MathsFurther Statistics 2 › Combinations of normal random variables

Combinations of normal random variables

Add independent normal variables and the result is normal again. Means add as you would expect. Variances add whether you are adding or subtracting, which is where the marks are lost.

Builds on The normal distribution and Discrete random variables and expectation.

IN THIS TOPIC

  • Write down the distribution of aX ± bY for independent normal X and Y.
  • Distinguish the sum of n independent copies from n times a single one.
  • Use the combined distribution to answer a probability question.

COMMON MISCONCEPTION

For independent X and Y, the variance of X − Y is Var(X) − Var(Y).

Means add, variances add

If X is N(μx, σx²) and Y is N(μy, σy²) independently, then any linear combination is normal too, with a mean that follows the signs. Neither of those two claims is printed anywhere in the booklet, so both are yours to memorise.

aX±bY is normal, with mean aμx±bμyaX ± bY \text{ is normal, with mean } aμ_{x} ± bμ_{y}NOT IN THE BOOKLET — LEARN IT

The variance is a different matter. It is printed, under Expectation algebra, for independent X and Y.

Var(aX±bY)=a2σx2+b2σy2\text{Var}(aX ± bY) = a^{2}σ_{x}^{2} + b^{2}σ_{y}^{2}IN THE FORMULAE BOOKLET

So the half of this result people most often get wrong is the half you can look up. The signs in the mean follow the combination. The variances always add and the coefficients always square. Subtracting cannot reduce uncertainty, since two independent sources of variation both contribute whichever way round you take them. Subtracting the variances would go negative whenever Y varied more than X, which is the quickest way to see it cannot be right.

Two independent normal variables and their sum: means add, variances add, and the spread growsN(20, 4) and N(18, 5)each on its ownaddN(38, 9)wider than eithervariances add, standard deviations do not
FIG. 1Two independent normal distributions and their sum: the mean lands where expected and the spread is wider than either.

WORKED EXAMPLE

Which of two is larger

X is N(20, 4) and Y is N(18, 5), independently. Find P(Y > X).

Work with the difference. X − Y is N(20 − 18, 4 + 5) = N(2, 9), with standard deviation 3.

P(Y > X) = P(X − Y < 0) = P(Z < (0 − 2)/3) = P(Z < −0.667).

That is 0.252 to three decimal places. Note the variance 9, and not 4 − 5.

Four bags, or one bag four times

Adding four independent copies of X gives variance 4σ², since each copy contributes its own. Multiplying one X by 4 gives variance 16σ², since the coefficient squares. Both have mean 4μ, so the distinction shows up only in the spread, and it is the single most common error in this topic.

The physical reading is worth holding on to. Four separate bags vary independently, so their errors partly cancel and the total is relatively more predictable than any one bag. Scaling a single bag up magnifies its error along with everything else.

Four bags added together against one bag multiplied by four: the same mean, twice the standard deviationfour bags addedsd 16one bag times foursd 32both have mean 2000 grams; only the spread differs
FIG. 2Four bags added against one bag multiplied by four: the same mean, and twice the standard deviation for the multiple.

GUIDED PRACTICE

Telling the two apart

Bags of flour have weight N(500, 64) grams. Find the distribution of the total weight of four bags, and of four times the weight of one bag.

Show the working

Four bags: mean 4 × 500 = 2000, variance 4 × 64 = 256, so N(2000, 256) with standard deviation 16.

Four times one bag: mean 2000, variance 4² × 64 = 1024, so N(2000, 1024) with standard deviation 32.

Same mean, double the spread. Independent errors partly cancel; a scaled-up single error does not.

ASSESSMENT FOCUS

  • Write the new distribution in full, N(mean, variance), before standardising anything.
  • Add the variances even for a difference, and square every coefficient first.
  • For 'is X bigger than Y', form X − Y and ask for the probability that it is positive.
  • State the independence assumption. Without it the variance rule does not hold.

CHECK YOURSELF

X is N(30, 9) and Y is N(10, 16), independently. State the distribution of 2X − 3Y.

Show a hint

Means follow the signs; variances add with squared coefficients.

Show the answer

Mean 2(30) − 3(10) = 30. Variance 4(9) + 9(16) = 36 + 144 = 180. So 2X − 3Y is N(30, 180).

For independent normals, aX ± bY is normal with mean aμx ± bμy and variance a²σx² + b²σy².

Variances add for a difference as well as a sum, and n independent copies give variance nσ² while n times one gives n²σ².

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

6 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the combinations of normal random variables questions page.

CHECK YOUR PROGRESS

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  • Write down the distribution of aX ± bY for independent normal X and Y.
  • Distinguish the sum of n independent copies from n times a single one.
  • Use the combined distribution to answer a probability question.

Open the full revision checklist to see every objective in the course in one place.