Linear combinations of random variables

• Means add and scale; variances square the coefficient and add, for independent variables, and never subtract. Then read the question for whether it means one item multiplied or several items added.
• Var(X − Y) = Var(X) + Var(Y) for independent X and Y; 2X has variance 4σ² while X1 + X2 has 2σ²; only a sum of independent Poisson variables is Poisson.

• 6.2.1 I can use — use, when solving problems, the results that (a) E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X) (b) E(aX + bY) = aE(X) + bE(Y) (c) Var(aX + bY) = a²Var(X) + b²Var(Y) for independent X and Y (d) if X has a normal distribution then so does aX + b (e) if X and Y have independent normal distributions then aX + bY has a normal distribution (f) if X and Y have independent Poisson distributions then X + Y has a Poisson distribution

Forward look to chapter 31. The mean of \(n\) independent observations is \(\tfrac{1}{n}(X_1 + \cdots + X_n)\). By (c) the total has variance \(n\sigma^2\), and by (a) dividing by \(n\) divides the variance by \(n^2\), giving \(\dfrac{\sigma^2}{n}\). Chapter 31 gives the sample mean its own name and distribution; its variance is nothing more than this result.

• “\(\text{Var}(X - Y) = \text{Var}(X) - \text{Var}(Y)\).” Correct \(\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y)\) for independent \(X\) and \(Y\). The coefficient of \(Y\) is \(-1\), and it is squared: \((-1)^2 = 1\). A difference is more variable than either variable on its own, never less; a subtraction can even produce a negative variance, which is impossible.
• “The total mass of two biscuits is \(2X\).” Correct \(2X\) is one biscuit's mass doubled, with variance \(4\sigma^2\). The total of two randomly chosen biscuits is \(X_1 + X_2\), two independent observations added, with variance \(2\sigma^2\). The means agree (\(2\mu\)); the variances do not, so the probabilities do not. Say in words which one the question describes before writing a number.
• “\(\text{Var}(3X - 2) = 9\,\text{Var}(X) - 2\).” Correct \(\text{Var}(3X - 2) = 9\,\text{Var}(X)\). The \(b\) vanishes from the variance: subtracting 2 moves every value by the same amount and does not change how spread out they are.
• “The standard deviation of \(X + Y\) is \(\sigma_X + \sigma_Y\).” Correct Variances add; standard deviations do not. \(\text{sd}(X + Y) = \sqrt{\sigma_X^2 + \sigma_Y^2}\): with \(\sigma_X = 3\) and \(\sigma_Y = 4\) that is \(5\), not \(7\).
• “\(X \sim \text{Po}(1.5)\), so \(2X \sim \text{Po}(3)\).” Correct \(2X\) has mean \(3\) but variance \(4 \times 1.5 = 6\); a Poisson variable has its variance equal to its mean, so \(2X\) is not Poisson. Only a sum of independent Poisson variables is Poisson. \(X - Y\) is not Poisson either: it can be negative.
• “Variances add” — used without saying why it is allowed. Correct Results (c), (e) and (f) hold for independent variables. Write “since \(X\) and \(Y\) are independent” beside the variance line; where the question does not say so, state it as the modelling assumption (different people's journey times, eggs chosen at random).
• “\(\text{Var}(3X - 2) = 3\,\text{Var}(X) - 2\).” Repair \(\text{Var}(3X - 2) = 9\,\text{Var}(X)\): the coefficient is squared and the constant disappears.
• “\(\text{Var}(3X - 2) = 9\,\text{Var}(X) - 2\).” Repair The \(-2\) shifts every value but does not change their spread. Drop it entirely.
• “\(\text{Var}(X - Y) = \text{Var}(X) - \text{Var}(Y)\).” Repair \((-1)^2 = 1\), so \(\text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y)\) for independent \(X\) and \(Y\). A variance found by subtraction can come out negative, which is impossible.
• “\(A - B \sim \text{N}(4,\ 9 - 16)\).” Repair \(A - B \sim \text{N}(4, 25)\): the variance of a difference is a sum.
• “The total mass of two biscuits is \(2X \sim \text{N}(40, 16)\).” Repair The total of two independent biscuits is \(X_1 + X_2 \sim \text{N}(40, 8)\). \(2X\) is one biscuit's mass doubled.
• “Six eggs: \(\text{Var} = 6^2 \times 9 = 324\).” Repair Six eggs are six independent observations: \(\text{Var} = 6 \times 9 = 54\). \(6^2\sigma^2\) is the variance of six times one egg.
• “\(\text{sd}(X + Y) = \text{sd}(X) + \text{sd}(Y)\).” Repair Variances add, not standard deviations: \(\text{sd}(X + Y) = \sqrt{\sigma_X^2 + \sigma_Y^2}\).
• “\(\text{P}(A > B)\): standardise \(50\) and \(46\) separately and compare.” Repair Form \(D = A - B\), write its distribution, and standardise \(0\): \(\text{P}(A > B) = \text{P}(D > 0)\).
• “\(X \sim \text{Po}(1.5)\), so \(2X \sim \text{Po}(3)\).” Repair \(2X\) has mean \(3\) and variance \(6\); it is not Poisson. Only \(X + Y\), for independent Poisson variables, is Poisson.
• “\(X - Y \sim \text{Po}(\lambda - \mu)\).” Repair \(X - Y\) can be negative and is not Poisson. Its mean is \(\lambda - \mu\) and its variance \(\lambda + \mu\).
• “\(\text{Var}(aX + bY) = a^2\,\text{Var}(X) + b^2\,\text{Var}(Y)\)”, used for a person's height and the same person's mass. Repair The result needs independence, and a person's height and mass are not independent. State the condition, and use the result only where it holds.
• “\(\text{Var}(F) = 1.8\,\text{Var}(C)\).” Repair \(\text{Var}(F) = 1.8^2\,\text{Var}(C) = 3.24\,\text{Var}(C)\).
• “\(\text{E}(X - Y) = \text{E}(X) + \text{E}(Y)\).” Repair The mean of a difference is the difference of the means. Only the variance adds.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. Chapter 31 uses result (c) for the variance of the mean of n observations: when you reach it, re-derive σ2/n from Var(X1 + … + Xn) = nσ2. Chapter 32 tests hypotheses using a single Poisson observation, which may be a count over several periods: re-answer worked example 6 there, since a count over three periods is the sum of three independent counts. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: later chapters use these methods without re-teaching them, and the syllabus says an individual examination question may involve ideas and methods from more than one section of the content for that paper, so nothing here is ever finished with.

