Other Distributions Calculator (binomial/Poisson/t/chi²)

Inspectable records

Ledger state

The current ledger marks this instrument verified clean within the source-bound verification corpus.

Area

Mathematics — arithmetic through advanced (algebra, matrices, calculus, sequences/series, number theory, combinatorics, graphing, complex numbers, geometry, logic, sets)

Tranche

tranche-6

Evidence ID

t6-133

Ledger rows

1

No ledger note is attached.

Review findings

The listed binomial and Poisson PMFs are standard and correct under their stated constraints. The overall specification needs review because Student's t and chi-squared PDFs, the numeric integration algorithm, convergence tolerance, tail treatment, and plotting semantics are not specified. The existing honest-label warning is appropriate, but it does not resolve the reproducibility and validation ambiguities.

Formula and scope findings

• The binomial PMF formula is correct, but the required constraints are omitted: n must be a nonnegative integer, k must be an integer with 0 <= k <= n, and 0 <= p <= 1.
• The Poisson PMF formula is correct, but the required constraints are omitted: lambda must satisfy lambda >= 0 and k must be a nonnegative integer. The lambda = 0, k = 0 case requires the convention 0^0 = 1 for the PMF limit.
• The Student's t and chi-squared density formulas are not provided, so the claimed PDF calculations cannot be independently verified from the specification.
• Numeric density integration is underspecified: no integration algorithm, finite-bound transformation, tail handling, singularity handling, quadrature step or tolerance is given. Therefore CDF results are not deterministic or reproducible.
• For chi-squared distributions, the domain x >= 0 must be enforced; the PDF is zero for x < 0 and may have an endpoint singularity when degrees of freedom are less than 2.
• For Student's t and chi-squared distributions, degrees of freedom must satisfy nu > 0.

Unit assumptions

• Binomial n and k are dimensionless integer counts; p is a unitless probability in [0,1].
• Poisson lambda is a dimensionless expected count for the stated observation interval; k is a dimensionless nonnegative integer count. If lambda is supplied as a rate, the interval duration must also be supplied and lambda = rate multiplied by duration.
• Student's t x is dimensionless, normally a standardized statistic; degrees of freedom nu is dimensionless and positive.
• Chi-squared x is dimensionless and nonnegative; degrees of freedom nu is dimensionless and positive.
• Reported PDF values have reciprocal units of x in a general continuous-variable interpretation, although these distributions conventionally use dimensionless standardized variables.
• CDF values and PMF values are dimensionless probabilities.

Boundary hardening

BREAK_CLAIMED · SELF_FLAGGED

The degrees-of-freedom input ν=0 produces multiple undefined operations in the Student's t density. The calculator must reject it rather than return a numeric PDF; accepting it without validation would be a failure. The discrete CDF convention also materially changes the result unless explicitly specified.

Errata and corrections

No applied erratum is attached to this unique-name record.

Inspectable example

4 review vectors are recorded. One source-bound sample is shown below.

{
  "inputs": {
    "distribution": "binomial",
    "n": 4,
    "p": 0.5,
    "k": 2,
    "plot_sample_k": [
      0,
      2,
      4
    ]
  },
  "expected_outputs": {
    "pmf": 0.375,
    "cdf": 0.6875,
    "mean": 2,
    "variance": 1,
    "plot_sample_points": [
      {
        "k": 0,
        "pmf": 0.0625
      },
      {
        "k": 2,
        "pmf": 0.375
      },
      {
        "k": 4,
        "pmf": 0.0625
      }
    ],
    "key_curve_features": {
      "support": "k in {0,1,2,3,4}; PMF is zero outside this support",
      "mode": "k=2",
      "symmetry": "PMF is symmetric about k=2"
    }
  },
  "working": "PMF at k=2: C(4,2) * 0.5^2 * (1-0.5)^(4-2) = 6 * 0.25 * 0.25 = 0.375. CDF through k=2: P(0)+P(1)+P(2) = [1*0.5^4] + [4*0.5^3*0.5] + [6*0.5^2*0.5^2] = 0.0625 + 0.25 + 0.375 = 0.6875. Mean = n*p = 4*0.5 = 2. Variance = n*p*(1-p) = 4*0.5*0.5 = 1. Plot samples: P(0)=C(4,0)*0.5^4=0.0625; P(2)=0.375; P(4)=C(4,4)*0.5^4=0.0625."
}

Related workflows

No source-defined multi-instrument workflow membership is recorded.

Sources and evidence

These pointers bind the page to evidence. They do not grant authority or collapse verification into deployment.

• VERIFIED_LEDGER.json · VERIFIED_CLEAN
• evidence-tranche6/tranche6-results.json
• BOUNDARY_HAZARD_CATALOG.json