Goal and scope
For a nonempty finite sample space S with nonnegative weights w_s summing to 1, prove E[c]=c, E[aX+b]=aE[X]+b and E[X+Y]=E[X]+E[Y] (X, Y on the same S, independence not assumed). Then on S={1,2,3} with w₁=1/2, w₂=1/4, w₃=1/4, X=(2,−2,0), Y=(1,3,−1), compute E[X], E[Y], E[X+Y] and E[−3X+5], show that X and Y are not independent while linearity still holds, and use E[XY]≠E[X]E[Y] to show that linearity does not give a product formula.
Stays inside the MKO's finite known probability model: weights nonnegative and summing exactly to 1, variables total on the same finite S, the general finite-sum properties allow the real coefficients a,b,c and normalized nonnegative weights of G1; the worked example uses integer coefficients and explicit rational weights, all computed exactly (a1, a2). The general properties are proved with finite sums; the instance serves only as verification and counterexample. Sample means, long-run convergence, infinite sample spaces and integrability are not treated; the canonical MKO and its English edition are not changed.
Main derivation
S01 · definition
Before
Z:S\to\mathbb{R}After
E[Z]=\sum_{s\in S}w_sZ(s)Conditions: w_s\ge0,\ \sum_s w_s=1
Rule: mko_expectation_def · Uses: G1, pointwise_rv
Leaf basis: mko_expectation_def + pointwise_rv
Define expectation as in the MKO: multiply the value at each outcome by its weight and add over all outcomes. aX+b, X+Y, XY and the constant c are random variables defined pointwise on the same S, so they all have expectations.
S02 · implies
Before
w_1=\tfrac12,\ w_2=\tfrac14,\ w_3=\tfrac14
After
w_s\ge0,\ \sum_s w_s=1
Conditions: none
Rule: numeric_evaluation · Uses: G2
Leaf basis: numeric_evaluation
Validate the instance's weights first: all three are positive and 1/2+1/4+1/4=1. The MKO asks for the weights to be checked before any expectation is computed.
S03 · equivalent
Before
E[c]
After
c
Conditions: \sum_s w_s=1
Rule: L_const · Uses: S01
Leaf basis: finite_sum_rules + mul_one
By Lemma L_const, the expectation of a constant is that constant. The proof is in S03.1; the key is that the weights sum to exactly 1.
S04 · equivalent
Before
E[aX+b]
After
aE[X]+b
Conditions: \sum_s w_s=1
Rule: L_affine · Uses: S01
Leaf basis: distributive + finite_sum_rules + def_unfold
By Lemma L_affine, the expectation of an affine transform is the affine transform of the expectation. The proof has three sub-steps: expand pointwise, split the sum, and fold back the definition with normalization.
S05 · equivalent
Before
E[X+Y]
After
E[X]+E[Y]
Conditions: X,Y\ \text{on the same}\ S
Rule: L_sum · Uses: S01
Leaf basis: distributive + finite_sum_rules + def_unfold
By Lemma L_sum, the expectation of a sum is the sum of the expectations. The proof uses only that X and Y are defined on the same S and no assumption about their joint behavior, so independence is not needed.
S06 · equivalent
Before
E[X]
After
\tfrac12\cdot2+\tfrac14\cdot(-2)+\tfrac14\cdot0=\tfrac12
Conditions: none
Rule: mko_expectation_def · Uses: S01, S02, G3
Leaf basis: mko_expectation_def + numeric_evaluation
Substitute into the definition: 1/2·2+1/4·(−2)+1/4·0=1−1/2+0=1/2. Note that 1/2 is not one of the possible values of X (X takes only 2, −2 and 0).
S07 · equivalent
Before
E[Y]
After
\tfrac12\cdot1+\tfrac14\cdot3+\tfrac14\cdot(-1)=1
Conditions: none
Rule: mko_expectation_def · Uses: S01, S02, G3
Leaf basis: mko_expectation_def + numeric_evaluation
Likewise: 1/2+3/4−1/4=1.
