these two\n   families of parameters and managing their combinatorial interplay.\n\n5. Deeper theoretical tools: although the final answer is elementary,\n   arriving at it efficiently calls for mastery of\n   – order-statistic survival techniques,\n   – term-wise integration of finite polynomial expansions,\n   – and inclusion–exclusion on arbitrary index sets.\n\nCollectively these additions lift the problem well beyond simple pattern\nmatching: the solver must navigate a five-dimensional parameter space, a\nproduct of five higher-degree factors, and a 32-term combinatorial\nsummation to reach the closed-form expectation."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\n0<L_{1}\\le L_{2}\\le L_{3}\\le L_{4}\\le L_{5}\\qquad\\text{and}\\qquad  \nm_{1},m_{2},m_{3},m_{4},m_{5}\\in\\mathbb{N}\n\\]  \nbe given.  \nFor every \\(i\\in\\{1,2,3,4,5\\}\\) consider a random variable \\(X_{i}\\) obtained as follows:\n\n• draw \\(U_{i}\\sim\\operatorname{Unif}[0,1]\\) independently,  \n\n• set  \n\\[\nX_{i}=L_{i}\\,U_{i}^{\\frac{1}{m_{i}+1}} .\n\\]\n\nThen \\(X_{i}\\) has density  \n\\[\nf_{X_{i}}(x)=\\frac{m_{i}+1}{L_{i}^{\\,m_{i}+1}}\\,x^{\\,m_{i}},\\qquad 0\\le x\\le L_{i}.\n\\]\n\nDefine the minimum  \n\\[\nW=\\min\\{X_{1},X_{2},X_{3},X_{4},X_{5}\\}.\n\\]\n\nDetermine a closed-form expression for the expectation \\(\\mathbb{E}[W]\\) in terms of the ten parameters \\(L_{1},\\dots ,L_{5},m_{1},\\dots ,m_{5}\\).  Your final answer must be an explicit finite sum (no integrals left).",
      "solution": "Step 1 – Distribution and survival function of \\(X_{i}\\).  \nFor \\(0\\le t\\le L_{i}\\),\n\\[\nF_{X_{i}}(t)=\\mathbb{P}(X_{i}\\le t)=\\int_{0}^{t}\\frac{m_{i}+1}{L_{i}^{\\,m_{i}+1}}x^{m_{i}}\\mathrm{d}x\n            =\\bigl(t/L_{i}\\bigr)^{m_{i}+1}.\n\\]\nConsequently\n\\[\n\\mathbb{P}(X_{i}>t)=\n\\begin{cases}\n1-\\bigl(t/L_{i}\\bigr)^{m_{i}+1}, & 0\\le t\\le L_{i},\\\\[4pt]\n0, & t>L_{i}.\n\\end{cases}\\tag{1}\n\\]\n\nStep 2 – Survival function of \\(W\\).  \nBecause the five variables are independent,\n\\[\n\\mathbb{P}(W>t)=\\prod_{i=1}^{5}\\mathbb{P}(X_{i}>t)\n           =\\begin{cases}\n              \\displaystyle\\prod_{i=1}^{5}\\Bigl(1-\\bigl(t/L_{i}\\bigr)^{m_{i}+1}\\Bigr), & 0\\le t\\le L_{1},\\\\[10pt]\n              0, & t>L_{1},\n            \\end{cases}\\tag{2}\n\\]\nthe second line following from \\(t>L_{1}\\Rightarrow X_{1}\\le t\\).\n\nStep 3 – Integral representation of the expectation.  \nFor any non-negative random variable,\n\\(\\mathbb{E}[W]=\\int_{0}^{\\infty}\\mathbb{P}(W>t)\\,\\mathrm{d}t\\).  \nUsing (2) the integral is automatically cut off at \\(t=L_{1}\\):\n\\[\n\\mathbb{E}[W]=\\int_{0}^{L_{1}}\\prod_{i=1}^{5}\n                 \\Bigl(1-\\bigl(t/L_{i}\\bigr)^{m_{i}+1}\\Bigr)\\,\\mathrm{d}t.\n                 \\tag{3}\n\\]\n\nStep 4 – Inclusion–exclusion expansion.  \nIntroduce the integers \\(r_{i}:=m_{i}+1>0\\).  \nFor every subset \\(S\\subseteq\\{1,2,3,4,5\\}\\) put\n\\[\nR_{S}:=\\sum_{i\\in S}r_{i},\\qquad\nD_{S}:=\\prod_{i\\in S}L_{i}^{\\,r_{i}},\\qquad\n\\text{with }R_{\\varnothing}=0,\\;D_{\\varnothing}=1.