Sums of independent variables | Year 12

01 / LEARN

Understand the idea.

Means add, independent variances add

Expectation is linear even without independence: $E (a X + b Y + c) = a E (X) + b E (Y) + c$ . For independent variables, variance is $a^{2} Var (X) + b^{2} Var (Y)$ . State independence before using this variance rule.

A difference still adds independent variances because the negative coefficient is squared. Add variances first, then take a square root for standard deviation. Subtracting a random quantity can increase uncertainty even while decreasing the mean.

Build the full distribution by pairing outcomes

To construct the distribution of an independent sum, list pairs of possible values. Multiply probabilities within each pair, then add the products for all pairs giving the same sum. These are different, mutually exclusive ways for that sum to occur.

Different variable names do not establish independence. Measurements affected by the same environment may move together. If dependence is present, pair probabilities and variance calculations require further information; multiplying marginal probabilities alone is unjustified.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

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.

Worked solution

1
Combine the means with their signed coefficients.
$E (T) = 2 E (X) - E (Y) = 2 (2) - 5 = - 1$
Linearity of expectation applies to the whole combination.
2
Combine the independent variances.
$Var (T) = 2^{2} (3) + {(- 1)}^{2} (4) = 12 + 4 = 16$
Independence removes a covariance term, while squared coefficients make both contributions nonnegative.
3
Find the standard deviation.
$SD (T) = \sqrt{16} = 4$
The square root is taken after adding variance contributions.

Answer

Mean −1; variance 16; standard deviation 4.

EXAMPLE 02

Two different success indicators

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

Worked solution

1
Find the probability of no successes.
$P (S = 0) = P (B = 0, C = 0) = (0.6) (0.3) = 0.18$
Independence allows the two failure probabilities to multiply.
2
Find the probability of two successes.
$P (S = 2) = P (B = 1, C = 1) = (0.4) (0.7) = 0.28$
Both indicators must equal one.
3
Add the two ways to obtain one success.
$P (S = 1) = (0.4) (0.3) + (0.6) (0.7) = 0.54$
The disjoint cases are success then failure, or failure then success.
4
Check the distribution.
$0.18 + 0.54 + 0.28 = 1$
The values $0, 1, 2$ account for all pairs; unequal success probabilities do not prevent constructing the sum.

Answer

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

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

Independence gives $Var (X - Y) = 2 + {(- 1)}^{2} (7) = 9$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet 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 $3 X - 2 Y + 5$ .
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 $3 X - 2 Y + 5$ .
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 $\frac{10}{5} = 2$ ; variance is $\frac{15}{25} = \frac{3}{5}$ .
$2; \frac{3}{5}$ .
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 $\frac{1}{36}$ , so the required probability is $\frac{4}{36}$ .
$\frac{1}{9}$ .
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 ↗

In this lesson

Understand the idea.

Means add, independent variances add

Build the full distribution by pairing outcomes

See the method in action.

A linear combination

Combine the means with their signed coefficients.

Combine the independent variances.

Find the standard deviation.

Two different success indicators

Find the probability of no successes.

Find the probability of two successes.

Add the two ways to obtain one success.

Check the distribution.

Try it yourself.

Check your understanding

Keep practising