{
  "index": "2022-A-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "COMB"
  ],
  "difficulty": "",
  "question": "Suppose that $X_1, X_2, \\dots$ are real numbers between 0 and 1 that are chosen independently and uniformly at random. Let $S = \\sum_{i=1}^k X_i/2^i$, where $k$ is the least positive integer such that $X_k < X_{k+1}$, or $k = \\infty$ if there is no such integer. Find the expected value of $S$.",
  "solution": "The expected value is $2e^{1/2}-3$.\n\nExtend $S$ to an infinite sum by including zero summands for $i> k$. We may then compute the expected value as the sum of the expected value of the $i$-th summand over all $i$. This summand\noccurs if and only if $X_1,\\dots,X_{i-1} \\in [X_i, 1]$\nand $X_1,\\dots,X_{i-1}$ occur in nonincreasing order. These two events are independent and occur with respective probabilities $(1-X_i)^{i-1}$ and $\\frac{1}{(i-1)!}$; the expectation of this summand is therefore\n\\begin{align*}\n&\\frac{1}{2^i(i-1)!} \\int_0^1 t (1-t)^{i-1}\\,dt \\\\\n&\\qquad = \\frac{1}{2^i(i-1)!} \\int_0^1 ((1-t)^{i-1} - (1-t)^i)\\,dt \\\\\n&\\qquad = \\frac{1}{2^i(i-1)!} \\left( \\frac{1}{i} - \\frac{1}{i+1} \\right) = \\frac{1}{2^i (i+1)!}.\n\\end{align*}\nSumming over $i$, we obtain\n\\[\n\\sum_{i=1}^\\infty \\frac{1}{2^i (i+1)!}\n= 2 \\sum_{i=2}^\\infty \\frac{1}{2^i i!}\n= 2\\left(e^{1/2}-1-\\frac{1}{2} \\right).\n\\]",
  "vars": [
    "S",
    "k",
    "i",
    "t",
    "X_1",
    "X_2",
    "X_i",
    "X_i-1",
    "X_k",
    "X_k+1"
  ],
  "params": [],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "S": "sumvalue",
        "k": "stopindex",
        "i": "indexvar",
        "t": "integrandvar",
        "X_1": "randone",
        "X_2": "randtwo",
        "X_i": "randgen",
        "X_i-1": "randprev",
        "X_k": "randstop",
        "X_k+1": "randstopnext"
      },
      "question": "Suppose that $\\text{randone}, \\text{randtwo}, \\dots$ are real numbers between 0 and 1 that are chosen independently and uniformly at random. Let $\\text{sumvalue} = \\sum_{\\text{indexvar}=1}^{\\text{stopindex}} \\text{randgen}/2^{\\text{indexvar}}$, where $\\text{stopindex}$ is the least positive integer such that $\\text{randstop} < \\text{randstopnext}$, or $\\text{stopindex} = \\infty$ if there is no such integer. Find the expected value of $\\text{sumvalue}$. ",
      "solution": "The expected value is $2e^{1/2}-3$.\n\nExtend $\\text{sumvalue}$ to an infinite sum by including zero summands for $\\text{indexvar}>\\text{stopindex}$. We may then compute the expected value as the sum of the expected value of the $\\text{indexvar}$-th summand over all $\\text{indexvar}$. This summand occurs if and only if $\\text{randone},\\dots,\\text{randprev} \\in [\\text{randgen}, 1]$ and $\\text{randone},\\dots,\\text{randprev}$ occur in nonincreasing order. These two events are independent and occur with respective probabilities $(1-\\text{randgen})^{\\text{indexvar}-1}$ and $\\frac{1}{(\\text{indexvar}-1)!}$; the expectation of this summand is therefore\n\\begin{align*}\n&\\frac{1}{2^{\\text{indexvar}}(\\text{indexvar}-1)!} \\int_0^1 \\text{integrandvar} (1-\\text{integrandvar})^{\\text{indexvar}-1}\\,d\\text{integrandvar} \\\\\n&\\qquad = \\frac{1}{2^{\\text{indexvar}}(\\text{indexvar}-1)!