Using SQL Server RAND Function Deep Dive


By:   |   Updated: 2020-02-11   |   Comments (3)   |   Related: More > T-SQL

Problem

Dr. Dallas Snider [1], a MSSQLTips author, mentioned that SQL Server RAND() function could generate random values uniformly distributed from 0 through 1, exclusive. He generated random numbers from the normal distribution with a specific mean and standard deviation [1]. In his other tip [2], he created a function to generate random values in a range. Using SQL Server RAND function for generating random variables from specific probability distributions is a popular topic on the web. Many professionals, for example Dwain Camps [3], have provided excellent solutions. With so many solutions available, we need to find one that fits our requirements. To select effective techniques and examine their performances, database professionals would like to know the principles behind these techniques.

Solution

The notion of "completely random" on an interval means every value in the interval has an equal chance to be chosen. If the SQL Server RAND() function returns a completely random number between 0 and 1, we say that the function has the uniform distribution on the interval (0,1), denoted by formula. Dwain [3] examined the distribution of uniform random numbers generated by SQL Server RAND() function. He concluded that the uniformity is not great, but probably good enough for noncritical purposes [3]. In this tip, we assume SQL server RAND() function can produce a completely random number between 0 and 1, exclusive.

The universality of the uniform, a remarkable characteristic of the uniform distribution, allows us to construct a random variable with any continuous distribution from a given formula [4]. This characteristic also has several other names, for example inverse transform sampling. We will use T-SQL to generate random numbers from continuous distributions and discrete distributions by using the universality of the uniform theorem.

This tip is organized as follows. In section 1, we explore concepts of random variable and distribution. Section 2 introduces three named discrete probability distributions. We look at two named continuous probability distributions in section 3. Next, in section 4, we give practice in the use of the universality of the uniform theorem for generating random variables from specific probability distributions. In this tip, T-SQL is used to produce random numbers, and R is used to create graphic representations. The sections build on each other. Information presented early in the tip will be helpful for generating random numbers in the later sections.

All the source codes used in this tip were tested with SQL Server Management Studio V17.4, Microsoft Visual Studio Community 2017, Microsoft R Client 3.4.3, and Microsoft ML Server 9.3 on Windows 10 Home 10.0 <X64>. The DBMS is Microsoft SQL Server 2017 Enterprise Edition (64-bit).

1 – Basic Concepts

The following T-SQL statements are used to randomly choose an integer value between 1 and 3, inclusive:

-- Declare and initialize variables 
DECLARE @lower_limit INT = 3 
DECLARE @upper_limit INT = 1 
-- Choose a random integer between @lower_limit and @upper_limit, inclusive 
SELECT ROUND((@upper_limit - @lower_limit) * RAND(CHECKSUM(NEWID())), 0) + @lower_limit 			

Intuitively, these T-SQL statements completely randomly selected one of these three integers.  In other words, all these integer numbers have an equal chance of being choosen. I wrote a script in R that ran these SQL statements 10,000 times and then loaded those chosen numbers into a data frame. I plotted a bar chart of the frequency distribution for these numbers to visualize the chance of occurrence of each integer number.

# Reset the compute context to your local workstation. 
rxSetComputeContext("local") 

# Define connection string to SQL Server 
sql.server.conn.string <- "Driver=SQL Server;Server=.Database=TutorialDB;Trusted_Connection={Yes}" 

# Execute T-SQL statements through an RxSqlServerData data source object 
data.source.object <- RxSqlServerData(sqlQuery = "                              
                    SELECT top 10000 
                       u.random_value, 
                       (ROUND((3-1)*u.random_value, 0)+1) as random_integer 
                    FROM sys.all_columns a1 
                    CROSS APPLY sys.all_columns a2 
                    CROSS APPLY 
                    ( 
                       SELECT RAND(CHECKSUM(NEWID())) as random_value 
                    ) u", 
                                connectionString = sql.server.conn.string) 

# Read the result into a data frame in the local R session. 
data.frame.random.numbers <- rxImport(data.source.object) 

# Create a Frequency Distribution Bar Chart quickly with the plot() function 
# when the categorical variable to be plotted is a factor or ordered factor. 
plot(as.factor(data.frame.random.numbers$random_integer), 
     main = "Frequency Distribution of Randomly Selected Integers Between 1 and 3", 
     xlab = "Random Integer", 
     ylab = "Number of Occurrence", 
     col = "blue") 

The bar chart, shown in Figure 1, illustrates that the chance of selecting integer number "2" is twice as many chances as other two integers. This concludes that the preceding SQL statements cannot generate completely random integers on an interval.

chart
Figure 1 Frequency Distribution of Random Integers Between 1 and 3 Generated by SQL Server RAND Function

Sometimes, we may be required to randomly pick a number from an integer array in which elements are not consecutive, furthermore, each element may have a different probability to be chosen. These tasks become complicated. I do not think we can tackle these tasks without understanding of random variables and their distributions. We will walkthrough these concepts in this tip.

1.1 Random variables

When we consider something random, mostly we mean all the people or things involved have an equal chance of being chosen. When we describe events as random, we mean that a definite plan or pattern is not followed [6]. The Collins online dictionary states that a variable is a factor that can change in quality, quantity, or size, which we must consider in a situation [7]. Putting all together these definitions, we can consider a random variable a symbol that can represent a random selected number, categories, etc. This kind of definition does not tell where randomness comes from. To make the notion of random variable both conceptually and technically correct, we define it as a function mapping the sample space to a set of real numbers [4]. Professor Jost also gave an informal definition: a random variable is a process for choosing a random number [5]. We often denote random variables by capital letters, for example formula. A random variable numerically describes outcomes of an experiment and reveals some aspects of the experiment.

