{
  "index": "1964-A-2",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "2. Find all continuous positive functions \\( f(x) \\), for \\( 0 \\leq x \\leq 1 \\), such that\n\\[\n\\begin{array}{c}\n\\int_{0}^{1} f(x) d x=1 \\\\\n\\int_{0}^{1} f(x) x d x=\\alpha \\\\\n\\int_{0}^{1} f(x) x^{2} d x=\\alpha^{2}\n\\end{array}\n\\]\nwhere \\( \\alpha \\) is a given real number.",
  "solution": "Solution. Multiply the first integral equation by \\( \\alpha^{2} \\), the second by \\( -2 \\alpha \\), the third by 1 , and then add to obtain\n\\[\n\\int_{0}^{1} f(x)(\\alpha-x)^{2} d x=0\n\\]\n\nBut this integral is clearly positive for any positive continuous function \\( f \\). Hence there are no functions satisfying the conditions of the problem.",
  "vars": [
    "f",
    "x"
  ],
  "params": [
    "\\\\alpha"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "function",
        "x": "variable",
        "\\alpha": "alphaparam"
      },
      "question": "2. Find all continuous positive functions \\( function(variable) \\), for \\( 0 \\leq variable \\leq 1 \\), such that\n\\[\n\\begin{array}{c}\n\\int_{0}^{1} function(variable) d variable=1 \\\\\n\\int_{0}^{1} function(variable) variable d variable=alphaparam \\\\\n\\int_{0}^{1} function(variable) variable^{2} d variable=alphaparam^{2}\n\\end{array}\n\\]\nwhere \\( alphaparam \\) is a given real number.",
      "solution": "Solution. Multiply the first integral equation by \\( alphaparam^{2} \\), the second by \\( -2 alphaparam \\), the third by 1 , and then add to obtain\n\\[\n\\int_{0}^{1} function(variable)(alphaparam-variable)^{2} d variable=0\n\\]\n\nBut this integral is clearly positive for any positive continuous function \\( function \\). Hence there are no functions satisfying the conditions of the problem."
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "lighthouse",
        "x": "sandcastle",
        "\\alpha": "telescope"
      },
      "question": "2. Find all continuous positive functions \\( lighthouse(sandcastle) \\), for \\( 0 \\leq sandcastle \\leq 1 \\), such that\n\\[\n\\begin{array}{c}\n\\int_{0}^{1} lighthouse(sandcastle) d sandcastle=1 \\\\\n\\int_{0}^{1} lighthouse(sandcastle) sandcastle d sandcastle=telescope \\\\\n\\int_{0}^{1} lighthouse(sandcastle) sandcastle^{2} d sandcastle=telescope^{2}\n\\end{array}\n\\]\nwhere \\( telescope \\) is a given real number.",
      "solution": "Solution. Multiply the first integral equation by \\( telescope^{2} \\), the second by \\( -2 telescope \\), the third by 1 , and then add to obtain\n\\[\n\\int_{0}^{1} lighthouse(sandcastle)(telescope-sandcastle)^{2} d sandcastle=0\n\\]\n\nBut this integral is clearly positive for any positive continuous function \\( lighthouse \\). Hence there are no functions satisfying the conditions of the problem."
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "nonpositive",
        "x": "immobile",
        "\\alpha": "lastvalue"
      },
      "question": "2. Find all continuous positive functions \\( nonpositive(immobile) \\), for \\( 0 \\leq immobile \\leq 1 \\), such that\n\\[\n\\begin{array}{c}\n\\int_{0}^{1} nonpositive(immobile) d immobile=1 \\\\\n\\int_{0}^{1} nonpositive(immobile) immobile d immobile=lastvalue \\\\\n\\int_{0}^{1} nonpositive(immobile) immobile^{2} d immobile=lastvalue^{2}\n\\end{array}\n\\]\nwhere \\( lastvalue \\) is a given real number.",
      "solution": "Solution. Multiply the first integral equation by \\( lastvalue^{2} \\), the second by \\( -2 lastvalue \\), the third by 1 , and then add to obtain\n\\[\n\\int_{0}^{1} nonpositive(immobile)(lastvalue-immobile)^{2} d immobile=0\n\\]\n\nBut this integral is clearly positive for any positive continuous function \\( nonpositive \\). Hence there are no functions satisfying the conditions of the problem."