• Chapter 29 · Probability & Statistics 2 · How it is assessed
• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Probability & Statistics 2 content, examined in Paper 6. Paper 6 (Probability & Statistics 2) is offered only as part of the A Level, where it is 20%. It assumes the whole of the Paper 5 content and the calculus of Paper 3. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A question on this section is a structured modelling question: a context (masses of items, lifetimes of components, a count of events in two periods) to be turned into a named combination such as T = X1 + … + X6 + B or D = A − B, its mean and variance written with the independence it rests on, and then a probability found with chapter 27 or chapter 28 methods. This chapter's own practice questions put two combinations that differ in one word into consecutive parts, and ask you to state whether a combination is Poisson, with a reason.
• MF19 gives E(X) = Σxp and Var(X) = Σx2p − {E(X)}2, the Poisson probability formula with mean and variance λ, and the normal distribution table. All six results of 6.2.1 must be known, and a solution writes the one it uses before the numbers, with the word independent wherever a variance is added.
• Write the variance of the combination exactly (58, 40, 7.29), take its square root once and carry it unrounded (7.6158) into the standardisation; give z to 3 decimal places, read Φ to 4 decimal places using the ADD columns, and round the probability once, to 3 significant figures. Every standardisation line must be visible: an unsupported calculator answer earns nothing.