In the experiment of rolling a balanced die, the sample space is {1 spot, 2 spots, 3 spots, 4 spots, 5 spots, 6 spots}. Let formula be the remainder after division of the number of spots by 3. Then, formula is a random variable with possible values 0, 1, and 2. Let formula be the quotient. The possible values of formula are 0, 1 and 2. Let formula be 1 if the remainder is 0 and 0 otherwise. Since the random variable formula indicates whether the number of spots is evenly divisible by 3, we also call this kind of random variable an indicator random variable. The mapping from the sample space to the set of real numbers is shown in Table 1.

Number of Spots (s) Possible Values of X Possible Values of Y Possible Values of I
1 1 0 0
2 2 0 0
3 0 1 1
4 1 1 0
5 2 1 0
6 0 2 1

Table 1 The Random Variables X, Y and I Defined on the Sample Space

Table 1 reveals the fact that multiple random variables can be defined on the same sample space, and one experiment outcome may map to several different real numbers. We have also observed that the randomness in a random variable is due to the random outcomes of an experiment. The probability of each outcome in the sample space determines the probability that the random variable will take on a value or an interval. In addition, we derived an equation for formula, formula and formula:

formula

where formula denotes the number of spots.

Then we write formula in terms of formula by rearranging the equation: formula. These two equations help to reinforce the understanding that random variable is a function mapping each outcome in sample space to a real number.

When a random variable can take on only a finite set of numeric values or an infinite sequence, we name this variable discrete random variable. The set of values is called the support of the discrete random variable.  On the other side, when a random variable can take any numeric values in an interval, possibly infinite, for example, formula, the variable is named as a continuous random variable.

1.2 Distributions

In the example of choosing one number from three consecutive integers, we expected to select a completely random integer number. However, the bar chart in Figure 1 argued that those chosen integer numbers are not uniformly distributed. The behavior of expected random numbers is different from those actual generated random numbers. We can describe this kind of difference by using the language of probability, that is distribution. The distribution of a random variable specifies the probabilities of all events associated with this random variable [4], consequently, describes the behavior of the random variable.

1.2.1 Probability Mass Function

For a discrete random variable, formula, the probability distribution can be expressed by a probability mass function (hereinafter referred to as the "PMF"), denoted by formula. This function provides the probability for each possible value of the random variable:

formula

where formula denotes an event that maps to a subset of the sample space.

In the experiment of rolling a balanced die, demonstrated in Table 1, formula is the remainder after division. The possible experiment outcomes, 1 spot and 4 spots, in the sample space map to the integer value 1 of the support of formula. Both subsets, {1 spot, 4 spots} and { 1 },  mean the same event formula. Note that formula is a function. We can write the event in another form: formula. Thus, we use naïve definition [4] to calculate the probability of the event:

formula

For completeness, we compute the probabilities of all possible values of formula. All these equations together determine the PMF of formula:

formula

formula

formula

formula

The PMF formula must satisfy the following two conditions:

  • PMF must be nonnegative for each value of the random variable;
  • The sum of the probabilities for each value of the random variable must equal to 1.

1.2.2 Cumulative Distribution Functions

In the rolling a die experiment, the random variable formula has 3 possible values: 0, 1 and 2. When the number of possible values increases, the probability of each value gets smaller, and the effect of noise increases [8]. This limit might be removed by using the cumulative distribution function (hereinafter referred to as the "CDF"). Unlike the PMF, which is defined for discrete random variables only, the CDF can describe both discrete random variables and continuous random variables.

The CDF of a random variable formula, denoted by formula or formula,  is the function given by:

formula

where formula denotes an event that is a subset of the sample space to which formula was assigned a real value less than or equal to formula.

Shown in Table 1, formula is the remainder after division. The event formula maps to the subset {1 spot, 3 spots, 4 spots, 6 spots} of the sample space. We use naïve definition [4] to calculate the probability of this event:

formula

An effective method for deriving a CDF is to convert a PMF to the CDF. The events formula, formula and formula are mutually exclusive events, thus we can use the addition rule:

formula

formula

formula

It is observed that a CDF for discrete random variables is not continuous. In addition, the PMF and CDF of a random variable contain the same information. If we have known the CDF of a discrete random variable, we can compute the PDM, for example,

formula

Continuous random variables differ from discrete random variables in that a continuous random variable formula that takes on any value in an interval. Because there are infinite number of values that formula could assume, the probability of formula taking on a specific value is 0, that is formula where formula is the specified value. No matter whether we describe a discrete random variable or a continuous variable, the CDF must satisfy the following three conditions [4]:

  • The CDF is an increasing function where: if formula, then formula;
  • The CDF is continuous from the right, namely, right-continuous;
  • The CDF converges to 0 and 1 in the limits;

Now we give the mathematical definitions of a continuous distribution and a continuous random variable: when a CDF of a random variable is differentiable, we consider that the random variable possesses a continuous distribution, and this random variable is a continuous random variable [4]. 

1.2.3 Probability Density Functions

A probability density function (hereinafter referred to as the "PDF"), denoted by formula determines a continuous random variable distribution and the CDF does too. They contain the same information about the continuous random variable distribution. We have already introduced CDF in section 1.2.2. Given a CDF formula of a continuous random variable formula, the PDF is the derivative of the CDF:

formula

where formula is in the support of formula.

Since the CDF is an increasing function and converges to 1 at the upper limit, the PDF of a continuous random variable must satisfy these two criteria:

  • PDF must be nonnegative;
  • PDF must integrate to 1 over the support of the random variable;

To find the probability of the event,  formula, we compute the area on the interval (a, b] under the probability density curve. Note that the endpoints of the interval have probability 0, therefore the probability will not change when we include or exclude the endpoints in the computation.

formula

On the other hand, we can derive a CDF from a known PDF:

formula

where formula is a dummy variable.

Let’s give a practice in use of CDF and PDF to describe a continuous random variable. We simulated a continuous random variable formula by using SQL function RAND(). The continuous random variable distribution can be determined by either the CDF or the PDF:

formula

formula

To find the probability of the random variable taking a value less than 0.5, we compute the area between 0 and 0.5 under the horizontal line formula, which is a rectangle area.

formula

When we generate completely random values by using SQL function RAND(), intuitively, the chance to get a value less than 0.5 is 50%. The intuition agrees with the mathematic calculation.

