4 Ijesha Close, Ilupeju, Lagos
+2347 086 296 002

Linear Combinations of Random Variables

Expectation and Variance of a Function of X
  • E(aX + b) = aE(X) + b
  • Var(aX + b) = a²Var(X)
Examples
  1. The random variable T has a mean of 5 and a variance of 16. Find the two pairs of values for the constants c and d such that E(cT + d) = 100 and Var(cT + d) = 144
    • Expand the expectation equation:
      • E(cT + d) = cE(T) + d = 100
      • ∴ 5c + d = 100
    • Expand the variance equation:
      • Var(cT + d) = c²Var(T) = 144
      • 16c² = 144
      • ∴ c = ±3
    • Use the first equation to find the two pairs:
      • when c = +3, d = 85
      • when c = -3, d = 115
Combinations of Random Variables
  • Expectations of combinations of random variables:
    • E(aX + bY) = aE(X) +  bE(Y)
  • Variance of combinations of independent random variables:
    • Var(aX + bY + c) = a²Var(X) + b²Var(Y)
    • Var(X ± Y) = Var(X) + Var(Y)
  • Combinations of identically distributed random variables having mean μ and variance σ²:
    • E(2X) = 2μ and E(X1) + E(X2) = 2μ
    • Var(2X) = 4σ² but Var(X1 + X2) = 2σ²
Examples
  1. It is given that X1 and X2 are independent, and E(X1) = E(X2) = μ, Var(X1) = Var(X2) = σ². Find E(X) and Var(X), where X = ½(X1 + X2)
    • Split the variance and expectation into individual components:
      • Var(½(X1 + X2)) = (½)²Var(X1) + (½)²Var(X2)
      • E(½(X1 + X2)) = ½E(X1) + ½E(X2)
    • Substitute the given values:
      • Var(½(X1 + X2)) = ¼σ² + ¼σ² = σ²
      • E(½(X1 + X2)) = ½μ + ½μ = μ
Expectation and Variance of Sample Mean
  • E(x̄) = μ
  • Var(x̄) = σ²/n
Examples
  1. The mean weight of a soldier may be taken to be 90 kg, and σ = 10 kg. 250 soldiers are on board an aircraft, find the expectation and variance of their weight. Hence, find μ and σ of the total weight of soldiers
    • Let X be the average weight, therefore:
      • E(x̄) = μ = 90
      • Var(x̄) = σ²/n10²/250 = 0.4 kg²
    • To find the μ of the total weight:
      • E(X1) + E(X2) + … + E(X250) = 250E(X) = 22500 kg
    • To find σ, find Var(X) first:
      • Var(X1) + … + Var(X250) = 250Var(X) = 2500 kg
      • Var(X) = σ² = 25000
      • ∴ σ = √25000 = 158.1 kg
Minimum 4 characters