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

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

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?

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

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?+?X_2?means?two observations?of?X?are taken and?added together
• 2X?and?X₁?+?X_2?both have the?same expected value?of?2E(X)
• 2X?and?X₁?+?X_2?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?+?X_2?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?+?E_2?+?E_3?+?E_4?+?E_5?+?E_6?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(X_1?+?X_2?+?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(X_1?+?X_2?+?X₃– 2Y?> 5)
  • STEP 2: Find the?mean and variance?of the?combined normal random?variable
    • μ?= E(X_1?+?X_2?+?X₃– 2Y) = E(X₁) + E(X₂) + E(X₃) - 2E(Y)
    • σ2 = Var(X_1?+?X_2?+?X₃– 2Y) = Var(X₁) + Var(X₂) + Var(X₃) + 22 Var(X₁)
  • STEP 3: Find the?required probability?using the?combined normal distribution
    • X_1?+?X_2?+?X₃– 2Y?~ N(μ,?σ2)
    • Use z-values and the table of values