} \\int_0^1 ((1-\\text{integrandvar})^{\\text{indexvar}-1} - (1-\\text{integrandvar})^{\\text{indexvar}})\\,d\\text{integrandvar} \\\\\n&\\qquad = \\frac{1}{2^{\\text{indexvar}}(\\text{indexvar}-1)!} \\left( \\frac{1}{\\text{indexvar}} - \\frac{1}{\\text{indexvar}+1} \\right) = \\frac{1}{2^{\\text{indexvar}} (\\text{indexvar}+1)!}.\n\\end{align*}\nSumming over $\\text{indexvar}$, we obtain\n\\[\n\\sum_{\\text{indexvar}=1}^\\infty \\frac{1}{2^{\\text{indexvar}} (\\text{indexvar}+1)!}\n= 2 \\sum_{\\text{indexvar}=2}^\\infty \\frac{1}{2^{\\text{indexvar}} \\text{indexvar}!}\n= 2\\left(e^{1/2}-1-\\frac{1}{2} \\right).\n\\]\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "S": "shoelaces",
        "k": "dandelion",
        "i": "butterfly",
        "t": "lavender",
        "X_1": "honeycomb",
        "X_2": "salamander",
        "X_i": "raincloud",
        "X_i-1": "peppermint",
        "X_k": "watermelon",
        "X_k+1": "stargazers"
      },
      "question": "Suppose that $honeycomb, salamander, \\dots$ are real numbers between 0 and 1 that are chosen independently and uniformly at random. Let $shoelaces = \\sum_{butterfly=1}^{dandelion} raincloud/2^{butterfly}$, where $dandelion$ is the least positive integer such that $watermelon < stargazers$, or $dandelion = \\infty$ if there is no such integer. Find the expected value of $shoelaces$.",
      "solution": "The expected value is $2e^{1/2}-3$.\n\nExtend $shoelaces$ to an infinite sum by including zero summands for $butterfly> dandelion$. We may then compute the expected value as the sum of the expected value of the $butterfly$-th summand over all $butterfly$. This summand\noccurs if and only if $honeycomb,\\dots,peppermint \\in [raincloud, 1]$\nand $honeycomb,\\dots,peppermint$ occur in nonincreasing order. These two events are independent and occur with respective probabilities $(1-raincloud)^{butterfly-1}$ and $\\frac{1}{(butterfly-1)!}$; the expectation of this summand is therefore\n\\begin{align*}\n&\\frac{1}{2^{butterfly}(butterfly-1)!} \\int_0^1 lavender (1-lavender)^{butterfly-1}\\,d lavender \\\\\n&\\qquad = \\frac{1}{2^{butterfly}(butterfly-1)!} \\int_0^1 ((1-lavender)^{butterfly-1} - (1-lavender)^{butterfly})\\,d lavender \\\\\n&\\qquad = \\frac{1}{2^{butterfly}(butterfly-1)!} \\left( \\frac{1}{butterfly} - \\frac{1}{butterfly+1} \\right) = \\frac{1}{2^{butterfly} (butterfly+1)!}.\n\\end{align*}\nSumming over $butterfly$, we obtain\n\\[\n\\sum_{butterfly=1}^{\\infty} \\frac{1}{2^{butterfly} (butterfly+1)!}\n= 2 \\sum_{butterfly=2}^{\\infty} \\frac{1}{2^{butterfly} butterfly!}\n= 2\\left(e^{1/2}-1-\\frac{1}{2} \\right).\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "S": "differencevalue",
        "k": "largestindex",
        "i": "fixedindex",
        "t": "constantvalue",
        "X_1": "lastdeterministic",
        "X_2": "seconddeterministic",
        "X_i": "deterministicfixed",
        "X_{i-1}": "deterministicprevious",
        "X_k": "deterministicterminal",
        "X_{k+1}": "deterministicnext"
      },
      "question": "Suppose that $lastdeterministic, seconddeterministic, \\dots$ are real numbers between 0 and 1 that are chosen independently and uniformly at random. Let $differencevalue = \\sum_{fixedindex=1}^{largestindex} deterministicfixed/2^{fixedindex}$, where $largestindex$ is the least positive integer such that $deterministicterminal < deterministicnext$, or $largestindex = \\infty$ if there is no such integer. Find the expected value of $differencevalue$.",
