LEARN · EXPLAIN · REVISE

Distribution means, variances and dependence

Read the idea, work independently, then explain what changed.

高三選擇性必修第三册(A版).pdf · 7.3 · PDF 67 / printed page 62

Revisit first: Conditions, partitions and posterior probability

TOPIC 01

Distribution means, variances and dependence

Compute weighted moments and apply shifts/scales with correct units and dependence conditions.

What you will be able to explain

Compute weighted moments and apply shifts/scales with correct units and dependence conditions.
Use nonnegative variance to reject impossible summaries and compare equal-mean distributions.

Average first, center before measuring spread

Two games can have the same average reward and very different risk.

Expectation averages values by probability; variance averages their squared deviations.

E(X)=\sum xp_x;\ \operatorname{Var}(X)=E(X^2)-E(X)^2

Expectation averages values by probability; variance averages their squared deviations.
$E[(X-\mu)^2]=E(X^2)-2\mu E(X)+\mu^2$
Expectation averages values by probability; variance averages their squared deviations.
$\operatorname{Var}(aX+b)=a^2\operatorname{Var}(X)$

When it applies

Probabilities normalize and are nonnegative; a sum variance includes covariance unless justified otherwise.

Counterexample: For Y=1−X, X+Y is constant even though each variable has positive variance.

Probability table, transformed outcomes, mean marker and squared-distance columns.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Predict the effects of scale a and shift b on a fair two-point distribution.

Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.

Mean a+b=1; variance a²=1. A shift changes location only.

Numerical data for the diagram; printable
Outcome	Probability	Squared deviation	Weighted deviation
0	0.5	1	0.5
2	0.5	1	0.5

Explain: Read the transformed mean and variance; a=0 combines both outcomes into one.

Transfer: Explain why a negative scale reverses positions but does not make variance negative.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.

01 / Foundation#Worked example

X takes 0,2 with probabilities 3/4,1/4. Find E(X).

X=(0,2);\ P=(3/4,1/4)

Finite-distribution probabilities must be nonnegative and sum to one; dependence is only as explicitly stated.

Skills and prerequisite lessons

Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.

Working and explanation

BUILD THE REASONING

Hint 1

Weight each value by its probability.

Hint 2

Start from this intermediate relation.

E(X)=0(3/4)+2(1/4)

Worked solution

Weight each value by its probability.
Establish and simplify this mathematical relation.
$E(X)=0(3/4)+2(1/4)$
Establish and simplify this mathematical relation.
$E(X)=1/2$
An unweighted average of possible values ignores probabilities.

The requested value is 0.5.

Checks and common pitfalls: An unweighted average of possible values ignores probabilities.

Think first. Reveal a hint when the class is ready.

02 / Standard#Worked example

X takes 0,2 with equal probabilities. Find Var(X) by centering.

X=(0,2);\ P=(1/2,1/2)

First step provided — complete the reasoning:Calculate the mean before squared deviations.

Finite-distribution probabilities must be nonnegative and sum to one; dependence is only as explicitly stated.

Skills and prerequisite lessons

Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.

Working and explanation

BUILD THE REASONING

Hint 1

Calculate the mean before squared deviations.

Hint 2

Start from this intermediate relation.

E(X)=1

Worked solution

Calculate the mean before squared deviations.
Establish and simplify this mathematical relation.
$E(X)=1$
Establish and simplify this mathematical relation.
$\operatorname{Var}(X)=(-1)^2/2+1^2/2=1$
Variance averages squared deviations rather than absolute deviations.

The requested value is 1.

Checks and common pitfalls: Variance averages squared deviations rather than absolute deviations.

Think first. Reveal a hint when the class is ready.

03 / Transfer#Worked example

X is a fair Bernoulli variable and Y=1−X. Find Var(X+Y), then explain why adding the two variances fails.

P(X=0)=P(X=1)=1/2;\ Y=1-X

Finite-distribution probabilities must be nonnegative and sum to one; dependence is only as explicitly stated.

Skills and prerequisite lessons

Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.

Working and explanation

BUILD THE REASONING

Hint 1

Inspect the sum before applying an independence formula.

