Goal and scope
In a valid finite probability model, prove Var(X)≥0, Var(X+c)=Var(X), Var(kX)=k²Var(X), and Var(X)=0 ⇔ X takes one common value at every positive-weight outcome. Then with S={1,2,3,4}, w₁=1/2, w₂=1/4, w₃=1/4, w₄=0 and X=(1,3,−1,10), compute μ=1, Var(X)=2, Var(X+5)=2 and Var(−3X)=18, and use Z=(2,2,2,7) to show that a different value at a zero-weight outcome does not prevent zero variance.
Stays inside the MKO: finite nonempty sample space, nonnegative normalized weights, a total real-valued function, the probability-distribution variance rather than a sample estimator, and a clear distinction between the positive-weight support and the whole sample space (a1, a2). All arithmetic is exact and rational. Linearity of expectation is cited from mko-expectation. Var(X+Y), covariance, standard deviation and the formula E[X²]−μ² are not treated; the canonical MKO and its English edition are not changed.
Main derivation
S01 · implies
Before
w=(\tfrac12,\tfrac14,\tfrac14,0)
After
w_s\ge0,\ \sum_s w_s=1,\ \mathrm{supp}=\{1,2,3\}Conditions: none
Rule: numeric_evaluation · Uses: G2, support_def
Leaf basis: numeric_evaluation + support_def
Validate the model first: the four weights are nonnegative and 1/2+1/4+1/4+0=1, so they are valid. The positive-weight support is {1,2,3}; outcome 4 belongs to the sample space S but has weight 0 and is not in the support. MKO assumption a2 asks for these two sets to be kept apart.
S02 · equivalent
Before
\mu=E[X]
After
\tfrac12\cdot1+\tfrac14\cdot3+\tfrac14\cdot(-1)+0\cdot10=1
Conditions: none
Rule: mko_expectation_lin · Uses: S01, G3, mko_expectation_lin
Leaf basis: mko_expectation_lin + numeric_evaluation
By the definition of expectation: 1/2+3/4−1/4+0=1. The value 10 at outcome 4 is multiplied by weight 0 and contributes 0.
S03 · equivalent
Before
X(s)-\mu\ \ (s=1,2,3,4)
After
0,\ 2,\ -2,\ 9
Conditions: none
Rule: numeric_evaluation · Uses: S02, G3
Leaf basis: numeric_evaluation
Compute the deviations pointwise: 1−1=0, 3−1=2, −1−1=−2, 10−1=9. The deviation is defined at every outcome, including the zero-weight outcome 4.
S04 · equivalent
Before
\mathrm{Var}(X)=\sum_s w_s(X(s)-\mu)^2After
\tfrac12\cdot0^2+\tfrac14\cdot2^2+\tfrac14\cdot(-2)^2+0\cdot9^2=2
Conditions: none
Rule: mko_variance_def · Uses: S03, S01, mko_variance_def
Leaf basis: mko_variance_def + power_laws + numeric_evaluation
Substitute into the definition: 0+1+1+0=2. The zero-weight outcome 4 has a large deviation 9, but multiplied by weight 0 it contributes 0.
S05 · implies
Before
w_s\ge0,\ (X(s)-\mu)^2\ge0
After
\mathrm{Var}(X)\ge0Conditions: none
Rule: square_nonneg · Uses: G1, mko_variance_def
Leaf basis: square_nonneg + order_basics + mul_zero + finite_sum_rules
General property. Each term t_s=w_s(X(s)−μ)²: the square is nonnegative; if w_s=0 then t_s=0; if w_s>0, multiplying 0≤(X(s)−μ)² by the positive w_s keeps the order, so t_s≥0. A finite sum of nonnegative numbers is nonnegative, so Var(X)≥0.
S06 · equivalent
Before
\mathrm{Var}(X+c)After
\mathrm{Var}(X)Conditions: none
Rule: L_shift · Uses: G1, mko_variance_def
Leaf basis: mko_expectation_lin + eq_replace
By Lemma L_shift, translation does not change variance. The proof is in S06.1 and S06.2: the mean also shifts by c, so the deviations are unchanged.
S07 · equivalent
Before
\mathrm{Var}(kX)After
k^2\,\mathrm{Var}(X)Conditions: none
Rule: L_scale · Uses: G1, mko_variance_def
Leaf basis: mko_expectation_lin + power_laws + finite_sum_rules
By Lemma L_scale, scaling by k multiplies the variance by k². The proof is in S07.1 to S07.3 and holds for every real k, including k=0 and negative k.
S08 · implies
Before
\mathrm{Var}(X+5),\ X+5=(6,8,4,15)After
\mathrm{Var}(X+5)=\mathrm{Var}(X)=2Conditions: none
Rule: L_shift · Uses: S06, S04, G3
Leaf basis: mko_variance_def + numeric_evaluation
By S06 (c=5), Var(X+5)=2. Direct check: X+5 has values 6, 8, 4, 15, mean 1/2·6+1/4·8+1/4·4+0·15=6=μ+5, deviations 0, 2, −2, 9 as before, so the variance is still 2.
