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

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

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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.

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 / 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.

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.