Hint 2

Start from this intermediate relation.

X+Y=1

Worked solution

Inspect the sum before applying an independence formula.
Establish and simplify this mathematical relation.
$X+Y=1$
Establish and simplify this mathematical relation.
$\operatorname{Var}(X+Y)=0;\ \operatorname{Cov}(X,Y)=-1/4$
The two variables are perfectly dependent, not independent.

The requested value is 0.

Checks and common pitfalls: The two variables are perfectly dependent, not independent.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Probabilities of X=0,1,2 are p,2p,p. Find p.

P=(p,2p,p)

Finite-distribution probabilities must be nonnegative and sum to one; dependence is only as explicitly stated.

Skills and prerequisite lessons

Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.

Working and explanation

BUILD THE REASONING

Hint 1

Use probability normalization.

Hint 2

Start from this intermediate relation.

p+2p+p=1

Worked solution

Use probability normalization.
Establish and simplify this mathematical relation.
$p+2p+p=1$
Establish and simplify this mathematical relation.
$p=1/4$
After normalization, check every probability is nonnegative.

The requested value is 0.25.

Checks and common pitfalls: After normalization, check every probability is nonnegative.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

X takes −1,2 with probabilities 2/3,1/3. Find E(X²).

X=(-1,2);\ P=(2/3,1/3)

Finite-distribution probabilities must be nonnegative and sum to one; dependence is only as explicitly stated.

Skills and prerequisite lessons

Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.

Working and explanation

BUILD THE REASONING

Hint 1

Square outcomes before averaging.

Hint 2

Start from this intermediate relation.

E(X^2)=1(2/3)+4(1/3)

Worked solution

Square outcomes before averaging.
Establish and simplify this mathematical relation.
$E(X^2)=1(2/3)+4(1/3)$
Establish and simplify this mathematical relation.
$E(X^2)=2$
E(X²) differs from the square of E(X)=0.

The requested value is 2.

Checks and common pitfalls: E(X²) differs from the square of E(X)=0.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

E(X)=3. Find E(2X−5).

E(X)=3

Finite-distribution probabilities must be nonnegative and sum to one; dependence is only as explicitly stated.

Skills and prerequisite lessons

Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.

Working and explanation

BUILD THE REASONING

Hint 1

Use linearity of expectation.

Hint 2

Start from this intermediate relation.

E(2X-5)=2E(X)-5

Worked solution

Use linearity of expectation.
Establish and simplify this mathematical relation.
$E(2X-5)=2E(X)-5$
Establish and simplify this mathematical relation.
$E(2X-5)=1$
A fixed offset shifts the mean directly.

The requested value is 1.

Checks and common pitfalls: A fixed offset shifts the mean directly.

Think first. Reveal a hint when the class is ready.

07 / Foundation#Your turn

Var(X)=3. Find Var(−2X+7).

\operatorname{Var}(X)=3

Finite-distribution probabilities must be nonnegative and sum to one; dependence is only as explicitly stated.

Skills and prerequisite lessons

Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.

Working and explanation

BUILD THE REASONING

Hint 1

Square the scale; ignore the constant shift for variance.

Hint 2

Start from this intermediate relation.

\operatorname{Var}(-2X+7)=(-2)^2\operatorname{Var}(X)

Worked solution

Square the scale; ignore the constant shift for variance.
Establish and simplify this mathematical relation.
$\operatorname{Var}(-2X+7)=(-2)^2\operatorname{Var}(X)$
Establish and simplify this mathematical relation.
$\operatorname{Var}(-2X+7)=12$
A negative scale cannot create negative variance.

The requested value is 12.

Checks and common pitfalls: A negative scale cannot create negative variance.

Think first. Reveal a hint when the class is ready.

08 / Standard#Your turn

A game pays 8 with probability 1/4 and zero otherwise. Entry costs 3. Find expected net gain.

P(W=8)=1/4;\ P(W=0)=3/4;\ G=W-3

Finite-distribution probabilities must be nonnegative and sum to one; dependence is only as explicitly stated.

Skills and prerequisite lessons

Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.

Working and explanation

BUILD THE REASONING

Hint 1

Subtract the fee from expected payment.

