RANDLAPLACE

Updated 2023-10-18 15:57:47.273000

Syntax

SELECT * FROM [westclintech].[wct].[RANDLAPLACE](
  <@Rows, int,>
 ,<@Location, float,>
 ,<@Scale, float,>)

Description

Use the table-valued function RANDLAPLACE to generate a sequence of random numbers from a LaPlace distribution with parameters @Location and @Scale.

Arguments

@Scale

the scale parameter. @Scale must be of the type float or of a type that implicitly converts to float.

@Location

the location parameter. @Shape must be of the type float or of a type that implicitly converts to float.

@Rows

the number of rows to generate. @Rows must be of the type int or of a type that implicitly converts to int.

Return Type

table

{"columns": [{"field": "colName", "headerName": "Name", "header": "name"}, {"field": "colDatatype", "headerName": "Type", "header": "type"}, {"field": "colDesc", "headerName": "Description", "header": "description", "minWidth": 1000}], "rows": [{"id": "d9a02b24-dde7-40da-86b5-4907819dfe96", "colName": "Seq", "colDatatype": "int", "colDesc": "A monotonically increasing sequence number"}, {"id": "29c8d09b-7d2e-42af-97a7-8f92d8423b89", "colName": "X", "colDatatype": "float", "colDesc": "The random variable"}]}

Remarks

@Scale must be greater than zero.

If @Shape is NULL then @Shape is set to 0.

If @Scale is NULL then @Scale is set to 1.

If @Rows is less than 1 then no rows are returned.

Examples

In this example we create a sequence 1,000,000 random numbers rounded to one decimal place from a Laplace distribution with @Location = 0 and @Scale =1, COUNT the results, paste then into Excel, and graph them.

SELECT X,
       COUNT(*) as [COUNT]
FROM
(
    SELECT ROUND(X, 1) as X
    FROM wct.RANDLAPLACE(   1000000, --@Rows
                            0,       --@Loations
                            1        --@Scale
                        )
) n
GROUP BY X
ORDER BY X;

This produces the following result.

In this example we generate 1,000,000 random numbers from a LaPlace distribution with @Shape of 5 and @Scale of 2. We calculate the mean, standard deviation, skewness, and excess kurtosis from the resultant table and compare th ose values to the expected values for the distribution.

DECLARE @size as int = 1000000;
DECLARE @location as float = -5;
DECLARE @scale as float = 4;
DECLARE @mean as float = @location;
DECLARE @var as float = 2 * POWER(@scale, 2);
DECLARE @stdev as float = SQRT(@var);
DECLARE @skew as float = 0;
DECLARE @kurt as float = 3;
SELECT stat,
       [RANDLAPLACE],
       [EXPECTED]
FROM
(
    SELECT x.*
    FROM
    (
        SELECT AVG(x) as mean_LAPLACE,
               STDEVP(x) as stdev_LAPLACE,
               wct.SKEWNESS_P(x) as skew_LAPLACE,
               wct.KURTOSIS_P(x) as kurt_LAPLACE
        FROM wct.RANDLAPLACE(@size, @location, @scale)
    ) n
        CROSS APPLY
    (
        VALUES
            ('RANDLAPLACE', 'avg', mean_LAPLACE),
            ('RANDLAPLACE', 'stdev', stdev_LAPLACE),
            ('RANDLAPLACE', 'skew', skew_LAPLACE),
            ('RANDLAPLACE', 'kurt', kurt_LAPLACE),
            ('EXPECTED', 'avg', @mean),
            ('EXPECTED', 'stdev', @stdev),
            ('EXPECTED', 'skew', @skew),
            ('EXPECTED', 'kurt', @kurt)
    ) x (fn_name, stat, val_stat)
) d
PIVOT
(
    sum(val_stat)
    FOR fn_name in ([RANDLAPLACE], [EXPECTED])
) P;

This produces the following result (your result will be different).

{"columns":[{"field":"stat"},{"field":"RANDLAPLACE","headerClass":"ag-right-aligned-header","cellClass":"ag-right-aligned-cell"},{"field":"EXPECTED","headerClass":"ag-right-aligned-header","cellClass":"ag-right-aligned-cell"}],"rows":[{"stat":"avg","RANDLAPLACE":"-4.99944519359237","EXPECTED":"-5"},{"stat":"kurt","RANDLAPLACE":"3.01552778472663","EXPECTED":"3"},{"stat":"skew","RANDLAPLACE":"-0.00631324114182914","EXPECTED":"0"},{"stat":"stdev","RANDLAPLACE":"5.65406406691652","EXPECTED":"5.65685424949238"}]}