Linear Combinations of Random Variables (Cambridge (CIE) A Level Maths: Probability & Statistics 2): Revision Note

Exam code: 9709

Dan Finlay

Written by: Dan Finlay

Reviewed by: Lucy Kirkham

Updated on

aX+b

  • If a and b are constants then the following results are true

    • E(aX + b) = aE(X) + b

    • Var(aX + b) = a² Var(X)

  • Note that the mean is affected by multiplication and addition whereas addition does not change the variance

  • The factor of   includes the squared because the values of X are squared in the calculation

    • You could try and use the first result and the formula for variance to verify the second result

  • Remember a subtraction can be written as an addition

    • X – b can be written as X + (-b)

  • And division can be written as a multiplication

    • xacan be written as 1aX

What does the distribution of aX + b look like?

  • A linear function is applied to each value of X

  • The graphical representation of aX + b is a linear transformation (a translation and a stretch) of the graphical representation of X

  • If X follows a normal distribution then aX + b will also follow a normal distribution

    • If X ~ N(μ, σ²) then aX + b ~ N(aμ + b, a²σ²)

  • If X follows a binomial, geometric or Poisson distribution then aX + b will no longer follow the same type of distribution

Worked Example

X is a random variable such that E(X)=5 and Var(X)=4.

Find the value of:

(i) E(3X+5)

(ii) Var(3X+5)

(iii) Var(2X)

Answer:

2-2-1-ax--b-we-solution-1-2

aX + bY

  • If X and Y are two random variables then X + Y is the random variable whose values are the sums of each pair containing one value of X and one value of Y

  • E(X + Y) = E(X) + E(Y)

    • this is true for any random variables X and Y

    • Note that E(X Y) = E(X) - E(Y) (see below for more information)

  • Var(X + Y) = Var(X) + Var(Y)

    • this is true if X and Y are independent

    • Note that Var(X - Y) = Var(X) + Var(Y) (see below for more information)

What does the distribution of X + Y  look like?

  • If X and Y are two independent Poisson distributions then X + Y is also a Poisson distribution

    • If X~Po(λ) and Y~Po(μ) then X+Y~Po(λ+μ)

  • If X and Y are two independent normal distributions then X + Y is also a normal distribution

    • If X~N(μ1,σ12) and Y~N(μ2,σ22) then X+Y~N(μ1+μ2,σ12+σ22)

What does the distribution of aX + bY look like?

  • If X and Y are random variables and a and b are two constants we can combine the results for aX + b and X + Y

  • E(aX + bY) = aE(X) + bE(Y)

    • this is true for any random variables X and Y

  • Var(aX + bY) = a²Var(X) + b²Var(Y)

    • this is true if X and Y are independent

  • Note that b is squared for the variance so we have

    • E(aX - bY) = aE(X) - bE(Y)

    • Var(aX - bY) = a²Var(X) + b²Var(Y)

      • Notice that the variances of aX + bY and aXbY are the same

  • If X and Y are two independent normal distributions then aX + bY is also a normal distribution

    • If X ~ N(μ1, σ1²) and Y ~ N(μ2, σ2²) then aX ± bY ~ N(1 ± 2, a²σ1² + b²σ2²)

  • Note that aX + bY is no longer Poisson even if X and Y are Poisson

    • This holds provided a and b are not 0 or 1

Worked Example

 X and Y are independent random variable such that

E(X) = 5 and Var (X) = 3

 E(Y)=2 and Var(Y)=4.

 Find the value of:

(i) E(2X+5Y)

(ii) Var(2X+5Y)

(iii) Var(4XY)

Answer:

2-2-1-ax--by-we-solution-1

Linear Combinations

For a given random variable X, what is the difference between 2X and X1 + X2?

  • 2X means one observation of X is taken and then doubled

  • X1 + X2 means two observations of X are taken and added together

  • 2X and X1 + X2 both have the same expected value of 2E(X)

  • 2X and X1 + X2 have different variances

    • Var(2X) = 2²Var(X) = 4Var(X)

    • Var(X1 + X2) = 2Var(X)

  • Imagine X could take the values 0 and 1

    • 2X could then take the values 0 and 2 (2 × 0 = 0 and 2 × 1 = 2)

    • X1 + X2 could then take the values 0, 1 and 2 (0 + 0 = 0, 0 +1 = 1, 1 + 1 = 2)

  • Sometimes questions may describe the variables in context

    • The mass of a carton of half a dozen eggs is the mass of the carton plus the mass of the 6 individual eggs and can be modelled using the random variable

    • C + E1 + E2 + E3 + E4 + E5 + E6 where

      • C is the mass of a carton

      • E is the mass of an egg

    • It is not C + 6E because the masses of the 6 eggs could be different

How do I use linear combinations of normal random variables to find probabilities?

  • If the random variables are normally distributed and independent you might be asked to find probabilities such as

    • P(X1 + X2 + X3 > 2Y + 5)

    • This could be given in words

      • Find the probability that the mass of three chickens (X) is more than 5 kg heavier than double the mass of a turkey (Y)

  • To solve these problems:

    • STEP 1: Rearrange the inequality to get all the random variables on one side

      • P(X1 + X2 + X3 – 2Y > 5)

    • STEP 2: Find the mean and variance of the combined normal random variable

      • μ = E(X1 + X2 + X3 – 2Y) = E(X1) + E(X2) + E(X3) - 2E(Y)

      • σ² = Var(X1 + X2 + X3 – 2Y) = Var(X1) + Var(X2) + Var(X3) + 2² Var(X1)

    • STEP 3: Find the required probability using the combined normal distribution

      • X1 + X2 + X3 – 2Y ~ N(μ, σ²)

      • Use z-values and the table of values

Worked Example

X~N(10, 42) and Y~N(5, 82)

Find P(3X>2Y+50)

Answer:

2-2-1-linear-combs-we-solution-1

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Dan Finlay

Author: Dan Finlay

Expertise: Portfolio 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

Reviewer: Lucy Kirkham

Expertise: Content Creator

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.