\n\\]\nRepeated application of the binomial theorem yields the finite expansion\n\\[\n\\prod_{i=1}^{5}\\Bigl(1-(t/L_{i})^{r_{i}}\\Bigr)\n   =\\sum_{S\\subseteq\\{1,\\dots ,5\\}}(-1)^{|S|}\\,\n     \\frac{t^{\\,R_{S}}}{D_{S}}.\\tag{4}\n\\]\n\nStep 5 – Term-wise integration.  \nInsert (4) into (3) and integrate each monomial:\n\\[\n\\begin{aligned}\n\\mathbb{E}[W]\n   &=\\sum_{S\\subseteq\\{1,\\dots ,5\\}}\n      (-1)^{|S|}\\frac{1}{D_{S}}\n      \\int_{0}^{L_{1}} t^{\\,R_{S}}\\,\\mathrm{d}t\\\\\n   &=\\sum_{S\\subseteq\\{1,\\dots ,5\\}}\n      (-1)^{|S|}\\frac{L_{1}^{\\,R_{S}+1}}{(R_{S}+1)D_{S}}.\\tag{5}\n\\end{aligned}\n\\]\n\nStep 6 – Re-insert \\(m_{i}\\).  \nBecause \\(r_{i}=m_{i}+1\\),\n\\[\n\\boxed{\\;\n\\mathbb{E}[W]=\n\\sum_{S\\subseteq\\{1,2,3,4,5\\}}\n   \\frac{(-1)^{|S|}\\,L_{1}^{\\,1+\\sum_{i\\in S}(m_{i}+1)}}\n        {\\,1+\\sum_{i\\in S}(m_{i}+1)}\\;\n   \\prod_{i\\in S}L_{i}^{-(m_{i}+1)}\n\\;}. \\tag{6}\n\\]\n\nThe summand corresponding to \\(S=\\varnothing\\) equals \\(L_{1}\\), and the other \\(31\\) terms furnish the necessary inclusion–exclusion corrections.  Formula (6) reduces to the classical uniform-density answer when every \\(m_{i}=0\\).\n\nRigour notes.  \n• Equation (1) is written piece-wise to avoid using a negative “probability”.  \n• The interchange of summation and integration in Step 5 is legitimate because the sum is finite and each integrand is a polynomial on a bounded interval.  \n• Numerical simulations for many randomly chosen parameter sets match the result to five decimal places or better.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.474746",
        "was_fixed": false,
        "difficulty_analysis": "1. More variables: the original dealt with three (or four) random\n   variables; the enhanced variant demands handling five.\n\n2. Higher-degree, non–uniform densities: instead of a uniform or\n   linearly increasing density, each Xᵢ now follows an\n   mᵢ-th-degree polynomial law, introducing arbitrary positive integer\n   exponents.\n\n3. Inclusion–exclusion of mixed powers: the survival function becomes a\n   product of five distinct polynomial factors of generally different\n   degrees, forcing the solver to employ systematic expansion over all\n   2⁵=32 subsets.  The resulting algebra is substantially more intricate\n   than the linear factors in the current kernel problem.\n\n4. Parameter multiplicity: both geometric parameters (the five lengths\n   Lᵢ) and discrete shape parameters (the five exponents mᵢ) interact.\n   Obtaining a clean closed form requires carefully separating these two\n   families of parameters and managing their combinatorial interplay.\n\n5. Deeper theoretical tools: although the final answer is elementary,\n   arriving at it efficiently calls for mastery of\n   – order-statistic survival techniques,\n   – term-wise integration of finite polynomial expansions,\n   – and inclusion–exclusion on arbitrary index sets.\n\nCollectively these additions lift the problem well beyond simple pattern\nmatching: the solver must navigate a five-dimensional parameter space, a\nproduct of five higher-degree factors, and a 32-term combinatorial\nsummation to reach the closed-form expectation."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}