    },
    "garbled_string": {
      "map": {
        "f": "qzxwvtnp",
        "x": "mfldzqre",
        "\\alpha": "tpshkcwe"
      },
      "question": "2. Find all continuous positive functions \\( qzxwvtnp(mfldzqre) \\), for \\( 0 \\leq mfldzqre \\leq 1 \\), such that\n\\[\n\\begin{array}{c}\n\\int_{0}^{1} qzxwvtnp(mfldzqre) d mfldzqre=1 \\\\\n\\int_{0}^{1} qzxwvtnp(mfldzqre) mfldzqre d mfldzqre=tpshkcwe \\\\\n\\int_{0}^{1} qzxwvtnp(mfldzqre) mfldzqre^{2} d mfldzqre=tpshkcwe^{2}\n\\end{array}\n\\]\nwhere \\( tpshkcwe \\) is a given real number.",
      "solution": "Solution. Multiply the first integral equation by \\( tpshkcwe^{2} \\), the second by \\( -2 tpshkcwe \\), the third by 1 , and then add to obtain\n\\[\n\\int_{0}^{1} qzxwvtnp(mfldzqre)(tpshkcwe-mfldzqre)^{2} d mfldzqre=0\n\\]\n\nBut this integral is clearly positive for any positive continuous function \\( qzxwvtnp \\). Hence there are no functions satisfying the conditions of the problem."
    },
    "kernel_variant": {
      "question": "Let n ≥ 2 be a fixed integer.  \nGiven a vector μ ∈ ℝⁿ and a real, symmetric, positive–definite matrix  \nΣ = (s_{ij})_{1≤i,j≤n}, find all continuous functions  \n\n  f : [−2,2]ⁿ → (0,∞)  \n\nthat satisfy the following six families of moment conditions  \n\n(1) ∫_{[−2,2]ⁿ} f(x) dx = 1,  \n\n(2) ∫_{[−2,2]ⁿ} x f(x) dx = μ,                                       (vector equality)  \n\n(3) ∫_{[−2,2]ⁿ} (x_i−μ_i)(x_j−μ_j) f(x) dx = s_{ij} for every 1≤i,j≤n,  \n\n(4) ∫_{[−2,2]ⁿ} (x_i−μ_i)²(x_j−μ_j)² f(x) dx = s_{ii}s_{jj} for every 1≤i,j≤n.  \n\nDetermine all such functions f (or prove that none exist).",
      "solution": "We interpret the integrals in probabilistic language.  \nLet X be the ℝⁿ-valued random variable with density f on the cube [−2,2]ⁿ.  \nConditions (1)–(4) translate to  \n\n (i) E[1] = 1  (normalisation),  \n (ii) E[X] = μ,  \n (iii) E[(X−μ)(X−μ)ᵗ] = Σ  (the covariance matrix of X is Σ),  \n (iv) E[(X_i−μ_i)²(X_j−μ_j)²] = s_{ii}s_{jj}.                         (1)\n\nStep 1:  Fourth–moment variance is zero.  \nFix an index i.  From (iii) we have Var(X_i) = s_{ii}>0 because Σ is positive-definite.  \nConsider the non–negative quantity\n\n Var((X_i−μ_i)²) = E[(X_i−μ_i)⁴] − (E[(X_i−μ_i)²])².    (2)\n\nTaking j = i in (1) gives E[(X_i−μ_i)⁴] = s_{ii}², while (iii) yields  \nE[(X_i−μ_i)²] = s_{ii}.  Hence (2) becomes Var((X_i−μ_i)²) = s_{ii}² − s_{ii}² = 0.\n\nTherefore (X_i−μ_i)² is almost surely constant:\n\n (X_i−μ_i)² = s_{ii} with probability 1.                                (3)\n\nStep 2:  Pairwise covariance of the squared centred coordinates is zero.  \nFor i ≠ j, using (1) again we obtain  \n\n Cov((X_i−μ_i)²,(X_j−μ_j)²)  \n  = E[(X_i−μ_i)²(X_j−μ_j)²] − E[(X_i−μ_i)²]E[(X_j−μ_j)²]  \n  = s_{ii}s_{jj} − s_{ii}s_{jj} = 0.\n\nBecause each of the two random variables is almost surely constant by (3), the covariance being zero is automatically satisfied; no new information arises.\n\nStep 3:  The coordinates themselves are a.s. constant up to sign.  \nEquation (3) implies |X_i−μ_i| = √{s_{ii}} almost surely, i.e.  \n\n X_i ∈ {μ_i − √{s_{ii}}, μ_i + √{s_{ii}}} with probability 1.         (4)\n\nHence every realisation of X lies in the 2ⁿ-point set  \n\n S := {μ ± (√{s_{11}},…,√{s_{nn}})}.                                (5)\n\nStep 4:  Continuity and strict positivity force degeneracy.  \nThe support of a strictly positive continuous density on [−2,2]ⁿ must be the whole cube, or at least must contain a non-empty open subset.  The finite set S in (5) has empty interior; therefore the only way a continuous f could concentrate its entire mass on S is for S to reduce to a single point.\n\nBut by positive-definiteness of Σ we have s_{ii}>0 for every i, so (5) contains at least 2ⁿ distinct points.  Contradiction.\n\nStep 5:  Conclusion.  \nNo strictly positive continuous function f on [−2,2]ⁿ can satisfy (1)–(4) for any positive-definite Σ.  Hence\n\n There does not exist any continuous function f:(−2,2)ⁿ→(0,∞) satisfying all the given conditions.\n\n(The only formal way to meet the moment equations is with a Dirac delta at the single point μ, which is not a positive continuous function.)",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.546992",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: The problem moves from a single real variable to an arbitrary-dimensional cube [−2,2]ⁿ, introducing vector and matrix moments.  \n2. Additional constraints: Besides first and second moments, full mixed fourth-order moments are imposed, greatly enlarging the system.  \n3. Advanced structures: The solver must use covariance matrices, fourth-moment tensors, and probabilistic variance arguments, not just scalar algebra.  \n4. Deeper theory: Proving impossibility hinges on non-elementary facts—variance non-negativity, properties of positive-definite matrices, and measure-theoretic support arguments.  \n5. Multiple interacting concepts: Linear algebra (positive-definite matrices), probability (moments, variance), analysis (continuity, support) all interplay.  \n\nAltogether, the enhanced variant requires several layers of reasoning—matrix inequalities, almost-sure statements, and topological support considerations—far beyond the single-variable quadratic trick that dispatched the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let n ≥ 2 be a fixed integer.  \nGiven a vector μ ∈ ℝⁿ and a real, symmetric, positive–definite matrix  \nΣ = (s_{ij})_{1≤i,j≤n}, find all continuous functions  \n\n  f : [−2,2]ⁿ → (0,∞)  \n\nthat satisfy the following six families of moment conditions  \n\n(1) ∫_{[−2,2]ⁿ} f(x) dx = 1,  \n\n(2) ∫_{[−2,2]ⁿ} x f(x) dx = μ,                                       (vector equality)  \n\n(3) ∫_{[−2,2]ⁿ} (x_i−μ_i)(x_j−μ_j) f(x) dx = s_{ij} for every 1≤i,j≤n,  \n\n(4) ∫_{[−2,2]ⁿ} (x_i−μ_i)²(x_j−μ_j)² f(x) dx = s_{ii}s_{jj} for every 1≤i,j≤n.  \n\nDetermine all such functions f (or prove that none exist).",
      "solution": "We interpret the integrals in probabilistic language.  \nLet X be the ℝⁿ-valued random variable with density f on the cube [−2,2]ⁿ.  \nConditions (1)–(4) translate to  \n\n (i) E[1] = 1  (normalisation),  \n (ii) E[X] = μ,  \n (iii) E[(X−μ)(X−μ)ᵗ] = Σ  (the covariance matrix of X is Σ),  \n (iv) E[(X_i−μ_i)²(X_j−μ_j)²] = s_{ii}s_{jj}.                         (1)\n\nStep 1:  Fourth–moment variance is zero.  \nFix an index i.  From (iii) we have Var(X_i) = s_{ii}>0 because Σ is positive-definite.  \nConsider the non–negative quantity\n\n Var((X_i−μ_i)²) = E[(X_i−μ_i)⁴] − (E[(X_i−μ_i)²])².    (2)\n\nTaking j = i in (1) gives E[(X_i−μ_i)⁴] = s_{ii}², while (iii) yields  \nE[(X_i−μ_i)²] = s_{ii}.  Hence (2) becomes Var((X_i−μ_i)²) = s_{ii}² − s_{ii}² = 0.\n\nTherefore (X_i−μ_i)² is almost surely constant:\n\n (X_i−μ_i)² = s_{ii} with probability 1.                                (3)\n\nStep 2:  Pairwise covariance of the squared centred coordinates is zero.  \nFor i ≠ j, using (1) again we obtain  \n\n Cov((X_i−μ_i)²,(X_j−μ_j)²)  \n  = E[(X_i−μ_i)²(X_j−μ_j)²] − E[(X_i−μ_i)²]E[(X_j−μ_j)²]  \n  = s_{ii}s_{jj} − s_{ii}s_{jj} = 0.\n\nBecause each of the two random variables is almost surely constant by (3), the covariance being zero is automatically satisfied; no new information arises.\n\nStep 3:  The coordinates themselves are a.s. constant up to sign.  \nEquation (3) implies |X_i−μ_i| = √{s_{ii}} almost surely, i.e.  \n\n X_i ∈ {μ_i − √{s_{ii}}, μ_i + √{s_{ii}}} with probability 1.         (4)\n\nHence every realisation of X lies in the 2ⁿ-point set  \n\n S := {μ ± (√{s_{11}},…,√{s_{nn}})}.                                (5)\n\nStep 4:  Continuity and strict positivity force degeneracy.  \nThe support of a strictly positive continuous density on [−2,2]ⁿ must be the whole cube, or at least must contain a non-empty open subset.  The finite set S in (5) has empty interior; therefore the only way a continuous f could concentrate its entire mass on S is for S to reduce to a single point.\n\nBut by positive-definiteness of Σ we have s_{ii}>0 for every i, so (5) contains at least 2ⁿ distinct points.  Contradiction.\n\nStep 5:  Conclusion.  \nNo strictly positive continuous function f on [−2,2]ⁿ can satisfy (1)–(4) for any positive-definite Σ.  Hence\n\n There does not exist any continuous function f:(−2,2)ⁿ→(0,∞) satisfying all the given conditions.\n\n(The only formal way to meet the moment equations is with a Dirac delta at the single point μ, which is not a positive continuous function.)",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.452866",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: The problem moves from a single real variable to an arbitrary-dimensional cube [−2,2]ⁿ, introducing vector and matrix moments.  \n2. Additional constraints: Besides first and second moments, full mixed fourth-order moments are imposed, greatly enlarging the system.  \n3. Advanced structures: The solver must use covariance matrices, fourth-moment tensors, and probabilistic variance arguments, not just scalar algebra.  \n4. Deeper theory: Proving impossibility hinges on non-elementary facts—variance non-negativity, properties of positive-definite matrices, and measure-theoretic support arguments.  \n5. Multiple interacting concepts: Linear algebra (positive-definite matrices), probability (moments, variance), analysis (continuity, support) all interplay.  \n\nAltogether, the enhanced variant requires several layers of reasoning—matrix inequalities, almost-sure statements, and topological support considerations—far beyond the single-variable quadratic trick that dispatched the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}