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CIE A-Level Maths Study Notes

5.2.1 Expectation of Linear Combinations

Understanding the expectation of linear combinations is crucial in the field of probability and statistics. This section is dedicated to exploring the concept of the expectation of linear combinations of random variables, a fundamental aspect of probability distributions and their practical applications.

Understanding the Linearity of Expectation

  • Basic Concept: The average value a random variable is expected to take.
  • Linearity Principle: The expected value of a sum is the sum of the expected values, even if the variables are not independent.

Formulas for Expectation

  • Single Variable: E(aX+b)=aE(X)+bE(aX + b) = aE(X) + b
  • Two Variables: E(aX+bY)=aE(X)+bE(Y)E(aX + bY) = aE(X) + bE(Y)

Applications in Distributions

  • Normal Distribution: Linear combinations of normal variables are normally distributed.
  • Poisson Distribution: The sum of independent Poisson variables is Poisson distributed.

Example Problems

Example 1: Single Variable

Determine the expected value of the expression 3X+43X+4, given that the expected value of XX is 55.

Given: E(X)=5E(X) = 5

Solution:

E(3X+4)=3E(X)+4E(3X+4) = 3E(X) + 4

E(3X+4)=3×5+4E(3X+4) = 3 \times 5 + 4

E(3X+4)=15+4E(3X+4) = 15 + 4

E(3X+4)=19E(3X+4) = 19

Graph: Illustration of E(3X+4)E(3X + 4).

Single Variable Graph

Example 2: Two Independent Variables

Calculate the expected value of 2X+3Y2X + 3Y, where XX and YY are independent variables with expected values of 33 and 44, respectively.

Given: E(X)=3,E(Y)=4E(X) = 3 , E(Y) = 4

Solution:

E(2X+3Y)=2E(X)+3E(Y)E(2X+3Y) = 2E(X) + 3E(Y)

E(2X+3Y)=2×3+3×4E(2X+3Y) = 2 \times 3 + 3 \times 4

E(2X+3Y)=6+12E(2X+3Y) = 6 + 12

E(2X+3Y)=18E(2X+3Y) = 18

Graph: Illustration of E(2X+3Y)E(2X + 3Y).

Two Independent Variables  Graph

Example 3: Poisson Distribution

Find the distribution of the sum ZZ, where ZZ is the sum of two independent Poisson processes XX and YY with rates λ1=2\lambda_1 = 2 and λ2=3\lambda_2 = 3.

Given: Poisson processes XX and YY with rates λ1=2\lambda_1 = 2 and λ2=3\lambda_2 = 3

Solution:

λZ=λ1+λ2λ_Z = λ_1 + λ_2

λZ=2+3λ_Z = 2 + 3

λZ=5λ_Z = 5

Graph: Poisson distribution for Z=X+YZ = X + Y.

Poisson Distribution Graph
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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