time-to-botec

Benchmark sampling in different programming languages
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geometric.js (2153B)


      1 /**
      2 * @license Apache-2.0
      3 *
      4 * Copyright (c) 2018 The Stdlib Authors.
      5 *
      6 * Licensed under the Apache License, Version 2.0 (the "License");
      7 * you may not use this file except in compliance with the License.
      8 * You may obtain a copy of the License at
      9 *
     10 *    http://www.apache.org/licenses/LICENSE-2.0
     11 *
     12 * Unless required by applicable law or agreed to in writing, software
     13 * distributed under the License is distributed on an "AS IS" BASIS,
     14 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
     15 * See the License for the specific language governing permissions and
     16 * limitations under the License.
     17 */
     18 
     19 'use strict';
     20 
     21 // MODULES //
     22 
     23 var floor = require( '@stdlib/math/base/special/floor' );
     24 var ln = require( '@stdlib/math/base/special/ln' );
     25 
     26 
     27 // MAIN //
     28 
     29 /**
     30 * Returns a pseudorandom number drawn from a geometric distribution.
     31 *
     32 * ## Proof
     33 *
     34 * Consider
     35 *
     36 * ```tex
     37 * N = \left \lfloor \ln (U) / \ln (1-p) \right \rfloor
     38 * ```
     39 *
     40 * where \\( U \\) is uniform on the interval \\((0,1)\\). Accordingly, \\(N\\) must be a nonnegative integer, and, for every \\( n \geq 0\\), the event \\(A_n = \left \{ N = n \right \}\\) is
     41 *
     42 * ```tex
     43 * A_n = \left \{(n+1) \ln (1-p) < \ln (U) \leq n \ln (1-p) \right \}
     44 * ```
     45 *
     46 * where \\(\ln (1-p) < 0\\). Thus,
     47 *
     48 * ```tex
     49 * A_n = \left \{(1-p)^{n+1} < U \leq (1-p)^n \right \}
     50 * ```
     51 *
     52 * For every \\(u < v\\) on the interval \\((0,1)\\),
     53 *
     54 * ```tex
     55 * P\left \[u < U \leq v\right \] = v - u
     56 * ```
     57 *
     58 * Hence,
     59 *
     60 * ```tex
     61 * P\left \[N = n \right \] = P\left \[A_n\right \] = (1-p)^n - (1-p)^{n+1} = (1-p)^n(1-(1-p)) = p(1-p)^n
     62 * ```
     63 *
     64 * which proves that \\(N\\) is a geometric random variable.
     65 *
     66 *
     67 * @private
     68 * @param {PRNG} rand - PRNG for uniformly distributed numbers
     69 * @param {Probability} p - success probability
     70 * @returns {NonNegativeInteger} pseudorandom number
     71 */
     72 function geometric( rand, p ) {
     73 	var u = rand();
     74 	if ( u === 0.0 ) {
     75 		// Drawing random variates from a PRNG (with period > 1) is effectively sampling without replacement. Thus, should not be possible to draw `0` twice in a row.
     76 		u = rand();
     77 	}
     78 	return floor( ln( u ) / ln( 1.0-p ) );
     79 }
     80 
     81 
     82 // EXPORTS //
     83 
     84 module.exports = geometric;