Maths › Further 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.
WHAT YOU PROBABLY THINK
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:
The signs in the mean follow the combination, but the variances always add and the coefficients always square. Subtracting cannot reduce uncertainty: two independent sources of variation both contribute, whichever way round you take them. The opening claim would make the variance negative whenever Y varied more than X, which is the quickest way to see it cannot be right.
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, 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.
YOUR TURN
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.
THE EXAM BIT
- 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 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
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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.
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.
No animated video for this topic yet; these notes stand alone.