2 – Discrete Probability Distributions

The distribution of the discrete random variable formula can be expressed with a PMF:

formula

where  formula denotes the support of the random variable formula, formula is an integer number, and formula is the size of the sample space. It is noting that formula could be infinite.

By arranging the elements of formula in an ascending order, then we derive the CDF:

formula

In this section we consider three common distribution functions of discrete random variables:  the discrete uniform distribution, the Bernoulli distribution and the binomial distribution.

2.1 The Discrete Uniform Distribution

The discrete uniform distribution is perhaps the simplest discrete probability distribution [9]. In the experiment of completely randomly choosing a number from a nonempty finite set of numbers, by adopting the notations used in book [4], let formula be the set, formula be the size of the set, and formula be the chosen number. The distribution of formula is a discrete uniform distribution with the parameter formula, denoted by formula. Then we have the PMF and CDF of this distribution:

formula

formula

where formula denotes a subset that contains all values in formula less than or equal to formula, and formula denotes the size of the subset.

The discrete uniform distribution is used very often in practice. The example given in the section 1 tried to generate a random number from the discrete uniform distribution: formula.

2.2 The Bernoulli Distribution

A Bernoulli trial is a random experiment that results in only two possible outcomes: success and failure. The probability of success formula is the same every time the experiment is conducted. The Bernoulli distribution, denoted by formula, arises from the realization of a single Bernoulli trial [10]. The Bernoulli random variable formula has a value of 1 when success occurs and 0 when failure occurs in a trial. The PMF is given by:

formula

formula

where formula.

Any random variables having this PMF possess the Bernoulli distribution with the parameter formula. Since every event has two states: occur, and fail to occur, a Bernoulli trial is naturally associated to every event. This kind of Bernoulli random variable, called the indicator random variable of the event, is often used to simplify problems in probability, therefore is very useful.

2.3 The Binomial Distribution

In the experiment of performing formula independent Bernoulli trials, the probability of success formula is the same in each trial. The number of successes in the experiment is a discrete random variable, denoted by formula. We call the distribution of formula the binomial distribution with parameters formula and formula, denoted by formula. In addition to the PMF and CDF, this is the third way to describe a distribution, which we use a story to explain how the distribution can arise [4]. Then we derive the PMF from the story:

formula

formula

To strengthen the understanding of the binomial distribution, let’s look at the example 6.3 in Stewart’s book [10]:

A grocery store receives a very large number of apples each day. Since past experience has shown that 5% of them will be bad, the manager chooses six at random for quality control purposes. Let formula be the random variable that denotes the number of bad apples in the sample. What values can formula assume? For each possible value of formula, what is the value of formula? (Stewart, 2009,p 119).

When the manager randomly selects an apple, the trial has two possible outcomes: a bad apple or a good apple. The purpose of the experiment is to detect bad apples. When a bad apple is detected, success occurs, and failure occurs otherwise. Since the manager chooses 6 apples from a very large amount that can be considered an indefinitely large, all 6 trials are independent and the probability of success for each trial is constant, that is 5%.

This experiment satisfies the following requirements of being a binomial distribution:

  1. All trials in the experiment are independent;
  2. Each trial has two possible outcomes: success and failure;
  3. The probability of success is the same in each trial;

Therefore, formula has a binomial distribution with parameters formula and formula, denoted by formula. The random variable formula is the random variable that denotes the number of bad apples in the sample. The variable formula can have any value in the set {0, 1, 2, 3, 4, 5, 6}. The PMF is given by:

formula

Table 2 shows all probabilities associated with the number of bad apples detected. The probability that the manager finds no bad apples is 0.735. The probability of finding 1 bad apple is 0.232. The chance to find all 6 bad apples is very low.

formula formula
0 0.735
1 0.232
2 0.031
3 2.14e-03
4 8.46e-05
5 1.78e-06
6 1.56e-08

Table 2 Binomial PMF for formula

3 – Continuous Probability Distributions

The probability distribution of a continuous random variable is called a continuous probability distribution. A continuous probability distribution differs from a discrete probability distribution in that the continuous probability distribution can be described by a PDF and the discrete probability distribution can be described by a PMF. In this section, we will introduce two well-known continuous distributions: the uniform distribution and the standard normal distribution.

3.1 The Uniform Distribution

We translate the notion "completely random" into a mathematical expression. The following PDF is used to describe a random variable that takes completely random values on a real number interval (a, b):

formula

A continuous random variable has the uniform distribution with parameters a and b on the interval (a, b) when it possesses this PDF. We denote this distribution by formula. The CDF is given by

formula

The formula is called standard uniform distribution which has simple PDF and CDF:

formula

formula

Figure 2 illustrates the PDF and the CDF of the formula. We have observed a special characteristic of the uniform distribution on this figure: partitioning the interval (0,1) into many small subintervals, the probability of choosing a random value in any subinterval is proportional to the length of the subinterval. The actual position of the subinterval does not affect the probability. For example, the chance to take a value in the subinterval (0.0, 0.6) is the same as in the subinterval (0.3, 0.9), and is three times as many as in the subinterval (0.5, 0.7).

chart
Figure 2 The Probability Density Function and The Cumulative Distribution Function of formula[4]

3.2 The Standard Normal Distribution

The normal distribution, which has the bell-shaped curve, plays a fundamental role in statistics. The standard normal distribution, denoted by formula, has mean 0 and variance 1. The normal distribution is extremely widely used in statistics due to the central limit theorem [10]. Since we want to generate test data from the standard normal distribution later, this section just simply gives the PDF and CDF of the standard normal distribution.