S08 · implies
Before
E[X+Y],\ X+Y=(3,1,-1)
After
E[X+Y]=\tfrac32=E[X]+E[Y]
Conditions: none
Rule: L_sum · Uses: S05, S06, S07, G3
Leaf basis: mko_expectation_def + numeric_evaluation
By S05, E[X+Y]=E[X]+E[Y]=1/2+1=3/2. Direct check: the pointwise values of X+Y are 3, 1, −1, and 1/2·3+1/4·1+1/4·(−1)=3/2, which agrees.
S09 · implies
Before
E[-3X+5],\ -3X+5=(-1,11,5)
After
E[-3X+5]=-3\cdot\tfrac12+5=\tfrac72
Conditions: none
Rule: L_affine · Uses: S04, S06, G3
Leaf basis: mko_expectation_def + numeric_evaluation
By S04 (a=−3, b=5): E[−3X+5]=−3·(1/2)+5=7/2. Direct check: pointwise values −1, 11, 5, and 1/2·(−1)+1/4·11+1/4·5=−1/2+4=7/2.
S10 · implies
Before
P(X=2,\,Y=1)=w_1=\tfrac12,\ P(X=2)P(Y=1)=\tfrac14
After
X,Y\ \text{not independent}Conditions: none
Rule: indep_def · Uses: G2, G3, indep_def
Leaf basis: indep_def + quantifier_negation + numeric_evaluation
Only outcome 1 has X=2 and only outcome 1 has Y=1, so P(X=2)=1/2 and P(Y=1)=1/2, while P(X=2 and Y=1)=w₁=1/2. The product 1/4≠1/2, so the definition of independence fails at x=2, y=1 and X, Y are not independent. Yet S08 holds: linearity does not need independence.
S11 · implies
Before
XY=(2,-6,0)
After
E[XY]=-\tfrac12\neq\tfrac12=E[X]E[Y]
Conditions: none
Rule: mko_expectation_def · Uses: S01, S06, S07, G3
Leaf basis: mko_expectation_def + pointwise_rv + numeric_evaluation
The pointwise product XY has values 2·1=2, (−2)·3=−6 and 0·(−1)=0, so E[XY]=1/2·2+1/4·(−6)+1/4·0=1−3/2=−1/2. But E[X]E[Y]=1/2·1=1/2. So linearity does not give E[XY]=E[X]E[Y]; an expectation can also be negative.
S12 · implies
Before
\tfrac{2+(-2)+0}{3}=0After
0\neq\tfrac12=E[X]
Conditions: none
Rule: mko_expectation_def · Uses: S06, G3
Leaf basis: mko_expectation_def + numeric_evaluation
Treating the three values as equally likely and taking the plain average gives 0, which differs from E[X]=1/2. The given weights are 1/2, 1/4, 1/4, not 1/3 each, so the plain average is not this model's expectation.
S13 · implies
Before
\text{S03 to S12}After
E[c]=c,\ E[aX+b]=aE[X]+b,\ E[X+Y]=E[X]+E[Y];\ E[XY]\neq E[X]E[Y]\ \text{here}Conditions: none
Rule: and_intro_elim · Uses: S03, S04, S05, S08, S09, S10, S11
Leaf basis: and_intro_elim
Summary: the three linearity properties hold in every finite valid model (S03 to S05); in the instance E[X]=1/2, E[Y]=1, E[X+Y]=3/2 and E[−3X+5]=7/2, X and Y are dependent while linearity still holds, and E[XY]=−1/2 differs from E[X]E[Y]=1/2.
Substeps and lemmas
S03.1 · equivalent
Before
\sum_{s\in S}w_s\,cAfter
c\sum_{s\in S}w_s=c\cdot 1=cConditions: \sum_s w_s=1
Rule: finite_sum_rules · Uses: S01, G1, finite_sum_rules
Leaf basis: finite_sum_rules + mul_comm + mul_one
By the finite-sum rule, factor out the constant c: Σ w_s c=cΣ w_s; then Σ w_s=1 from G1 gives c·1=c. If the weights did not sum to 1, the result would be c times the total, not c. This step proves Lemma L_const.
