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RANDPARETO Function (Gnumeric)

Statistical Intermediate Gnumeric Gnumeric 1.1.6

RANDPARETO(a,b) uses positive shape a and nonnegative minimum/scale b. For positive b, the ideal mean exists only when a>1 and the variance is finite only when a>2.

Verification status: See the evidence and test scope described in this guide. How formula examples are checked

Quick answer: RANDPARETO(a,b) uses positive shape a and nonnegative minimum/scale b. For positive b, the ideal mean exists only when a>1 and the variance is finite only when a>2.

Syntax

=RANDPARETO(a,b)

Availability and evidence

  • Introduction: Gnumeric 1.1.6 NEWS explicitly lists this new function. This release-history evidence is separate from a native test of that old version.
  • Source support: Registered in the Gnumeric 1.12.62 random plugin. No retirement is established; the last functioning native application version is unverified.
  • Evidence dates: Official Gnumeric source retained October 13–14, 2026 was reviewed October 15. Mathematical controls were independently recalculated October 15. No native application, compiled Gnumeric helper or random generator was executed.

Parameters and source mapping

Both arguments are required. The helper rejects a≤0 or b<0 with #NUM!. After those checks, b=0 returns 0 immediately. For b>0 it retries a zero unit driver U and computes b*U^(-1/a).

This is a Pareto Type I model with lower bound b, not a shifted model starting at 0. For x≥b>0 the ideal survival probability is (b/x)^a. The first argument controls tail heaviness: smaller positive a gives a heavier tail. The second sets the scale and lower boundary.

Complete worksheet setup

Create ParetoControls. Put shape 3 in B2 and minimum 10 in B3. In D2 enter =RANDPARETO($B$2,$B$3). The result changes; no fixed expected draw is supplied.

In F2 enter =B2*B3/(B2-1) for the ideal mean. It gives 15. In F3 enter =B2*B3^2/((B2-1)^2*(B2-2)) for the ideal variance, 75. These formulas require a>1 and a>2, respectively, and must not be filled indiscriminately into rows with smaller shapes.

Deterministic transformation controls

Put 0.125,0.5 and 0.875 in A7:A9. In B7 enter =$B$3*A7^(-1/$B$2) and fill down. The independent references are:

Supplied control inputs Independent reference result
U=.125 20.0
U=.5 12.59921049895
U=.875 10.45515917149

The U=0.5 row is the ideal median, approximately 12.5992104989, below the mean 15. A lower-tail probability p would use U=1-p; the source mapping decreases as U increases.

The supplied U=0.125 gives 20 because its cube root is 0.5 and the reciprocal doubles the lower bound 10. This simple exact control helps catch swapped shape and scale arguments.

Advanced: quantify a heavy tail

For the original parameters, the ideal probability of exceeding 20 is (10/20)^3, or 0.125. The probability of exceeding 100 is (10/100)^3, or 0.001. These are mathematical tail probabilities, not frequencies observed from D2.

Doubling b from 10 to 20 doubles every same-driver value, doubles the ideal mean from 15 to 30, and multiplies the variance from 75 to 300. To show same-draw scaling, multiply an existing random cell by 2 instead of comparing two separately called formulas.

Advanced: respect moment-existence boundaries

With a=2 and b=10, the ideal mean is 20 but the variance is not finite. With a=1 and b=10 there is no finite mean. The random-generation function can still accept these positive shapes; an undefined population moment is not the same as an invalid generator argument.

A finite collection of machine outputs can have a numerical sample average and sample variance. That does not make the corresponding ideal population moments finite. Do not validate a shape 1 model by insisting that a long sample average approach a nonexistent finite mean.

Advanced: a logarithmic transformation

For an ideal Pareto variable X, LN(X/b) is exponential with scale 1/a. In the supplied-driver worksheet, enter =LN(B7/$B$3) in C7 and fill down. Algebraically it equals =-LN(A7)/$B$2.

For the exact U=0.125 case, this logarithmic excess is LN(2), approximately 0.69314718056. The ideal mean of this transformed quantity is 1/3 for the original shape 3, even though X itself has mean 15. Nonlinear transformations do not preserve means by simply transforming the mean.

Boundary and error controls

=RANDPARETO(3,0) returns 0 before requesting a unit driver. =RANDPARETO(0,0) still gives #NUM!, because shape validation precedes the zero-scale branch. Negative shape or negative b also gives #NUM!.

The ideal positive-scale model has an unbounded upper tail and lower boundary b. A finite unit grid and arithmetic rounding do not guarantee every mathematical value or exact boundary behavior. No maximum machine output or empirical tail-quality claim is made.

Recalculation and evidence limits

The plugin registers this function as volatile, so recalculation can replace its result. There is no worksheet seed argument in the signature shown here. Exact GUI triggers, saved-file replay and cross-application transfer were not tested. Two calls do not promise the same underlying driver; repeated references to one random cell do not create independent observations.

The supplied driver values and mathematical summaries are deterministic teaching controls. They are not captured random samples. Ideal continuous distributions differ from a finite-resolution generator: rounding, underflow, overflow and endpoint treatment can matter. Non-finite arithmetic is converted to #NUM! by the numeric result constructor. Explicit finite numeric inputs are used throughout; text/blank coercion and locale-sensitive parsing are outside this verification.