Linear Transformations of Random Variables

Sometimes, it is necessary to apply a linear transformation to a random variable. This lesson explains how to make a linear transformation and how to compute the mean and variance of the result.

What is a Linear Transformation?

A linear transformation is a change to a variable characterized by one or more of the following operations: adding a constant to the variable, subtracting a constant from the variable, multiplying the variable by a constant, and/or dividing the variable by a constant.

When a linear transformation is applied to a random variable, a new random variable is created. To illustrate, let X be a random variable, and let m and b be constants. Each of the following examples show how a linear transformation of X defines a new random variable Y.

  • Adding a constant: Y = X + b
  • Subtracting a constant: Y = X - b
  • Multiplying by a constant: Y = mX
  • Dividing by a constant: Y = X/m
  • Multiplying by a constant and adding a constant: Y = mX + b
  • Dividing by a constant and subtracting a constant: Y = X/m - b

Note: Suppose X and Z are variables, and the correlation between X and Z is equal to r. If a new variable Y is created by applying a linear transformation to X, then the correlation between Y and Z will also equal r.

How Linear Transformations Affect the Mean and Variance

Suppose a linear transformation is applied to the random variable X to create a new random variable Y. Then, the mean and variance of the new random variable Y are defined by the following equations.

Y = mX + b       and       Var(Y) = m2 * Var(X)

where m and b are constants, Y is the mean of Y, X is the mean of X, Var(Y) is the variance of Y, and Var(X) is the variance of X.

Note: The standard deviation (SD) of the transformed variable is equal to the square root of the variance. That is, SD(Y) = sqrt[ Var(Y) ].

Test Your Understanding of This Lesson

Problem 1

The average salary for an employee at Acme Corporation is $30,000 per year. This year, management awards the following bonuses to every employee.

  • A Christmas bonus of $500.
  • An incentive bonus equal to 10 percent of the employee's salary.

What is the mean bonus received by employees?

(A) $500
(B) $3,000
(C) $3,500
(D) None of the above.
(E) There is not enough information to answer this question.

Solution

The correct answer is C. To compute the bonus, management applies the following linear transformation to the each employee's salary.

Y = mX + b
Y = 0.10 * X + 500

where Y is the transformed variable (the bonus), X is the original variable (the salary), m is the multiplicative constant 0.10, and b is the additive constant 500.

Since we know that the mean salary is $30,000, we can compute the mean bonus from the following equation.

Y = mX + b
Y = 0.10 * $30,000 + $500 = $3,500

 

Problem 2

The average salary for an employee at Acme Corporation is $30,000 per year, with a variance of 4,000,000. This year, management awards the following bonuses to every employee.

  • A Christmas bonus of $500.
  • An incentive bonus equal to 10 percent of the employee's salary.

What is the standard deviation of employee bonuses?

(A) $200
(B) $3,000
(C) $40,000
(D) None of the above.
(E) There is not enough information to answer this question.

Solution

The correct answer is A. To compute the bonus, management applies the following linear transformation to the each employee's salary.

Y = mX + b
Y = 0.10 * X + 500

where Y is the transformed variable (the bonus), X is the original variable (the salary), m is the multiplicative constant 0.10, and b is the additive constant 500.

Since we know the variance of employee salaries, we can compute the variance of employee bonuses from the following equation.

Var(Y) = m2 * Var(X) = (0.1)2 * 4,000,000 = 40,000

where Var(Y) is the variance of employee bonuses, and Var(X) is the variance of employee salaries.

And finally, since the standard deviation is equal to the square root of the variance, the standard deviation of employee bonuses is equal to the square root of 40,000 or $200.