The PDF and the CDF of the standard normal distribution are given by

formula

formula

Conventionally, the Greek letterformula denotes the standard normal PDF, formula denotes the standard normal CDF, and formula denotes the standard normal random variable.

4 – Random Number Generator

We have now seen five well-known distributions: the discrete uniform distribution, the Bernoulli distribution, the binomial distribution, the uniform distribution and the standard normal distribution. In this section, random numbers from these distributions will be generated by using SQL server function RAND(). Techniques used in this section can generate random numbers from other distributions as well. Other programming languages can also use these techniques to simulate random variates as long as the uniform pseudo random number generator is available in the programming environment.

4.1 Universality of the Uniform

The universality of the uniform, possessed by the uniform distribution, allows us to create a random variable from any continuous distribution. On the other hand, we can transform any continuous distribution into a formula [4]. This property also has several other names, for example inverse transform sampling. We can produce random values from both continuous distributions and discrete distributions by using the universality of the uniform theorem.

The universality of the uniform theorem is the foundation to these random number generation techniques covered in this tip. To precisely present this theorem, I quote it from Hwang & Blitzstein’s book [4]:

(Universality of the Uniform). Let formula be a CDF which is a continuous function and strictly increasing on the support of the distribution. This ensures that the inverse function formula exists, as a function from (0, 1) to formula. We then have the following results.

  1. Let formula and formula(U). Then formula is an r.v. with CDF formula.
  2. Let formula be an r.v. with CDF formula. Then formula.( Blitzstein & Hwang, 2015, p205) 

In their book, "r.v" denotes random variable; formula denotes the inverse function of CDF. We have already known the definition of the CDF:

formula

where formula is a real number and formula.

The function returns the cumulative probability of which the random variable formula takes a value less than or equal to the real number formula. In other words, given a real number formula, the function returns a value of cumulative probability, denoted by formula. Then the inverse function of CDF can be written as

formula

The inverse function returns a real number formula when the cumulative probability value formula is known. The inverse function tells us what value formula would make the probability of the event formula be formula. For example, a snowblower manufacture investigates engine failure times. They can use the inverse function of a CDF to compute the time by which 3% of engines would fail. In this example, the cumulative probability 3% was passed to the inverse function and the failure time was returned.

The first part of the theorem states that, by using the formula, we can create a random variable formula from any continuous distribution. Let’s prove this:

  1. Assume an arbitrary continuous distribution with CDF formula, and the inverse function of CDF formula exists;
  2. Construct a random variable formula by plugging formula into formula, i.e., formula;
  3. The CDF of the formula distribution is defined by formula;
  4. Plug formula into this definition:formula
  5. Since formula is an increasing function, the inequality formula is equivalent to the inequality formula;
  6. The CDF of the formula distribution from step 5 can be re-written as: formula
  7. Since formula has formula, the probability is proportional to the length of the subinterval. That means formula. Namely, formula since formula;
  8. Obtain this equation: formula; that means the CDF of formula distribution has the CDF formula;

Through step 1 to step 8, we have proved that the distribution of the random variable formula possesses the CDF formula, which is the arbitrary distribution, therefore we prove the first part of the theorem.

The second part of the theorem states formula has the standard uniform distribution, i.e., formula. We have already known that the values of formula fall within the range [0,1], but we are not sure all possible values are uniformly distributed on the range.

To prove this part, we will use a fact that a function of a random variable is a random variable [4]. Since formula is a random variable and formula is a function, formula is a random variable. It is noting that a function of discrete random variables is still discrete. The second part of this theorem does not work for discrete random variables.

Let a random variable formula. Then, we verify that formula has the standard uniform distribution:

  1. Assume a continuous distribution with CDF formula, and the inverse function of CDF formula exists;
  2. Define the random variable formula and the CDF of formula is formula;
  3. Plug formula into  formula, then formula;
  4. Since formula is an increasing function, the inequality formula is equivalent to the inequality  formula;
  5. The CDF of formula from step 3 can be re-written as formula
  6. Since formula is CFD of formula, i.e. formula, then formula
  7. From steps 5 and 6, we have known: formula
  8. The CDF formula describes distribution formula, therefore we proved the second part of the theorem;

The proof helps to strengthen our understanding of the theorem. In the rest of this tip, we will demonstrate a method to generate random variables from specified distributions. The method can be summarized into 4 steps:

  1. Derive the inverse function formula of a specified CDF formula;
  2. Produce a random number from the standard uniform distribution, formula;
  3. Plug the random number into the inverse function formula;
  4. Return generated random number formula, which has the CDF formula;

4.2 Discrete Random Number Generator

The inverse function of the CDF for a discrete function is discontinuous. Assuming formula, the CDF is given by

formula

formula

formula

Figure 3 illustrates the CDF for the discrete function. The point (2, 0.6) implicates how we define the inverse function: given formula, we then find formula. By defining formula in terms of CDF, we obtained the inverse function in the following equation. This expression demonstrates a simple search algorithm.

formula

chart
Figure 3 The Cumulative Distribution Function for formula

4.2.1 Generating a Random Number from the Discrete Uniform Distribution

Let’s revisit the example: generating a completely random integer number in the range [1,3].

Step 1. Derive the inverse function formula of the CDF formula.  We have already derived the inverse function:

formula

Step 2. Produce a random number from formula. We use the following SQL statement to generate a random value:

SELECT RAND(CHECKSUM(NEWID())) AS Random_Number 

The query returned a random number: 0.351873119639275. If you run this query, you would obtain a different value from mine.

Step 3. Plug the value, formula = 0.351873119639275, into the inverse function formula:

formula

Step 4. Return the generated random number formula. The random number generator returns formula.

I wrote a script to examine whether the random numbers generated are distributed as formula. In the script, I ran this 4-steps procedure 10,000 times, then loaded the results into a data frame in R. I plotted a bar chart of frequency distribution to visualize the probability of choosing each integer in the set {1, 2, 3}.