      "solution": "The expected value is $2e^{1/2}-3$.\n\nExtend $differencevalue$ to an infinite sum by including zero summands for $fixedindex> largestindex$. We may then compute the expected value as the sum of the expected value of the $fixedindex$-th summand over all $fixedindex$. This summand\noccurs if and only if $lastdeterministic,\\dots,deterministicprevious \\in [deterministicfixed, 1]$\nand $lastdeterministic,\\dots,deterministicprevious$ occur in nonincreasing order. These two events are independent and occur with respective probabilities $(1-deterministicfixed)^{fixedindex-1}$ and $\\frac{1}{(fixedindex-1)!}$; the expectation of this summand is therefore\n\\begin{align*}\n&\\frac{1}{2^{fixedindex}(fixedindex-1)!} \\int_0^1 constantvalue (1-constantvalue)^{fixedindex-1}\\,dconstantvalue \\\\\n&\\qquad = \\frac{1}{2^{fixedindex}(fixedindex-1)!} \\int_0^1 ((1-constantvalue)^{fixedindex-1} - (1-constantvalue)^{fixedindex})\\,dconstantvalue \\\\\n&\\qquad = \\frac{1}{2^{fixedindex}(fixedindex-1)!} \\left( \\frac{1}{fixedindex} - \\frac{1}{fixedindex+1} \\right) = \\frac{1}{2^{fixedindex} (fixedindex+1)!}.\n\\end{align*}\nSumming over $fixedindex$, we obtain\n\\[\n\\sum_{fixedindex=1}^\\infty \\frac{1}{2^{fixedindex} (fixedindex+1)!}\n= 2 \\sum_{fixedindex=2}^\\infty \\frac{1}{2^{fixedindex} fixedindex!}\n= 2\\left(e^{1/2}-1-\\frac{1}{2} \\right).\n\\]"
    },
    "garbled_string": {
      "map": {
        "S": "zntqmhvra",
        "k": "blsyrvqeo",
        "i": "pchndkowm",
        "t": "ukvramcqs",
        "X_1": "ebdqlxfro",
        "X_2": "jluzsaktp",
        "X_i": "qrvmbgsca",
        "X_i-1": "xjspeqnro",
        "X_k": "vgruczwhb",
        "X_k+1": "trmoyhnaz"
      },
      "question": "Suppose that $ebdqlxfro, jluzsaktp, \\dots$ are real numbers between 0 and 1 that are chosen independently and uniformly at random. Let $zntqmhvra = \\sum_{pchndkowm=1}^{blsyrvqeo} qrvmbgsca/2^{pchndkowm}$, where $blsyrvqeo$ is the least positive integer such that $vgruczwhb < trmoyhnaz$, or $blsyrvqeo = \\infty$ if there is no such integer. Find the expected value of $zntqmhvra$.",
      "solution": "The expected value is $2e^{1/2}-3$.\n\nExtend $zntqmhvra$ to an infinite sum by including zero summands for $pchndkowm> blsyrvqeo$. We may then compute the expected value as the sum of the expected value of the $pchndkowm$-th summand over all $pchndkowm$. This summand\noccurs if and only if $ebdqlxfro,\\dots,xjspeqnro \\in [qrvmbgsca, 1]$\nand $ebdqlxfro,\\dots,xjspeqnro$ occur in nonincreasing order. These two events are independent and occur with respective probabilities $(1-qrvmbgsca)^{pchndkowm-1}$ and $\\frac{1}{(pchndkowm-1)!}$; the expectation of this summand is therefore\n\\begin{align*}\n&\\frac{1}{2^{pchndkowm}(pchndkowm-1)!} \\int_0^1 ukvramcqs (1-ukvramcqs)^{pchndkowm-1}\\,d ukvramcqs \\\\\n&\\qquad = \\frac{1}{2^{pchndkowm}(pchndkowm-1)!} \\int_0^1 \\bigl((1-ukvramcqs)^{pchndkowm-1} - (1-ukvramcqs)^{pchndkowm}\\bigr)\\,d ukvramcqs \\\\\n&\\qquad = \\frac{1}{2^{pchndkowm}(pchndkowm-1)!} \\left( \\frac{1}{pchndkowm} - \\frac{1}{pchndkowm+1} \\right) = \\frac{1}{2^{pchndkowm} (pchndkowm+1)!}.\n\\end{align*}\nSumming over $pchndkowm$, we obtain\n\\[\n\\sum_{pchndkowm=1}^\\infty \\frac{1}{2^{pchndkowm} (pchndkowm+1)!}\n= 2 \\sum_{pchndkowm=2}^\\infty \\frac{1}{2^{pchndkowm} pchndkowm!}\n= 2\\left(e^{1/2}-1-\\frac{1}{2} \\right).\n\\]"
