Linear Combinations of Random Variables (College Board AP® Statistics)

Study Guide

Dan Finlay

Written by: Dan Finlay

Reviewed by: Lucy Kirkham

Linear combinations of random variables

What is a linear combination of random variables?

  • A linear combination of random variables is where multiples of given random variables are added together

  • If X and Y are random variables, then a linear combination is of the form W equals a X plus b Y

    • where a and b are real constants

  • Linear combinations can model real-life scenarios

    • e.g. on a given day, X laptops are sold at $400 each and Y desktops are sold at $600 each

      • the total revenue for a given day, R, can therefore be modeled as R equals 400 X plus 600 Y

How do I find the mean of a linear combination of random variables?

  • To find the mean of a linear combination of random variables

    • treat the original means just like the values of the variables

      • multiply each mean by the relevant constant

      • then add together

  • If W equals a X plus b Y then mu subscript W equals a mu subscript X plus b mu subscript Y

  • If W equals a X minus b Y then mu subscript W equals a mu subscript X minus b mu subscript Y

  • This can be extended to multiple random variables

    • If Y equals a subscript 1 X subscript 1 plus-or-minus a subscript 2 X subscript 2 plus-or-minus... plus-or-minus a subscript n X subscript n then mu subscript Y equals a subscript 1 mu subscript 1 plus-or-minus a subscript 2 mu subscript 2 plus-or-minus... plus-or-minus a subscript n mu subscript n

How do I find the variance of a linear combination of random variables?

  • To find the variance of a linear combination of independent random variables

    • multiply each of the original variances by the relevant constant squared

    • then add together

  • If W equals a X plus b Y then sigma subscript W squared equals a squared sigma subscript X squared plus b squared sigma subscript Y squared

  • If W equals a X minus b Y then sigma subscript W squared equals a squared sigma subscript X squared plus b squared sigma subscript Y squared

    • The terms are always added within the formula

      • This is because the square of any negative number is always positive

  • This can be extended to multiple independent random variables

    • If Y equals a subscript 1 X subscript 1 plus-or-minus a subscript 2 X subscript 2 plus-or-minus... plus-or-minus a subscript n X subscript n then sigma subscript Y squared equals a subscript 1 squared sigma subscript 1 squared plus a subscript 2 squared sigma subscript 2 squared plus... plus a subscript n squared sigma subscript n squared

Examiner Tips and Tricks

The formula for the variance of a linear combination only works if the random variables are independent. If you use this formula in an exam question, be sure to state the condition of independence.

How do I find the standard deviation of a linear combination of random variables?

  • To find the standard deviation of a linear combination of independent random variables

    • take the positive square root of the variance

  • If W equals a X plus-or-minus b Y then sigma subscript W equals square root of a squared sigma subscript X squared plus b squared sigma subscript Y squared end root

Examiner Tips and Tricks

Remember that sigma subscript W not equal to open vertical bar a close vertical bar sigma subscript X plus open vertical bar b close vertical bar sigma subscript Y.

Formulas for the standard deviation do not look nice because the square root function is not a linear operator, square root of a plus b end root not equal to square root of a plus square root of b.

If you are asked to find the standard deviation of a linear combination, always find the variance first.

Is the distribution of adding two observations of X the same as the distribution of 2X?

  • If you add two independent observations of the random variable X then this distribution is denoted X subscript 1 plus X subscript 2

    • where X subscript 1 is the random variable for the first observation

    • and X subscript 2 is the random variable for the second observation

    • both X subscript 1 and X subscript 2 have the same distribution as X

  • If you double one observation of the random variable X then this distribution is denoted 2 X

  • The distribution of X subscript 1 plus X subscript 2 is not the same as the distribution of 2 X

    • The means of both are equal

      • mu subscript X subscript 1 plus X subscript 2 end subscript equals mu subscript 2 X end subscript equals 2 mu subscript X

    • However, the variances are not equal

      • sigma subscript X subscript 1 plus X subscript 2 end subscript superscript 2 equals sigma subscript X subscript 1 end subscript superscript 2 plus sigma subscript X subscript 2 end subscript superscript 2 equals 2 sigma subscript X superscript 2

      • sigma subscript 2 X end subscript superscript 2 equals 2 squared sigma subscript X superscript 2 equals 4 sigma subscript X superscript 2

Worked Example

An exam contains two sections, Section I and Section II, both of which are out of 100 points.

X is the random variable for the number of points awarded to a randomly selected student in Section I and Y is the random variable for the number of points awarded to that student in Section II.

The number of points awarded in each section can be assumed to be independent.

The mean and standard deviation of the two variables are given in the table.

Variable

Mean

Standard deviation

Number of points in Section I (X)

62.5

10.7

Number of points in Section II (Y)

74.1

8.4

The overall score is calculated by adding 40% of the number of points in Section I and 60% of the number of points in Section II.

Let the score of a randomly selected student be represented by the random variable S. The score is calculated using the formula S equals 0.4 X plus 0.6 Y.

(a) Calculate the expected value of S.

Answer:

This is a linear combination of random variables

If W equals a X plus b Y then mu subscript W equals a mu subscript X plus b mu subscript Y

Use the formula and the means in the table to calculate the mean of the scores

table row cell mu subscript S end cell equals cell 0.4 mu subscript X plus 0.6 mu subscript Y end cell row blank equals cell 0.4 open parentheses 62.5 close parentheses plus 0.6 open parentheses 74.1 close parentheses end cell row blank equals cell 69.46 end cell end table

The expected value of S is 69.46 points

(b) Calculate the standard deviation of S.

Answer:

First, find the variance of S

If W equals a X plus b Y then sigma subscript W squared equals a squared sigma subscript X squared plus b squared sigma subscript Y squared

Use the formula and standard deviations in the table to calculate the variance of the scores

State the condition of independence

X subscript 1 and X subscript 2 are independent, therefore

table row cell sigma subscript S superscript 2 end cell equals cell 0.4 squared sigma subscript X superscript 2 plus 0.6 squared sigma subscript Y superscript 2 end cell row blank equals cell 0.4 squared times 10.7 squared plus 0.6 squared times 8.4 squared end cell row blank equals cell 43.72 end cell end table

Take the positive square root to find the standard deviation

table row cell sigma subscript S end cell equals cell square root of 43.72 end root end cell row blank equals cell 6.612... end cell end table

The standard deviation of S is 6.61

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Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.

Lucy Kirkham

Author: Lucy Kirkham

Expertise: Head of STEM

Lucy has been a passionate Maths teacher for over 12 years, teaching maths across the UK and abroad helping to engage, interest and develop confidence in the subject at all levels.Working as a Head of Department and then Director of Maths, Lucy has advised schools and academy trusts in both Scotland and the East Midlands, where her role was to support and coach teachers to improve Maths teaching for all.