# Linear combinations of random variables > CIE A-Level Mathematics · 9709 > Source: https://www.owlsprep.com/study/cie-9709-u4-linear-combinations-of-random-variables/ This module covers expectation and variance rules for linear combinations of random variables, including independent normal variable applications, aligned to CIE 9709 exam mark schemes. **Prerequisites:** [Basic expectation and variance for single random variables](https://www.owlsprep.com/study/cie-9709-u4-expectation-variance-single-variables/); [Normal distribution properties and standardisation](https://www.owlsprep.com/study/cie-9709-u4-normal-distribution-intro/) ## Learning objectives - Calculate the expected value of any linear combination of two or more random variables, with no requirement for independence - Compute the variance of linear combinations for both independent and non-independent random variables, accounting for covariance - Apply linear combination rules to solve problems involving sums and differences of normally distributed variables - Distinguish between linear combinations of multiple independent variables and multiples of a single repeated variable ## How often is this tested? Based on OwlsPrep's analysis of official CIE 9709 S2 past papers (2016–2025): **Linear combinations of random variables** appears **64 times** in the last 10 years — **15%** of all S2 questions (64 of 428). Most-tested forms: Continuous random variables: PDF, CDF; verifying validity (60), E(X), Var(X), median, mode for continuous RVs (59). ## Linearity of Expectation for Linear Combinations **Linearity of Expectation** — This identity holds for *any* pair of random variables, no matter if they are dependent or independent, for all real constants a, b, c. *Notation:* $E[aX + bY + c] = aE[X] + bE[Y] + c$ This rule extends naturally to any number of variables. For a sum of $n$ random variables $X_1, X_2, ... X_n$ with weights $a_1, a_2 ... a_n$, the expected value is $E[\sum a_i X_i] = \sum a_i E[X_i]$. No additional conditions apply, making this one of the most robust identities in probability theory. **Worked example:** Given $E[X] = 5$, $E[Y] = 3$, calculate the value of $E[4X - 2Y + 7]$ 1. Apply linearity of expectation directly, no independence checks required 2. $$E[4X - 2Y +7] = 4E[X] - 2E[Y] +7$$ 3. $$= 4(5) - 2(3) +7 = 20 -6 +7 = 21$$ **Check your understanding** Test your understanding of linearity of expectation: 1. Which of these values is equal to $E[2A + 3B - 4]$ if $E[A] = 2$, $E[B] = 1$? - 3 - 5 - 7 - 9 *Why:* Calculation: 2*2 + 3*1 -4 = 4 +3 -4 = 3, no dependence between A and B affects the result. > **Exam tip:** CIE exam mark schemes award full 2/3 marks for correct substitution of values into the linearity of expectation formula, no working beyond the final calculation is required unless stated. ## Variance of Linear Combinations $$Var(aX + bY) = a^2 Var(X) + b^2 Var(Y) + 2ab \text{Cov}(X,Y)$$ If X and Y are independent, their covariance $Cov(X,Y) = 0$, so the formula simplifies to $Var(aX + bY) = a^2 Var(X) + b^2 Var(Y)$. Note that variance never scales linearly with constants: all coefficients are squared, and the constant offset $c$ added to the combination has no impact on variance, as it does not change the spread of the distribution. > **Common Exam Trick** > > For independent variables, the variance of a difference $Var(X-Y) = Var(X) + Var(Y)$, not $Var(X) - Var(Y)$. The minus sign is squared so it disappears. **Worked example:** Given $Var(X) = 9$, $Var(Y) = 16$, X and Y are independent. Calculate $Var(3X - 2Y + 10)$ 1. Drop the constant offset 10 first, as it does not affect variance 2. Square all coefficients, sum the individual variances multiplied by their squared coefficients 3. $$Var(3X -2Y) = 3^2 Var(X) + (-2)^2 Var(Y)$$ 4. $$= 9*9 + 4*16 = 81 + 64 = 145$$ > **Exam tip:** You must explicitly state 'X and Y are independent' in your working to justify dropping the covariance term, to avoid losing 1 mark in the exam. ## Linear Combinations of Independent Normal Variables **Reproductive Property of