# Reset the compute context to your local workstation. 
rxSetComputeContext("local") 

# Define connection string to SQL Server 
sql.server.conn.string <- "Driver=SQL Server;Server=.;Database=TutorialDB;Trusted_Connection={Yes}" 

# Create an RxSqlServerData data source object 
data.source.object <- RxSqlServerData(sqlQuery = " 
    SELECT top 10000 
       x.random_value, 
      CASE 
         WHEN x.random_value <= (1.0/3.0) THEN 1 
         WHEN x.random_value > (1.0/3.0) and x.random_value <= (2.0/3.0) THEN 2 
         WHEN x.random_value > (2.0/3.0) and x.random_value <= 1.0 THEN 3 
      END as random_integer 
    FROM sys.all_columns a1 
    CROSS APPLY sys.all_columns a2 
    CROSS APPLY 
    ( 
       SELECT RAND(CHECKSUM(NEWID())) as random_value 
    ) x                             
                                ", 
                                connectionString = sql.server.conn.string) 
# Read the data into a data frame in the local R session. 
data.frame.random.numbers <- rxImport(data.source.object) 

# Create a Frequency Distribution Bar Chart quickly with the plot() function 
# when the categorical variable to be plotted is a factor or ordered factor. 
plot(as.factor(data.frame.random.numbers$random_integer), main = "Frequency Distribution of Random Integers", 
     xlab = "Random Integer", 
     ylab = "Number of Occurrence", 
     col = "blue") 

I have obtained a bar chart as shown in Figure 3. Comparing to Figure 1, the chance of selecting one number from these three integer numbers was almost equally likely.

chart
Figure 4 Frequency Distribution of Generated Random Integers

4.2.2 Generating a Random Number from the Bernoulli Distribution

Now let’s look at a case where events are not equally likely to occur. We are asked to generate sample data from a Bernoulli distribution, formula.

Step 1. Derive the inverse function formula of the CDF formula.  The formula is written as

formula

formula

Then, the inverse function of this CDF is given by

formula

Step 2. Produce a random number from formula. We used the following SQL statement to generate a random value, and the query returned a random number: 0.418275981900669:

SELECT RAND(CHECKSUM(NEWID())) AS Random_Number 

Step 3. Plug the value, formula = 0. 418275981900669, into the inverse function formula:

formula

Step 4. Return the generated random number formula. The random number generator returns formula.

To verify the generated random numbers are from formula, I ran the 4-step procedure 10,000 times through a R script. I plotted a bar chart of frequency distribution to visualize the chances of generating the number 0 and 1.

# Reset the compute context to your local workstation. 
rxSetComputeContext("local") 

# Define connection string to SQL Server 
sql.server.conn.string <- "Driver=SQL Server;Server=.;Database=TutorialDB;Trusted_Connection={Yes}" 

# Create an RxSqlServerData data source object 
data.source.object <- RxSqlServerData(sqlQuery = "                                      
            SELECT top 10000 
            x.random_value, 
            CASE 
               WHEN x.random_value <= 0.6 THEN 0 
               WHEN x.random_value > 0.6 and x.random_value <= 1.0 THEN 1 
            END as random_integer 
            FROM sys.all_columns a1 
            CROSS APPLY sys.all_columns a2 
            CROSS APPLY 
            ( 
               SELECT RAND(CHECKSUM(NEWID())) as random_value 
            ) x                               
                                ", 
                                connectionString = sql.server.conn.string) 

# Read the data into a data frame in the local R session. 
data.frame.random.numbers <- rxImport(data.source.object) 

# Create a Frequency Distribution Bar Chart quickly with the plot() function 
# when the categorical variable to be plotted is a factor or ordered factor. 
plot(as.factor(data.frame.random.numbers$random_integer), main = "Frequency Distribution of Random Integers", 
     xlab = "Random Integer", 
     ylab = "Number of Occurrence", 
     col = "blue") 

The bar chart, shown in Figure 5, illustrates that the chance of success, i.e. formula, is 40%, and the chance of failure, i.e. formula, is 60%. The sample data has the desired distribution.

chart
Figure 5 Frequency Distribution of Generated Random Integers

4.2.3 Generating a Random Number from the Binomial Distribution

Deriving an inverse function formula of a CDF formula is the critical step in this 4-step process. Sometimes, it may be difficult to find the inverse function of a CDF. This section introduces a search algorithm to find the value of random variable formula when given a value of CDF.

Assuming formula are in the support of formula, and they are in an ascending order. Because the CDF formula is an increasing function, this inequality exists:

formula

Through the method of computing CDF from PMF, the above inequality can be written as:

formula

Then, given PMF, we can find the value of formula that makes formula satisfy the inequality.

Let’s look at the grocery store example in the section 2.3. To simplify the calculation, we use the data in Table 2.

Step 1. Derive the inverse function formula. Inspired by Matt’s article [11], I wrote a Pseudo-code to explain how to numerically compute the returning values of the inverse function:

Input: the lookup value "u", a list of PMF values "probability list", a list of sequential random numbers "random number list". //Each element in the probability list corresponds the element in the random list.
Output: the value of random variable.

discrete inverse function ("u", "probability list", "random number list")
{
   FOR each position number in the probability list 
      IF the position number is 1 THEN
         Do Nothing
      ELSE
         F1 = sum of values in the probability list from 1 to the position number less 1
         F2 = sum of values in the probability list from 1 to the position number 
         IF (the value of "u" is greater than F1) 
            and (the value of "u" is less than or equal to F2) THEN 
               REUTN the element in the random number list based on the position
         END IF
      END IF
   END FOR 
}

Step 2. Generate a random number from formula. We used the following SQL statement to generate a random value and the query returned a random number: 0.782271114545155:

SELECT RAND(CHECKSUM(NEWID())) AS Random_Number 

Step 3. Plug the value, formula= 0.782271114545155, into the inverse function formula:

formula

formula

Since formula,  we have formula

The approach to find the returning value of the inverse function is analogous to Figure 3, thus

formula

Step 4. Return the generated random number formula. The random number generator returns formula.