Hint 2

Start from this intermediate relation.

E(W)=8/4=2

Worked solution

Subtract the fee from expected payment.
Establish and simplify this mathematical relation.
$E(W)=8/4=2$
Establish and simplify this mathematical relation.
$E(G)=2-3=-1$
An expected loss is an average, not a claim that every play loses the same amount.

The requested value is -1.

Checks and common pitfalls: An expected loss is an average, not a claim that every play loses the same amount.

Think first. Reveal a hint when the class is ready.

09 / Standard#Your turn

Independent X,Y have variances 2 and 5. Find Var(X−Y).

X\perp Y;\ \operatorname{Var}(X)=2;\ \operatorname{Var}(Y)=5

X and Y are independent with finite variances; otherwise include covariance.

Skills and prerequisite lessons

Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.

Working and explanation

BUILD THE REASONING

Hint 1

The minus sign is squared, and independence removes covariance.

Hint 2

Start from this intermediate relation.

\operatorname{Var}(X-Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)

Worked solution

The minus sign is squared, and independence removes covariance.
Establish and simplify this mathematical relation.
$\operatorname{Var}(X-Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)$
Establish and simplify this mathematical relation.
$\operatorname{Var}(X-Y)=7$
Subtracting variables does not mean subtracting variances.

The requested value is 7.

Checks and common pitfalls: Subtracting variables does not mean subtracting variances.

Think first. Reveal a hint when the class is ready.

10 / Standard#Your turn

The proposed probabilities are 0.7,0.4,−0.1. Explain why they fail.

P=(0.7,0.4,-0.1)

Finite-distribution probabilities must be nonnegative and sum to one; dependence is only as explicitly stated.

Skills and prerequisite lessons

Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.

Working and explanation

BUILD THE REASONING

Hint 1

Check both axioms, not just normalization.

Hint 2

Start from this intermediate relation.

0.7+0.4-0.1=1

Worked solution

Check both axioms, not just normalization.
Establish and simplify this mathematical relation.
$0.7+0.4-0.1=1$
Establish and simplify this mathematical relation.
$-0.1<0$
A correct total cannot repair an invalid cell.

They sum to one but include a negative probability.

Checks and common pitfalls: A correct total cannot repair an invalid cell.

Reasoning checklist · self / teacher assessment

Check both axioms, not just normalization.
Give the intermediate relations and explain the conclusion with its restrictions.
A correct total cannot repair an invalid cell.

Think first. Reveal a hint when the class is ready.

11 / Standard#Your turn

Explain what Var(X)=0 implies for a finite distribution.

\operatorname{Var}(X)=0

Finite-distribution probabilities must be nonnegative and sum to one; dependence is only as explicitly stated.

Skills and prerequisite lessons

Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.

Working and explanation

BUILD THE REASONING

Hint 1

A sum of nonnegative weighted squares can be zero only when every positive-weight square vanishes.

Hint 2

Start from this intermediate relation.

\sum_xp_x(x-\mu)^2=0

Worked solution

A sum of nonnegative weighted squares can be zero only when every positive-weight square vanishes.
Establish and simplify this mathematical relation.
$\sum_xp_x(x-\mu)^2=0$
Establish and simplify this mathematical relation.
$p_x>0\implies x=\mu$
Zero-probability support labels do not contradict the almost-sure claim.

X equals its mean with probability one.

Checks and common pitfalls: Zero-probability support labels do not contradict the almost-sure claim.

Reasoning checklist · self / teacher assessment

A sum of nonnegative weighted squares can be zero only when every positive-weight square vanishes.
Give the intermediate relations and explain the conclusion with its restrictions.
Zero-probability support labels do not contradict the almost-sure claim.

Think first. Reveal a hint when the class is ready.

12 / Transfer#Your turn

Can E(X)=2 and E(X²)=3 hold? Prove or refute.

E(X)=2;\ E(X^2)=3

Finite-distribution probabilities must be nonnegative and sum to one; dependence is only as explicitly stated.

Skills and prerequisite lessons

Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.

Working and explanation

BUILD THE REASONING

Hint 1

Use nonnegativity of variance as a feasibility condition.