S09 · implies
Before
\mathrm{Var}(-3X),\ -3X=(-3,-9,3,-30)After
\mathrm{Var}(-3X)=(-3)^2\cdot2=18Conditions: none
Rule: L_scale · Uses: S07, S04, G3
Leaf basis: mko_variance_def + numeric_evaluation
By S07 (k=−3), Var(−3X)=9·2=18. Direct check: −3X has values −3, −9, 3, −30, mean −3=−3μ, deviations 0, −6, 6, −27, and variance 1/4·36+1/4·36+0·729=18. A negative factor does not make the variance negative.
S10 · implies
Before
t_s\ge0,\ \sum_s t_s=0
After
t_s=0\ \ \forall s
Conditions: none
Rule: L_nonneg_sum_zero · Uses: finite_sum_rules, order_basics
Leaf basis: order_basics + finite_sum_rules + contradiction
Prepare Lemma L_nonneg_sum_zero for the zero-variance characterization: if a finite sum of nonnegative numbers is zero, every term is zero. The proof is in S10.1 and S10.2.
S11 · equivalent
Before
\mathrm{Var}(X)=0After
X(s)=\mu\ \ \forall s\in\mathrm{supp}Conditions: none
Rule: L_var_zero · Uses: G1, mko_variance_def, support_def
Leaf basis: L_nonneg_sum_zero + zero_product + square_nonneg + case_analysis
By Lemma L_var_zero, zero variance means exactly that X equals the mean at every positive-weight outcome. The condition involves only the support; values at zero-weight outcomes are unrestricted. The proof is in S11.1 (⇒) and S11.2 (⇐).
S12 · equivalent
Before
\mathrm{Var}(X)=0After
\exists c\ \forall s\in\mathrm{supp}:\ X(s)=cConditions: none
Rule: L_const_support · Uses: S11
Leaf basis: L_var_zero + exists_intro + finite_sum_rules
Rewrite S11 in the MKO's words: zero variance means exactly that X takes one common value at all positive-weight outcomes. This is Lemma L_const_support, proved in S12.1 and S12.2.
S13 · implies
Before
Z=2\ \text{on}\ \mathrm{supp}=\{1,2,3\},\ Z(4)=7After
\mathrm{Var}(Z)=0Conditions: none
Rule: L_const_support · Uses: S12, S01, G3
Leaf basis: L_const_support + numeric_evaluation
Z equals 2 throughout the support {1,2,3}, so by S12 Var(Z)=0, even though Z is not constant on all of S (Z(4)=7). Direct check: μ_Z=2·(1/2+1/4+1/4)+7·0=2 and Var(Z)=0+0+0+0·25=0.
S14 · implies
Before
X(1)=1\neq3=X(2),\ 1,2\in\mathrm{supp}After
\mathrm{Var}(X)\neq0Conditions: none
Rule: contrapositive · Uses: S12, S01, G3, S04
Leaf basis: contrapositive + L_const_support
X takes the different values 1 and 3 at two outcomes of the support, so it has no common value there; by the contrapositive of the ⇒ direction of S12, Var(X)≠0. This agrees with Var(X)=2>0 from S04.
S15 · implies
Before
\text{S04 to S14}After
\mathrm{Var}\ge0,\ \mathrm{Var}(X+c)=\mathrm{Var}(X),\ \mathrm{Var}(kX)=k^2\mathrm{Var}(X),\ \mathrm{Var}=0\iff\text{const on supp}Conditions: none
Rule: and_intro_elim · Uses: S04, S05, S06, S07, S08, S09, S12, S13
Leaf basis: and_intro_elim
Summary: the four general properties are proved; in the instance μ=1, Var(X)=2, Var(X+5)=2, Var(−3X)=18, and Var(Z)=0 although Z differs at a zero-weight outcome.
Substeps and lemmas
S06.1 · equivalent
Before
E[X+c]
After
\mu+c
Conditions: none
Rule: mko_expectation_lin · Uses: G1, mko_expectation_lin
Leaf basis: mko_expectation_lin
By linearity of expectation (a=1, b=c): E[X+c]=E[X]+c=μ+c. This needs the weights to sum to 1.
S06.2 · equivalent
Before
\sum_s w_s\bigl((X(s)+c)-(\mu+c)\bigr)^2
After
\sum_s w_s(X(s)-\mu)^2
Conditions: none
Rule: eq_replace · Uses: S06.1, eq_replace
Leaf basis: eq_replace + add_assoc + add_comm + add_inverse + add_zero
For each s: (X(s)+c)−(μ+c)=X(s)−μ (associativity, commutativity and inverses for addition). Each term is replaced by an equal term, so the sum is unchanged. S06.1 and S06.2 together prove Lemma L_shift.