I wrote a SQL script to run the 4-step procedure 10,000 times, then computed the mean and variance of these generated numbers.

USE TutorialDB 
GO
 
CREATE TABLE #generated_random_numbers ( 
  random_numbernumeric(16,14) 
); 
GO 

DECLARE @random_value TABLE 
( 
  posint, 
  kint, 
  pnumeric(16,14) 
) 

DECLARE @random_list_size int = 7, 
@look_count int = 2, 
@unif numeric(16,14) = RAND(CHECKSUM(NEWID())), 
@rtn_random_value int = 0 
insert into @random_value 
values  (1, 0,0.7350919), 
(2,1,0.2321343), 
(3,2,0.03054398), 
(4,3,0.002143437), 
(5,4,8.460938e-05), 
(6,5,1.78125e-06), 
(7,6,1.5625e-08) 

WHILE @look_count <= @random_list_size 
BEGIN 
   IF (@unif > (SELECT SUM(p) FROM @random_value WHERE pos <= @look_count -1) 
   AND 
   @unif <= (SELECT SUM(p) FROM @random_value WHERE pos <= @look_count)) 
  
   SELECT @rtn_random_value = k FROM @random_value WHERE pos <= @look_count 

   SET @look_count = @look_count + 1 
END
 
INSERT INTO #generated_random_numbers VALUES(@rtn_random_value) 
GO 10000    

SELECT  COUNT(*)AS [Sample Size], 
AVG(random_number)AS [Expectation], 
VAR(random_number)AS [Variance] 
FROM #generated_random_numbers 
GO 

SELECT CAST(random_number as INT)AS  [Random Number], 
   COUNT(*)*1.0/10000.0AS[Relative Frequency] 
FROM #generated_random_numbers 
GROUP BY random_number 
ORDER BY random_number 
GO 

DROP TABLE #generated_random_numbers; 
GO 

In theory, the expectation and variance of binomial distribution formula are given by

formula

formula

Comparing the statistics of the generated numbers, shown in Figure 6, to the theoretical computation results, they are close. The relative frequency table in Figure 6 does not include random numbers 4, 5 and 6. The probabilities to produce these numbers are very low. We need to increase the sample size in order to generate all possible values. The results in Figure 6 indicate that the sample data approximately has formula.

query result set
Figure 6 The Screenshot of Sample Statistics Calculated by T-SQL

4.3 Continuous Random Number Generator

We have given some practice in use of the universality of the uniform theorem to generate discrete variables from specific distributions. The critical step is to derive an inverse or near-inverse function of CDF. When the CDF of a continuous variable is invertible, we continue to use the 4-step method. Tim [12] provides a list of inverse functions of CDFs. We must consider other approaches when we cannot express the inverse function by using the standard elementary functions. In this section, we first walkthrough an example when the inverse function of CDF exists, then we generate a random variable from a distribution without a closed-form inverse function of the CDF.

4.3.1 Generating a Random Number from the Uniform Distribution

We are going to generate test data from formula. the CDF is given by

formula

Step 1. Derive the inverse function formula of the CDF formula:

formula

Step 2. Produce a random number from formula. We used the following SQL statement to generate a random value and the query returned a random number: 0.220432802523734:

SELECT RAND(CHECKSUM(NEWID())) AS Random_Number 

Step 3. Plug the value, formula= 0.220432802523734, into the inverse function formula:

formula

Step 4. Return the generated random number formula. The random number generator returns formula.

To examine the generated random numbers are from formula, I wrote a script to execute the 4-step procedure 10,000 times, then loaded the results into a data frame in R. I plotted a histogram to illustrate the frequency distribution.

# Reset the compute context to your local workstation. 
rxSetComputeContext("local") 
# Define connection string to SQL Server 
sql.server.conn.string <- "Driver=SQL Server;Server=.;Database=TutorialDB;Trusted_Connection={Yes}" 
# Create an RxSqlServerData data source object 
data.source.object <- RxSqlServerData(sqlQuery = "                                     
                                SELECT top 10000 
                                x.random_value, 
                            x.random_value * 27.0 + 12.0 as random_number 
                                FROM sys.all_columns a1 
                            CROSS APPLY sys.all_columns a2 
                            CROSS APPLY 
                            ( 
                            SELECT RAND(CHECKSUM(NEWID())) as random_value 
                            ) x                                
                                ", 
                                connectionString = sql.server.conn.string) 
# Read the data into a data frame in the local R session. 
data.frame.random.numbers <- rxImport(data.source.object) 
# Create a Histogram of Random Numbers 
hist(data.frame.random.numbers$random_number, main = "The Histogram of Random Numbers", 
     breaks = seq(12,39,by=3), 
     xlab = "Random Number", 
     ylab = "Frequency", 
     col = "blue") 

Figure 7 indicates that these random numbers are approximately uniformly distributed in the range [12, 27].

chart
Figure 7 The Histogram of Random Numbers

4.3.2 Generating a Random Number from the Standard Normal Distribution

The standard normal distribution, denoted by formula, is widely used probability distribution. Both PDF and CDF, given in the section 3.2, are not closed-form expression. We are going to use the Box-Muller transformation [13] for generating a standard normal variable:

  1. Generate two continuous random variables from the standard uniform distribution: formula and formula;
  2. Create two functions of these two standard uniform random variables:

formula

formula

  1. Obtain two identically independent standard normal distributions:

formula

formula

Several approaches [14,15] have proven the Box-Muller transformation. I prefer to the explanation in chapter 8 of book [4], which is easier to understand.

I have run a script to examine these random numbers generating fromformula. The script ran these 3 steps 10,000 times. I plotted a histogram to illustrate the frequency distribution of these random numbers.