S04.1 · equivalent
Before
\sum_{s}w_s\,(aX(s)+b)After
\sum_{s}\bigl(a\,w_sX(s)+b\,w_s\bigr)Conditions: none
Rule: distributive · Uses: S01, pointwise_rv, distributive
Leaf basis: distributive + mul_comm + mul_assoc + eq_replace
By the pointwise definition, (aX+b)(s)=aX(s)+b; for each s, distributivity and commutativity give w_s(aX(s)+b)=a·w_sX(s)+b·w_s. Each term is replaced by an equal expression.
S04.2 · equivalent
Before
\sum_{s}\bigl(a\,w_sX(s)+b\,w_s\bigr)After
a\sum_{s}w_sX(s)+b\sum_{s}w_sConditions: none
Rule: finite_sum_rules · Uses: S04.1, finite_sum_rules
Leaf basis: finite_sum_rules
A finite sum splits into two sums (Σ(u_s+v_s)=Σu_s+Σv_s), and constants factor out (Σ c·u_s=cΣu_s).
S04.3 · equivalent
Before
a\sum_{s}w_sX(s)+b\sum_{s}w_sAfter
aE[X]+b\cdot1=aE[X]+b
Conditions: \sum_s w_s=1
Rule: def_unfold · Uses: S04.2, S01, G1
Leaf basis: def_unfold + mul_one
The first sum is E[X] by definition; the second equals 1 by G1. S04.1 to S04.3 together prove Lemma L_affine. The constant b is multiplied by the total weight 1, not by the number of outcomes.
S05.1 · equivalent
Before
\sum_{s}w_s\,(X(s)+Y(s))After
\sum_{s}\bigl(w_sX(s)+w_sY(s)\bigr)Conditions: X,Y\ \text{on the same}\ S
Rule: distributive · Uses: S01, pointwise_rv, distributive
Leaf basis: distributive + pointwise_rv + eq_replace
Since X and Y live on the same S, (X+Y)(s)=X(s)+Y(s) is defined for every s; then apply distributivity to each term. If they were defined on different sample spaces, the pointwise sum would not make sense.
S05.2 · equivalent
Before
\sum_{s}\bigl(w_sX(s)+w_sY(s)\bigr)After
\sum_{s}w_sX(s)+\sum_{s}w_sY(s)=E[X]+E[Y]Conditions: none
Rule: finite_sum_rules · Uses: S05.1, finite_sum_rules, S01
Leaf basis: finite_sum_rules + def_unfold
Split the finite sum; by definition the two sums are E[X] and E[Y]. S05.1 and S05.2 together prove Lemma L_sum; P(X=x and Y=y) never appears in the argument, so it does not matter whether X and Y are independent.
Error controls and negative controls
Do not average the values when equal likelihood is not given
One misconception identified in the MKO is averaging all values without looking at the weights. The plain average of the values 2, −2, 0 of X is 0, but the weights are 1/2, 1/4, 1/4 and E[X]=1/2 (S12). Expectation is a weighted sum: each outcome's value is multiplied by its own probability.
Linearity neither needs nor implies independence
The other misconception is to infer independence or a product formula from linearity. S10 shows X and Y are not independent, yet E[X+Y]=E[X]+E[Y] holds; S11 shows E[XY]=−1/2≠E[X]E[Y]=1/2. Linearity is only the splitting of a finite sum and has nothing to do with the joint distribution.
The constant term is multiplied by the total weight 1
The b in E[aX+b] comes from bΣw_s=b·1 (S04.3). Dropping b gives −3/2, and multiplying b by the number of outcomes 3 gives −3/2+15=27/2; both are wrong, and the correct value is 7/2.
- E[X]=(2+(−2)+0)/3=0 →
reject — The weights are not equal; E[X]=1/2·2+1/4·(−2)+1/4·0=1/2. - E[X+Y]=E[X]+E[Y] holds here, so X and Y are independent →
reject — Linearity holds for any variables on the same model; P(X=2 and Y=1)=1/2≠1/4=P(X=2)P(Y=1), so X and Y are not independent. - E[XY]=E[X]E[Y]=1/2 →
reject — Direct computation gives E[XY]=1/2·2+1/4·(−6)+1/4·0=−1/2. - E[−3X+5]=−3E[X]=−3/2 →
reject — The constant term was dropped: E[−3X+5]=−3/2+5=7/2.