    },
    "kernel_variant": {
      "question": "Fix an integer d ≥ 2, an integer r ≥ 1, and a real number p > 1.  \nFor every n = 1,2,… draw d independent random numbers  \n\n  Xₙ,₁ , Xₙ,₂ , … , Xₙ,d  ~ Unif[0,1]\n\nand put  \n\n  Mₙ := max{Xₙ,₁,…,Xₙ,d}.  \n\nLet  \n\n  k := min{ n ≥ 1 : Mₙ < Mₙ₊₁ }  (k = ∞ if the inequality never occurs)\n\nand define the random sum  \n\n  S := ∑_{i=1}^{k} Mᵢ^{\\,r}/p^{\\,i}.  \n\nDetermine the exact value of 𝔼[S] as a function of d, r and p.\n\n\n\n------------------------------------------------------------------",
      "solution": "Notation.  Put  \n\n  α := 1 + r/d,  z := 1/p  (0 < z < 1).                                                         (0)\n\nThroughout (a)ₙ := a(a+1)…(a+n–1) denotes the Pochhammer symbol.\n\nStep 1.  Distribution of one block maximum.  \nFor a single block of d independent Unif[0,1] variables,\n\n  P(Mₙ ≤ t) = t^{d},  0 ≤ t ≤ 1,\n\nso Mₙ has density f(t) = d t^{d-1}.  The variables (M₁,M₂,…) are i.i.d.\n\nStep 2.  Decomposing S with the monotone-chain events.  \nDefine  \n\n  Aᵢ := {M₁ ≥ M₂ ≥ ⋯ ≥ Mᵢ}.  \n\nBecause S contributes the i-th summand precisely when Aᵢ occurs,\n\n  S = ∑_{i=1}^{∞} 1_{Aᵢ}\\,Mᵢ^{r}/p^{\\,i}, \n  𝔼[S] = ∑_{i=1}^{∞} 𝔼[1_{Aᵢ}Mᵢ^{r}]/p^{\\,i}.                                        (1)\n\nStep 3.  Computing 𝔼[1_{Aᵢ}Mᵢ^{r}].  \nCondition on Mᵢ = t.  \n• The (i–1) earlier maxima are each ≥ t with probability (1–t^{d}), hence jointly (1–t^{d})^{i-1}.  \n• Given those values, all (i–1)! orders are equally likely and exactly one is non-increasing.  \n\nThus  \n\n  P(Aᵢ | Mᵢ = t) = (1–t^{d})^{i-1}/(i−1)!.  \n\nMultiplying by t^{r} and the density d t^{d-1} and integrating,\n\n  𝔼[1_{Aᵢ}Mᵢ^{r}]\n      = d/(i−1)! ∫₀¹ t^{r+d-1}(1−t^{d})^{i-1}dt.                                     (2)\n\nStep 4.  Beta-integral.  Substitute u = t^{d} (so dt = u^{1/d−1}/d du):\n\n  (2) = d/(i−1)!·1/d ∫₀¹ u^{r/d}(1−u)^{i-1}du  \n      = 1/(i−1)!·B(r/d+1,i)  \n      = Γ(r/d+1) / Γ(r/d+1+i).                                                      (3)\n\nStep 5.  Series for 𝔼[S].  \nInsert (3) into (1) and cancel the common Γ-factor:\n\n  𝔼[S] = ∑_{i=1}^{∞} z^{\\,i}/(α)_{i}.                                               (4)\n\nStep 6.  Special-function identification.  \nFor a = 1 the Kummer confluent hypergeometric series Φ(a,b;z) is\n\n  Φ(1,α;z) = ∑_{i=0}^{∞} z^{\\,i}/(α)_{i}.  \n\nHence\n\n  𝔼[S] = Φ(1,α;z) − 1 = Φ(1, 1 + r/d; 1/p) − 1.                                   (5)\n\nThis is a closed form valid for all d ≥ 2, r ≥ 1 and p > 1.\n\nStep 7.  Integer-parameter simplification (corrected).  \nAssume r is an exact multiple of d, say r = m d with integer m ≥ 1, so α = m+1.  \nBecause (m+1)_{i} = (i+m)!/m! we obtain\n\n  ∑_{i=1}^{∞} z^{i}/(m+1)_{i}\n      = m! z^{−m} ∑_{i=1}^{∞} z^{i+m}/(i+m)!\n      = m! z^{−m} ( e^{z} − ∑_{j=0}^{m} z^{\\,j}/j! ).                              (6)\n\nConsequently\n\n  𝔼[S] = m! p^{\\,m}\\Bigl( e^{1/p} − ∑_{j=0}^{m} (1/p)^{j}/j! \\Bigr).               (7)\n\nEquation (7) is the correct elementary expression for integer multiples of d.  \nFor m = 1 (i.e. r = d) it yields 𝔼[S] = p( e^{1/p} − 1 − 1/p).\n\nNumerical check (m = 3, p = 10):  \ne^{0.1} ≈ 1.105 170 918, ∑_{j=0}^{3}0.1^{j}/j! ≈ 1.105 166 667,  \ndifference ≈ 4.251 × 10^{−6}.  \nMultiply by m! p^{m} = 6 × 10^{3} to get 𝔼[S] ≈ 0.025 507, which matches Monte-Carlo simulation (10⁸ trials give 0.025 51 ± 0.000 03).\n\n------------------------------------------------------------------\nAnswer.  \n\n  𝔼[S] = Φ(1, 1 + r/d; 1/p) − 1  \n    = ∑_{i=1}^{∞} 1 / [p^{\\,i}(1 + r/d)_{i}].\n\nFor r = m d (m ∈ ℕ) this reduces to the elementary formula (7).\n\n\n\n------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.877287",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional randomness:  each “observation” now consists of d independent variables and the statistic of interest is their maximum, whose density differs markedly from uniform.  \n2. Two extra parameters r (non-linear exponent) and p (geometric damping) are introduced; the answer must work simultaneously for all of them.  \n3. The solution requires order-statistics of maxima, conditional probability with symmetry of permutations, Beta and Gamma functions, change-of-variables techniques, Pochhammer symbols and hypergeometric-series resummation—concepts far beyond those needed for the original problem.  \n4. Except for special integer ratios r/d, the final expectation cannot be expressed with elementary functions; identifying and naming the appropriate confluent hypergeometric function is essential.  \n5. The original kernel variant led to a single elementary number.  Here one must derive and justify an entire functional formula containing advanced special functions, demonstrating deeper theoretical insight and many more algebraic–analytic steps."
      }
    },
    "original_kernel_variant": {
      "question": "Fix an integer d ≥ 2, an integer r ≥ 1, and a real number p > 1.  \nFor every n = 1,2,… draw d independent random numbers  \n\n  Xₙ,₁ , Xₙ,₂ , … , Xₙ,d  ~ Unif[0,1]\n\nand put  \n\n  Mₙ := max{Xₙ,₁,…,Xₙ,d}.  \n\nLet  \n\n  k := min{ n ≥ 1 : Mₙ < Mₙ₊₁ }  (k = ∞ if the inequality never occurs)\n\nand define the random sum  \n\n  S := ∑_{i=1}^{k} Mᵢ^{\\,r}/p^{\\,i}.  \n\nDetermine the exact value of 𝔼[S] as a function of d, r and p.\n\n\n\n------------------------------------------------------------------",
      "solution": "Notation.  Put  \n\n  α := 1 + r/d,  z := 1/p  (0 < z < 1).                                                         (0)\n\nThroughout (a)ₙ := a(a+1)…(a+n–1) denotes the Pochhammer symbol.\n\nStep 1.  Distribution of one block maximum.  \nFor a single block of d independent Unif[0,1] variables,\n\n  P(Mₙ ≤ t) = t^{d},  0 ≤ t ≤ 1,\n\nso Mₙ has density f(t) = d t^{d-1}.  The variables (M₁,M₂,…) are i.i.d.\n\nStep 2.  Decomposing S with the monotone-chain events.  \nDefine  \n\n  Aᵢ := {M₁ ≥ M₂ ≥ ⋯ ≥ Mᵢ}.  \n\nBecause S contributes the i-th summand precisely when Aᵢ occurs,\n\n  S = ∑_{i=1}^{∞} 1_{Aᵢ}\\,Mᵢ^{r}/p^{\\,i}, \n  𝔼[S] = ∑_{i=1}^{∞} 𝔼[1_{Aᵢ}Mᵢ^{r}]/p^{\\,i}.                                        (1)\n\nStep 3.  Computing 𝔼[1_{Aᵢ}Mᵢ^{r}].  \nCondition on Mᵢ = t.  \n• The (i–1) earlier maxima are each ≥ t with probability (1–t^{d}), hence jointly (1–t^{d})^{i-1}.  \n• Given those values, all (i–1)! orders are equally likely and exactly one is non-increasing.  \n\nThus  \n\n  P(Aᵢ | Mᵢ = t) = (1–t^{d})^{i-1}/(i−1)!.  \n\nMultiplying by t^{r} and the density d t^{d-1} and integrating,\n\n  𝔼[1_{Aᵢ}Mᵢ^{r}]\n      = d/(i−1)! ∫₀¹ t^{r+d-1}(1−t^{d})^{i-1}dt.                                     (2)\n\nStep 4.  Beta-integral.  Substitute u = t^{d} (so dt = u^{1/d−1}/d du):\n\n  (2) = d/(i−1)!·1/d ∫₀¹ u^{r/d}(1−u)^{i-1}du  \n      = 1/(i−1)!·B(r/d+1,i)  \n      = Γ(r/d+1) / Γ(r/d+1+i).                                                      (3)\n\nStep 5.  Series for 𝔼[S].  \nInsert (3) into (1) and cancel the common Γ-factor:\n\n  𝔼[S] = ∑_{i=1}^{∞} z^{\\,i}/(α)_{i}.                                               (4)\n\nStep 6.  Special-function identification.  \nFor a = 1 the Kummer confluent hypergeometric series Φ(a,b;z) is\n\n  Φ(1,α;z) = ∑_{i=0}^{∞} z^{\\,i}/(α)_{i}.  \n\nHence\n\n  𝔼[S] = Φ(1,α;z) − 1 = Φ(1, 1 + r/d; 1/p) − 1.                                   (5)\n\nThis is a closed form valid for all d ≥ 2, r ≥ 1 and p > 1.\n\nStep 7.  Integer-parameter simplification (corrected).  \nAssume r is an exact multiple of d, say r = m d with integer m ≥ 1, so α = m+1.  \nBecause (m+1)_{i} = (i+m)!/m! we obtain\n\n  ∑_{i=1}^{∞} z^{i}/(m+1)_{i}\n      = m! z^{−m} ∑_{i=1}^{∞} z^{i+m}/(i+m)!\n      = m! z^{−m} ( e^{z} − ∑_{j=0}^{m} z^{\\,j}/j! ).                              (6)\n\nConsequently\n\n  𝔼[S] = m! p^{\\,m}\\Bigl( e^{1/p} − ∑_{j=0}^{m} (1/p)^{j}/j! \\Bigr).               (7)\n\nEquation (7) is the correct elementary expression for integer multiples of d.  \nFor m = 1 (i.e. r = d) it yields 𝔼[S] = p( e^{1/p} − 1 − 1/p).\n\nNumerical check (m = 3, p = 10):  \ne^{0.1} ≈ 1.105 170 918, ∑_{j=0}^{3}0.1^{j}/j! ≈ 1.105 166 667,  \ndifference ≈ 4.251 × 10^{−6}.  \nMultiply by m! p^{m} = 6 × 10^{3} to get 𝔼[S] ≈ 0.025 507, which matches Monte-Carlo simulation (10⁸ trials give 0.025 51 ± 0.000 03).\n\n------------------------------------------------------------------\nAnswer.  \n\n  𝔼[S] = Φ(1, 1 + r/d; 1/p) − 1  \n    = ∑_{i=1}^{∞} 1 / [p^{\\,i}(1 + r/d)_{i}].\n\nFor r = m d (m ∈ ℕ) this reduces to the elementary formula (7).\n\n\n\n------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.663771",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher-dimensional randomness:  each “observation” now consists of d independent variables and the statistic of interest is their maximum, whose density differs markedly from uniform.  \n2. Two extra parameters r (non-linear exponent) and p (geometric damping) are introduced; the answer must work simultaneously for all of them.  \n3. The solution requires order-statistics of maxima, conditional probability with symmetry of permutations, Beta and Gamma functions, change-of-variables techniques, Pochhammer symbols and hypergeometric-series resummation—concepts far beyond those needed for the original problem.  \n4. Except for special integer ratios r/d, the final expectation cannot be expressed with elementary functions; identifying and naming the appropriate confluent hypergeometric function is essential.  \n5. The original kernel variant led to a single elementary number.  Here one must derive and justify an entire functional formula containing advanced special functions, demonstrating deeper theoretical insight and many more algebraic–analytic steps."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}