# Reset the compute context to your local workstation. 
rxSetComputeContext("local") 

# Define connection string to SQL Server 
sql.server.conn.string <- "Driver=SQL Server;Server=.;Database=TutorialDB;Trusted_Connection={Yes}" 

# Create an RxSqlServerData data source object 
data.source.object <- RxSqlServerData(sqlQuery = "                                     
    SELECT top 10000 
    sqrt(-2.0*log(random_value_2))*cos(2.0*pi()*(random_value_1)) as random_number_x, 
    sqrt(-2.0*log(random_value_2))*sin(2.0*pi()*(random_value_1)) as random_number_y 
    FROM sys.all_columns a1 
    CROSS APPLY sys.all_columns a2 
    CROSS APPLY 
    ( 
    SELECT RAND(CHECKSUM(NEWID())) as random_value_1, 
    RAND(CHECKSUM(NEWID())) as random_value_2 
    ) x                              
                                ", 
                                connectionString = sql.server.conn.string) 

# Read the data into a data frame in the local R session. 
data.frame.random.numbers <- rxImport(data.source.object) 

# Create a a standard normal distribution 
densitySamps <- seq(-4, 4, by = 0.01) 

# 2 figures arranged in 1 row and 2 columns 
par(mfrow = c(1, 2)) 

# Create a Histogram of Random Numbers 
hist(data.frame.random.numbers$random_number_x, main = "The Histogram of Random Variable X",     
     xlab = "X",freq = FALSE) 

# Add a standard normal density curve 
lines(densitySamps, dnorm(densitySamps), lwd = 2, col = "red") 

# Create a Histogram of Random Numbers 
hist(data.frame.random.numbers$random_number_y, main = "The Histogram of Random Variable Y", 
     xlab = "Y", freq = FALSE) 

# Add a standard normal density curve 
lines(densitySamps, dnorm(densitySamps), lwd = 2, col = "red") 

Figure 8 illustrates that the random variables approximately have the standard normal distribution. 

chart
Figure 8 The Histograms of two Random Variables X and Y

Summary

In this tip, we have explored random variables and their distributions. We have learned the universality of the uniform theorem, and Box-Muller transformation, as well as how to create random values from a specific distribution. We used R language to execute SQL statements for generating random numbers and applied visualization techniques to examine frequency distributions of these random numbers.

The universality of the uniform theorem has been proved in this tip so that the 4-step method was provided to generate a random variable from any specific distribution. We also demonstrated numerical methods to find inverse functions of cumulative distribution functions.

References

[1] Snider, D. (2016, March 29). SQL Server T-SQL Code to Generate a Normal Distribution. Retrieved from https://www.mssqltips.com/sqlservertip/4233/sql-server-tsql-code-to-generate-a-normal-distribution/.

[2] Snider, D. (2014, August 04). Create Your Own RANDBETWEEN Function in T-SQL. Retrieved from https://www.mssqltips.com/sqlservertip/3297/create-your-own-randbetween-function-in-tsql/.

[3] Camps, D. (2012, July 03). Generating Non-uniform Random Numbers with SQL. Retrieved from https://www.sqlservercentral.com/articles/generating-non-uniform-random-numbers-with-sql/.

[4] Blitzstein, K. J. & Hwang, J. (2015). Introduction to Probability. Boca Raton, FL: CRC Press.

[5] Jost, S. (2017). CSC 423: Data Analysis and Regression. Retrieved from DePaul University Website: http://facweb.cs.depaul.edu/sjost/csc423/

[6] Collins online dictionary. Random. Retrieved from https://www.collinsdictionary.com/dictionary/english/random.

[7] Collins online dictionary. Variable. Retrieved from https://www.collinsdictionary.com/dictionary/english/variable.

[8] Downey, B. A. (2014, October).  Think Stats, 2nd Edition. Sebastopol, CA: O’Reilly Media.

[9] Guttman, I., & Gupta, C. B. (2013). Statistics and Probability with Applications for Engineers and Scientists. Hoboken, NJ: John Wiley & Sons.  

[10] Stewart, J. W. (2009). Probability, Markov Chains, Queues, and Simulation. Princeton, NJ: Princeton University Press.  

[11] Bonakdarpour, M. (2016, February 02). Inverse Transform Sampling. Retrieved from fiveMinuteStats: https://stephens999.github.io/fiveMinuteStats/inverse_transform_sampling.html/.

[12] Hargreaves, T. (2019, February). Generating Normal Random Variables - Part 1: Inverse Transform Sampling. Retrieved from T-Tested: https://www.ttested.com/generating-normal-random-variables-part-1/.

[13] Box, G. E. P.& Muller, Mervin E. (1958). A Note on the Generation of Random Normal Deviates. Ann. Math. Statist. 29 (1958), no. 2, 610--611. doi:10.1214/aoms/1177706645. https://projecteuclid.org/euclid.aoms/1177706645

[14] Goodman, J. (2005, August). Chapter 2: Simple Sampling of Gaussians. Retrieved from Courant Institute: https://www.math.nyu.edu/faculty/goodman/teaching/MonteCarlo2005/notes/GaussianSampling.pdf.

[15] Vafa, K. (2017, February). The Box-Muller Transform. Retrieved from keyonvafa.com: http://keyonvafa.com/box-muller-transform/.

Next Steps


Last Updated: 2020-02-11


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About the author
MSSQLTips author Nai Biao Zhou Nai Biao Zhou is a Senior Software Developer with 20+ years of experience in software development, specializing in Data Warehousing, Business Intelligence, Data Mining and solution architecture design.

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Monday, February 17, 2020 - 8:41:19 AM - Jeff Moden Back To Top

Ah... understood.  The way you wrote the first part of the article, it sounded like you were trying to prove that RAND() by itself does not provide an even distribution and we bothn know that would be incorrect.

Thanks again, for the article.


Friday, February 14, 2020 - 3:58:24 PM - Nai Biao Zhou Back To Top

Hello Jeff,

Thank you to read this tip and add your comments.