Normal Distribution** — Any linear combination of independent normally distributed random variables is also normally distributed. If $X \sim N(\mu_X, \sigma_X^2)$ and $Y \sim N(\mu_Y, \sigma_Y^2)$ and X and Y are independent, then $aX + bY + c \sim N(a\mu_X + b\mu_Y + c, a^2\sigma_X^2 + b^2\sigma_Y^2)$. This property is unique to a small set of common distributions, and is tested frequently in CIE Paper 6 and Paper 7. **Worked example:** The mass of a small apple $A \sim N(120, 15^2)$ g, the mass of a banana $B \sim N(150, 20^2)$ g. Find the probability that the total mass of 1 apple and 1 banana is less than 300g. 1. Define the total mass $T = A + B$ 2. $$E[T] = 120 + 150 = 270$$ 3. $$Var(T) = 15^2 + 20^2 = 225 + 400 = 625, \sigma_T = 25$$ 4. So $T \sim N(270, 25^2)$ 5. $$P(T < 300) = \Phi\left(\frac{300 - 270}{25}\right) = \Phi(1.2) = 0.8849$$ *Calculator:* allowed ## Distinguishing $aX$ vs $X_1 + X_2 + ... + X_a$ This is the single most commonly tested trick in this topic. If you take a single random variable X and multiply it by 2, you get $2X$, with variance $4Var(X)$. If you take two independent copies of X, $X_1$ and $X_2$, each with the same distribution as X, their sum $X_1 + X_2$ has variance $Var(X_1) + Var(X_2) = 2Var(X)$. These are completely different distributions, and exam questions explicitly test this distinction. **Comparing methods** Side by side comparison of the two cases: - **Single variable multiplied by 2: $2X$** — Every value of X is doubled, so the spread of the distribution is doubled - Pros: Simple scaling of the original distribution - Cons: Variables are perfectly correlated, not independent - **Sum of 2 independent copies: $X_1 + X_2$** — Two separate independent observations from the same distribution, summed - Pros: No covariance term, variance adds linearly - Cons: Distribution shape is different from scaled X **Worked example:** Given $X \sim N(10, 4)$, find the difference between $Var(3X)$ and $Var(X_1 + X_2 + X_3)$ where $X_1,X_2,X_3$ are independent copies of X. 1. $$Var(3X) = 3^2 Var(X) = 9 * 4 = 36$$ 2. $$Var(X_1 + X_2 + X_3) = Var(X) + Var(X) + Var(X) = 4 + 4 + 4 = 12$$ 3. Difference = 36 - 12 = 24 > **Exam tip:** Read the question wording carefully: 'the total of 3 independent observations' means sum of 3 variables, 'three times the value of a single observation' means 3X. ## Common pitfalls - **Wrong:** Calculating $Var(aX)$ as $aVar(X)$ instead of $a^2 Var(X)$ - Why it fails: Forgetting variance measures squared spread, so scaling a variable by a scales its variance by $a^2$ - Correct: Always square the constant coefficient when calculating variance of a scaled variable - **Wrong:** Calculating $Var(X-Y)$ as $Var(X) - Var(Y)$ - Why it fails: Treating variance like expectation, where the minus sign carries through - Correct: Square the negative coefficient, so $Var(X-Y) = Var(X) + Var(Y)$ for independent variables - **Wrong:** Treating $2X$ as equivalent to $X_1 + X_2$ - Why it fails: Confusing scaling a single variable with summing two independent copies - Correct: Check question wording: if it refers to multiple separate observations, sum their variances directly - **Wrong:** Including constant offsets in variance calculations - Why it fails: Assuming adding a fixed number to a distribution changes its spread - Correct: Ignore all constant terms when calculating variance, they only affect the expected value - **Wrong:** Applying the normal reproductive property to dependent variables - Why it fails: Forgetting the rule only applies to independent normal variables - Correct: Explicitly confirm independence is stated in the question before using the simplified variance formula ## Cheatsheet | Operation | Expectation | Variance (independent variables) | Variance (dependent variables) | | --- | --- | --- | --- | | $aX + b$ | $aE[X] + b$ | $a^2 Var(X)$ | $a^2 Var(X)$ | | $aX + bY$ | $aE[X] + bE[Y]$ | $a^2 Var(X) + b^2 Var(Y)$ | $a^2 Var(X) + b^2 Var(Y) + 2ab Cov(X,Y)$ | | $X_1 + X_2 + ... + X_n$ | $\sum E[X_i]$ | $\sum Var(X_i)$ | $\sum Var(X_i) + 2\sum_{i