Linear Combinations of Random Variables (Cambridge (CIE) A Level Maths): Revision Note

aX+b

How are the mean and variance of X related to the mean and variance of 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 a²  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

  • can be written as 

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

aX + bY

How are the means and variances of X and Y related to the mean and variance of X + Y ?

• 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 and then 

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

  • If and then 

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 aX – bY 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(aμ₁ ± bμ₂, 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

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