Hint 2

Start from this intermediate relation.

\operatorname{Var}(X)=3-2^2=-1

Worked solution

Use nonnegativity of variance as a feasibility condition.
Establish and simplify this mathematical relation.
$\operatorname{Var}(X)=3-2^2=-1$
Establish and simplify this mathematical relation.
$\operatorname{Var}(X)\ge0$
Moment summaries must satisfy structural constraints, not only look plausible separately.

No; the implied variance is −1.

Checks and common pitfalls: Moment summaries must satisfy structural constraints, not only look plausible separately.

Reasoning checklist · self / teacher assessment

Use nonnegativity of variance as a feasibility condition.
Give the intermediate relations and explain the conclusion with its restrictions.
Moment summaries must satisfy structural constraints, not only look plausible separately.

Think first. Reveal a hint when the class is ready.

13 / Transfer#Your turn

Compare X=2 always with Y taking 0,4 equally. They have the same mean; which has larger variance, and by how much?

X=2;\ P(Y=0)=P(Y=4)=1/2

Finite-distribution probabilities must be nonnegative and sum to one; dependence is only as explicitly stated.

Skills and prerequisite lessons

Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.

Working and explanation

BUILD THE REASONING

Hint 1

Compare centered distributions, not just their means.

Hint 2

Start from this intermediate relation.

E(X)=E(Y)=2;\ \operatorname{Var}(X)=0

Worked solution

Compare centered distributions, not just their means.
Establish and simplify this mathematical relation.
$E(X)=E(Y)=2;\ \operatorname{Var}(X)=0$
Establish and simplify this mathematical relation.
$\operatorname{Var}(Y)=[(0-2)^2+(4-2)^2]/2=4$
An equal expected value does not imply equal dispersion.

The requested value is 4.

Checks and common pitfalls: An equal expected value does not imply equal dispersion.

Think first. Reveal a hint when the class is ready.

End-of-lesson check and correction

Review your latest checked answers and explanations. A draft change requires a fresh check. Written work needs your self-assessment or a teacher’s review.

Enable JavaScript for a summary of your local work.

Choose a foundation skill to revisit ↗

Teacher preparation and assessment

40-minute lesson

Minutes	Focus	Teaching and student action
5	Predict	Predict what happens to variance after multiplying all values by −2 and adding 7.
8	Build the idea	Sample response: shift does not affect spread; the scale squares to four.
10	Guided comparison	Sample response: X+1−X=1, so the sum has variance zero.
12	Independent practice and discussion	Compare equal-mean games and audit proposed moments for feasibility.
5	Exit and next step	Exit: distinguish E(X²) from E(X)² and state the independence requirement for adding variances.

Question sequence

Predict what happens to variance after multiplying all values by −2 and adding 7.
Sample response: X+1−X=1, so the sum has variance zero.

Board development

Average first, center before measuring spread
$E(X)=\sum xp_x;\ \operatorname{Var}(X)=E(X^2)-E(X)^2$

Expected correct and incorrect responses

Correct sample: 1
Anticipated wrong response: 2 — Uses the second raw moment without centering.
Anticipated wrong response: 11 — Scales variance linearly and adds a constant shift.
Using unweighted means for unequal probabilities.
Adding variances without checking covariance.

Follow-up questions

Follow-up: Center before squaring or subtract the squared mean from the second moment. Then explain why 2 fails.
Follow-up: Square the scale and remove the constant shift. Then explain why 11 fails.
Explain why a negative scale reverses positions but does not make variance negative.

Remediation

Sample response: shift does not affect spread; the scale squares to four.
Return to the first example, identify the missing condition, then solve one fresh task without viewing its solution.

Exit check

Exit: distinguish E(X²) from E(X)² and state the independence requirement for adding variances.

Assessment checklist and sample responses

Representation and conditions
For probabilities 3/4,1/4 on 0,2, E(X)=0·3/4+2·1/4=1/2.
Connected mathematical reasoning
For fair 0,2, the mean is 1 and both squared deviations are 1, so variance is 1.
Interpretation and boundary check
If Y=1−X, X+Y=1 and variance is zero; independence is false.