Sums of independent variables | Year 12

01 / LEARN

The idea, step by step

Means add, independent variances add

For any variables, $E (a X + b Y + c) = a E (X) + b E (Y) + c$ . This mean rule does not require independence. Under independence, $Var (a X + b Y) = a^{2} Var (X) + b^{2} Var (Y)$ .

Subtracting a random quantity does not subtract uncertainty. A difference has the sum of independent variances because both coefficients are squared. Add variances, then take a square root; do not add standard deviations.

Build the full distribution by pairing outcomes

If X and Y are independent, the probability of a pair is the product P(X = x)P(Y = y). List the pairs that give each required sum, multiply for each pair, then add. Different pairs can lead to the same sum.

Independence needs a reason from the process, not just different variable names. Two measurements from the same fluctuating environment may be dependent. Without the required dependence information, the independent variance calculation cannot be justified.

A common trap

Var(X − Y) = Var(X) − Var(Y).

Instead: For independent variables, the coefficient −1 is squared, so variances add. Dependence would require additional information.

02 / WORKED EXAMPLES

See the thinking

Example 1

A linear combination

Independent X and Y have means 2 and 5 and variances 3 and 4. Find the mean and variance of T = 2X − Y.

E(T) = 2(2) − 5 = −1.
Independence permits variance addition with squared coefficients.
Var(T) = 2²(3) + (−1)²(4) = 16.

Mean −1; variance 16; standard deviation 4.

Example 2

Two different success indicators

Independent B and C equal 1 with probabilities 0.4 and 0.7. Find the distribution of S = B + C.

P(S = 0) = 0.6 × 0.3 = 0.18.
P(S = 2) = 0.4 × 0.7 = 0.28.
Exactly one success has probability 0.4 × 0.3 + 0.6 × 0.7 = 0.54. Check that all three sum to 1.

S = 0, 1, 2 with probabilities 0.18, 0.54, 0.28.

Check your understanding

Independent X and Y have variances 2 and 7. Find Var(X − Y).

Show answer

The independent variances add, even for a difference.

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Independent X and Y have means 4, 7 and variances 1, 9. Find the mean and variance of X + Y.
2 marks
Using the same X and Y, find the mean and variance of 3X − 2Y + 5.
3 marks
Five independent copies of X each have mean 2 and variance 3. Find the mean and variance of their sum.
2 marks
For those five independent copies, find the mean and variance of their average.
3 marks
Independent fair dice are rolled. Let X and Y be their scores. Find P(X + Y = 5).
2 marks
A supplier gives E(X) = 8, E(Y) = 3, Var(X) = 4 and Var(Y) = 2, but does not state independence. Which of E(X + Y) = 11 and Var(X + Y) = 6 is justified?
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Independent X and Y have means 4, 7 and variances 1, 9. Find the mean and variance of X + Y.
2 marks
E(X + Y) = 4 + 7.
Var(X + Y) = 1 + 9 by independence.
11 and 10.
Using the same X and Y, find the mean and variance of 3X − 2Y + 5.
3 marks
Mean = 3(4) − 2(7) + 5 = 3.
Variance = 9(1) + 4(9) = 45.
Mean 3; variance 45.
Five independent copies of X each have mean 2 and variance 3. Find the mean and variance of their sum.
2 marks
Add five means: 5(2).
Add five independent variances: 5(3).
Mean 10; variance 15.
For those five independent copies, find the mean and variance of their average.
3 marks
The average is one fifth of the sum.
Its mean is 10/5 = 2; variance is 15/25 = 3/5.
$2; 3 / 5$ .
Independent fair dice are rolled. Let X and Y be their scores. Find P(X + Y = 5).
2 marks
The ordered pairs are (1,4), (2,3), (3,2), (4,1).
Each has probability 1/36, so the required probability is 4/36.
$1 / 9$ .
A supplier gives E(X) = 8, E(Y) = 3, Var(X) = 4 and Var(Y) = 2, but does not state independence. Which of E(X + Y) = 11 and Var(X + Y) = 6 is justified?
3 marks
Expectation is additive without independence.
The variance calculation also needs independence or suitable dependence information.
Only the mean statement is justified.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: discrete random variables and probability distributions, expected value and variance, transformations and independent sums, Bernoulli and binomial models. Original supporting lessons and exercises follow the approved Ascent chapter structure.

Official curriculum source ↗

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