Edexcel IAL - Statistics 3- 1.1 Linear Combinations of Independent Normal Random Variables- Study notes - New syllabus
Edexcel IAL – Statistics 3- 1.1 Linear Combinations of Independent Normal Random Variables -Study notes- New syllabus
Edexcel IAL – Statistics 3- 1.1 Linear Combinations of Independent Normal Random Variables -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 1.1 Linear Combinations of Independent Normal Random Variables
Distribution of Linear Combinations of Independent Normal Random Variables
If random variables are normally distributed, then any linear combination of those variables is also normally distributed, provided the variables are independent. This result is fundamental and is widely used in probability and statistics.
Statement of the Result
Let
\( \mathrm{X \sim N(\mu_X, \sigma_X^2)} \)
\( \mathrm{Y \sim N(\mu_Y, \sigma_Y^2)} \)
where \( X \) and \( Y \) are independent random variables.
Then, for constants \( a \) and \( b \), the linear combination
\( \mathrm{Z = aX + bY} \)
is also normally distributed.
Mean of the Linear Combination
Using the properties of expectation:
\( \mathrm{E(Z) = aE(X) + bE(Y)} \)
\( \mathrm{E(Z) = a\mu_X + b\mu_Y} \)
Variance of the Linear Combination
Because \( X \) and \( Y \) are independent, the variance is
\( \mathrm{Var(Z) = a^2 Var(X) + b^2 Var(Y)} \)
\( \mathrm{Var(Z) = a^2 \sigma_X^2 + b^2 \sigma_Y^2} \)
There are no cross terms because of independence.
Generalisation
If \( X_1, X_2, \dots, X_n \) are independent normal random variables and
\( \mathrm{Z = a_1 X_1 + a_2 X_2 + \dots + a_n X_n} \)
then
\( \mathrm{Z \sim N\!\left(\sum a_i \mu_i,\; \sum a_i^2 \sigma_i^2 \right)} \)
This result may be used directly without proof.
Important Conditions
- Each random variable must be normally distributed
- The variables must be independent
If the variables are not independent, this result does not apply in this simple form.
Interpretation
- Sums and differences of independent normal variables are normal
- Scaling a normal variable changes the mean and variance but preserves normality
Example :
Let
\( \mathrm{X \sim N(10,\,4)} \), \( \mathrm{Y \sim N(6,\,1)} \)
where \( X \) and \( Y \) are independent.
Find the distribution of \( \mathrm{Z = X + Y} \).
▶️ Answer/Explanation
Mean:
\( \mathrm{E(Z) = E(X) + E(Y) = 10 + 6 = 16} \)
Variance:
\( \mathrm{Var(Z) = Var(X) + Var(Y) = 4 + 1 = 5} \)
Conclusion:
\( \mathrm{Z \sim N(16,\,5)} \)
Example :
Let
\( \mathrm{X \sim N(50,\,9)} \), \( \mathrm{Y \sim N(20,\,4)} \)
where \( X \) and \( Y \) are independent.
Find the distribution of \( \mathrm{W = 2X – Y} \).
▶️ Answer/Explanation
Mean:
\( \mathrm{E(W) = 2E(X) – E(Y) = 2(50) – 20 = 80} \)
Variance:
\( \mathrm{Var(W) = 2^2 Var(X) + (-1)^2 Var(Y)} \)
\( \mathrm{Var(W) = 4(9) + 1(4) = 40} \)
Conclusion:
\( \mathrm{W \sim N(80,\,40)} \)
Example :
The masses of apples and oranges are independent and normally distributed.
\( \mathrm{A \sim N(180,\,25)} \) g, \( \mathrm{O \sim N(120,\,16)} \) g
One apple and one orange are selected at random.
Find the distribution of the total mass \( \mathrm{T} \).
▶️ Answer/Explanation
Let \( \mathrm{T = A + O} \).
Mean:
\( \mathrm{E(T) = 180 + 120 = 300} \)
Variance:
\( \mathrm{Var(T) = 25 + 16 = 41} \)
Conclusion:
\( \mathrm{T \sim N(300,\,41)} \)