A key point of this tip is to introduce the universality of the uniform theorem, which allows us to create a random variable from any continuous distributions.

To use the universality of the uniform, we should have a uniform distribution. In this tip, I assumed the SQL function RAND can generate random values from the uniform distribution. In other words, I assumed the values generated by the function is evenly distributed.

Dwain Camps investigated the RAND function in this article:   https://www.sqlservercentral.com/articles/generating-non-uniform-random-numbers-with-sql

In this tip, I did not investigate how the values generated by the RAND function were distributed. I just assumed the RAND function could generate the uniform distribution.

This tip did not provide proof that RAND is not evenly distributed. On the contrast, this tip assumed that RAND was evenly distributed.

I used the example in the tip to show a situation that random values we created may not satisfy our requirements. Then, I introduced how to create random variables by using the universality of the uniform theorem. Like in a cooking class, I show overcooked food first, then introduce some techniques to cook food appropriately. I revisit the problem in Section 4.2.1, where I gave a mathematical formula:

formula

We then can write this formula by using SQL functions, for example, Round, Ceiling, Floor or other T-SQL techniques. The result is shown in Figure 4.

 chart

Your comments on the script I wrote are correct. I spent some time on designing this example so that readers can easily discover the flaw in the script. That is the reason I used only three numbers 1, 2 and 3. The distribution is obviously uneven. In reality, when we wrote codes to generate random numbers, we may just simply run the script to check if the script will return unpredicted values. The example shows that this may not be accurate. We should also check the distribution as well.

If you still have any question on this, please let me know.

Best Regards,

Nai Biao Zhou


Tuesday, February 11, 2020 - 10:07:07 PM - Jeff Moden Back To Top

I really appreciate the time and effort that went into this article and my hat is off to you for that.  Unfortunately, your “proof” that RAND is not evenly distributed is flawed because your formula to generate random integers from 1 to 3 is incorrect.  Let’s see what I mean.

First of all, your “range” formula is incorrect.  There are 3 integers in the domain of 1, 2, and 3.  Your formula of (3-1) results in “2”, not “3”.

If we consider the full range and domain of RAND, it returns values >= 0.0 thru 0.999999999999999.  If we multiply those two limiting values by the range of “2”, we end up with a possible domain of values from 0.0 thru 1.99999999999998.

From there, the formula applys the ROUND function to 0 places.  Let’s see what happens there.

For values < than .5, ROUND will round down for a rounded value of “0”.  .5 is only ¼ of the range of values between 0.0 and 1.99999999999998 and so, out of 10,000 random values, only about 2,500 will have a value of “0”.

Similarly, for values >= 1.5, ROUND will round up for a rounded value of “2”.  The values >= 1.5 is only ¼ of the range of values between 0.0 and 1.99999999999998 and so, out of 10,000 random values, only about 2,500 will have a value of “2”.

All of the remaining possible values will be >= 0.5 and 1.5.  That’s a full ½ of all the possible values from 0.0 and 1.99999999999998 and, so, out of 10,000 random values, about 5,000 will have the value of “1”.

After that, you add a “1” to those highly skewed values and you end up with about ¼ of the values being “1”, ¼ of the values being “3”, and all the rest, which is ½ of all the values returned, being a “2”.

The likely reason you made the formula the way you did is because you rounded instead of truncating and you not only got values of 1, 2, and 3 but the rounding also cause spurious values of “4”.  If you had used the 3rd operand of ROUND to truncate rather than round, you would have been ok provided that you also remembered to calculate the range of integers correctly.

With all of that in mind, here’s some code to demonstrate what I’m talking about.  Except for the fact that I declared some variables to use rather than embedding numeric constants in the code and I also used the 3rd operand (unknown to most) of ROUND to truncate rather than round, you’ll find that the distribution of values is VERY even.  Logically it has to be because the distribution of NEWID() values is also incredibly evenly distributed.

Here’s the commented code.  I hope this forum preserves the formatting but, if it doesn’t, I did try.

--===== Declare and assign the limits of the desired range of integers.
DECLARE  @LoDomainLimit INT = 1
        
,@HiDomainLimit INT = 3
;
--===== Correctly calculate the range of integers in the domain.
DECLARE  @DomainRange   INT = @HiDomainLimit-@LoDomainLimit+1
;
--===== If the Temp Table already exists, drop it to make it easier.
     
IF OBJECT_ID('tempdb..#MyHead','u') IS NOT NULL
   
DROP TABLE #MyHead
;
--===== Randomly calculate integers within the limits of the domain.
     
-- NOTE THE MOSTLY UNKNOWN 3rd OPERAND OF ROUND.
     
-- It should be a CONVERT(INT,n) instead because we want to return INTeger values.
 
SELECT TOP (10000) 
         u
.random_value
        
,FullDomainValue = @DomainRange * u.random_value
        
,random_integer = ROUND(@DomainRange * u.random_value ,0,1) + @LoDomainLimit
 
--INTO #MyHead
   
FROM       sys.all_columns a1 
  
CROSS APPLY sys.all_columns a2 
  
CROSS APPLY ( 
                
SELECT RAND(CHECKSUM(NEWID())) as random_value 
              
) u
;

If we uncomment the “INTO #MyHead” in the code above and run it again, it will create and populate the “#MyHead” table so that we can run check the counts to determine the distribution.  Like this…

--===== Display the distribution and total count.
 
SELECT random_integer, Occurences = COUNT(*)
   
FROM #MyHead
  
GROUP BY random_integer WITH ROLLUP
;

Since it IS random data (and no longer has an artificial skew), you get numbers similar to the following and will vary every time you all the code.  The line with the NULL on it is from the ROLLUP and is the total count of all rows, just as a sanity check.

The bottom line here is that your “proof” that RAND is not evenly distributed is flawed and you should work with the owners of this fine forum to make a correction.  I’ve not checked the rest of what you’ve done but, I have to say it again, you put a great effort into and my hat’s off to you.



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