S07.1 · equivalent
Before
E[kX]
After
k\mu
Conditions: none
Rule: mko_expectation_lin · Uses: G1, mko_expectation_lin
Leaf basis: mko_expectation_lin
By linearity of expectation (a=k, b=0): E[kX]=kμ.
S07.2 · equivalent
Before
(kX(s)-k\mu)^2
After
k^2(X(s)-\mu)^2
Conditions: none
Rule: power_laws · Uses: S07.1, distributive, power_laws
Leaf basis: distributive + power_laws + mul_comm
First factor out k by distributivity: kX(s)−kμ=k(X(s)−μ); then use (ab)²=a²b². The deviation is multiplied by k, and squaring brings out k², so a negative sign disappears.
S07.3 · equivalent
Before
\sum_s w_s\,k^2(X(s)-\mu)^2
After
k^2\sum_s w_s(X(s)-\mu)^2=k^2\,\mathrm{Var}(X)Conditions: none
Rule: finite_sum_rules · Uses: S07.2, finite_sum_rules
Leaf basis: finite_sum_rules + def_unfold
The constant k² factors out of the finite sum, and what remains is Var(X) by definition. S07.1 to S07.3 together prove Lemma L_scale. For k=0 this gives Var(0)=0, consistent with the variance of a constant variable.
S10.1 · implies
Before
t_r>0\ \text{for some}\ rAfter
\sum_s t_s=t_r+\sum_{s\neq r}t_s\ge t_r>0Conditions: t_s\ge0
Rule: order_basics · Uses: finite_sum_rules, order_basics
Leaf basis: finite_sum_rules + order_basics
Assume for the moment that some term t_r>0. Split the sum into t_r plus the remaining terms; the remaining terms are nonnegative, so is their sum, and the total is ≥t_r>0.
S10.2 · implies
Before
\sum_s t_s>0\ \text{and}\ \sum_s t_s=0After
\neg\exists r:\ t_r>0,\ \text{so}\ t_s=0Conditions: none
Rule: contradiction · Uses: S10.1, contradiction
Leaf basis: contradiction + order_basics
The conclusion of S10.1 contradicts the hypothesis Σ t_s=0, so no t_r>0. Each t_s is ≥0 and not >0, so by trichotomy t_s=0. S10.1 and S10.2 together prove Lemma L_nonneg_sum_zero.
S11.1 · implies
Before
\mathrm{Var}(X)=0,\ s\in\mathrm{supp}After
X(s)=\mu
Conditions: w_s>0
Rule: L_nonneg_sum_zero · Uses: S05, S10, zero_product, square_nonneg
Leaf basis: L_nonneg_sum_zero + zero_product + square_nonneg
(⇒) By S05 each term t_s=w_s(X(s)−μ)²≥0, and their sum Var(X)=0, so by L_nonneg_sum_zero every t_s=0. For s∈supp, w_s>0 so w_s≠0, and the zero-product property gives (X(s)−μ)²=0; since a square is zero exactly when its base is zero, X(s)=μ.
S11.2 · implies
Before
X(s)=\mu\ \ \forall s\in\mathrm{supp}After
\mathrm{Var}(X)=\sum_s 0=0Conditions: none
Rule: case_analysis · Uses: G1, support_def, case_analysis
Leaf basis: case_analysis + support_def + mul_zero + finite_sum_rules
(⇐) Each s is either in the support or not. In the support: X(s)−μ=0 and t_s=w_s·0=0. Outside: w_s=0 and t_s=0·(X(s)−μ)²=0. Every term is 0, so the sum is 0. S11.1 and S11.2 together prove Lemma L_var_zero.
S12.1 · implies
Before
\mathrm{Var}(X)=0After
X(s)=\mu\ \ \forall s\in\mathrm{supp}\ \Rightarrow\ c=\muConditions: none
Rule: L_var_zero · Uses: S11, exists_intro
Leaf basis: L_var_zero + exists_intro
(⇒) By the ⇒ direction of L_var_zero, X equals μ throughout the support; taking c=μ gives the common value.
S12.2 · implies
Before
X(s)=c\ \ \forall s\in\mathrm{supp}After
\mu=\sum_{s\in\mathrm{supp}}w_s c+\sum_{s\notin\mathrm{supp}}0\cdot X(s)=c,\ \text{so}\ \mathrm{Var}(X)=0Conditions: none
Rule: L_var_zero · Uses: S11, G1, support_def
Leaf basis: L_var_zero + finite_sum_rules + support_def + mul_zero
(⇐) Split the sum for μ into the parts inside and outside the support: outside, every weight is 0 and contributes 0, which also makes the weights inside sum to 1; so μ=c·1=c. Then X(s)=c=μ for every s∈supp, and the ⇐ direction of L_var_zero gives Var(X)=0. S12.1 and S12.2 together prove Lemma L_const_support.