Construct finite independent sums and compare repeated copies with scaling

Identical distributions and independence are separate

X₁ and X₂ are identically distributed if each has the same probabilities for the same values. They are independent if each joint pair probability equals the product of its two marginal probabilities. Equal marginal distributions alone do not imply independence. For more than two copies, assume mutual independence

Build the sum from ordered outcomes

Suppose each copy takes 0 with probability 3/4 and 2 with probability 1/4. For independent copies, the four ordered pairs have probabilities 9/16,3/16,3/16,1/16. Pairs (0,2) and (2,0) are distinct and both give sum 2

Using the displayed independent-pair table, find Pr(X₁+X₂=2)

For independent X₁ and X₂, each taking 0 with probability 3/4 and 2 with probability 1/4, find Pr(X₁+X₂=4)

Sum equal totals, not the possible values alone

For independent X₁ and X₂, each taking 0 with probability 3/4 and 2 with probability 1/4, their sum takes 0,2,4 with probabilities 9/16,6/16,1/16. Its masses total 1. Equal numbers of ordered pairs do not imply equal probability when the original values have unequal probabilities

Twice the same draw is different

Let X take 0 with probability 3/4 and 2 with probability 1/4, and let X₁ and X₂ be independent copies of X. T=2X takes only 0 and 4, with probabilities 3/4 and 1/4. It cannot take value 2. S₂=X₁+X₂ can take 2. In T, the two written copies of X always agree; they are not independent copies of this nonconstant variable

If X takes 0 with probability 3/4 and 2 with probability 1/4, and T=2X, find Pr(T=2)

Two variables both have Pr(0)=3/4 and Pr(2)=1/4. Is their sum distribution determined by this alone?

Construct and justify a full table, ungraded

For independent copies of X with values −1,1,3 and probabilities 1/4,1/2,1/4, construct all nine ordered pairs and the complete distribution of their sum. Explain why probabilities are multiplied within pairs and added between pairs

Model: group nine pair outcomes

Each of two independent copies takes −1, 1 and 3 with probabilities 1/4, 1/2 and 1/4. The pair probabilities by rows are (1,2,1)/16, (2,4,2)/16 and (1,2,1)/16. Grouping their sums gives −2,0,2,4,6 with probabilities 1/16,4/16,6/16,4/16,1/16. Independence gives the products; disjoint ordered outcomes give the additions. The grouped masses total 16/16

For independent X₁ and X₂, each taking −1, 1 and 3 with probabilities 1/4, 1/2 and 1/4, find Pr(X₁+X₂=2)

General convolution and more copies

For a finite support A, for each possible first-copy value, multiply its probability by the probability that the second copy supplies the remaining total, then add the products, treating values outside A as having probability zero. Repeat the operation for further mutually independent copies. Starting with probability 1 at total 0, add one copy at each step; add every product contributing to the same total

A third copy remains a separate draw

For three independent copies taking 0 or 2 with probabilities 3/4,1/4, totals 0,2,4,6 have probabilities 27/64,27/64,9/64,1/64. For total 4, exactly two copies equal 2; there are three ordered positions for the remaining 0

For three independent copies each taking 0 with probability 3/4 and 2 with probability 1/4, find Pr(X₁+X₂+X₃=4)

Expectation adds even with dependence

For any finite joint distribution, E(X+Y)=E(X)+E(Y): sum (x+y)p(x,y) and separate the x and y terms using their marginal probabilities. Independence is not required. Thus n identical marginals give E(Sₙ)=nμ

Independent variances add

For mutually independent copies with mean μ and variance σ², Var(Sₙ)=nσ² and SD(Sₙ)=√n σ. In contrast Var(nX)=n²σ² and SD(nX)=nσ for a positive integer n. Both means equal nμ. The difference disappears when σ²=0 or n=1

If E(X)=3/2 and Var(X)=5/4, find Var(X₁+...+X₄) for independent copies

If Var(X)=5/4, find Var(4X)

Prove the variance rule, ungraded

For independent finite X and Y, prove Var(X+Y)=Var(X)+Var(Y), explicitly identifying where independence is used. Extend to mutually independent copies and compare with nX

Model: the centred cross term

Set U=X−E(X) and V=Y−E(Y). Expanding gives E[(U+V)²]=E(U²)+2E(UV)+E(V²). Independence factors the finite joint sum: E(UV)=E(U)E(V)=0. Thus variances add. Repeating this argument for mutually independent copies yields nσ². For nX, the centred variable is n(X−μ), giving n²σ² instead; the repeated terms are not independent

Test a dependence claim, ungraded

Let X take −1 and 1 with equal probabilities. Compare Y=X and Y=−X. Show that all marginals have mean 0 and variance 1 while the two sum variances differ. Explain why identical marginals cannot justify variance addition

Model: two exact counterexamples

Let X take −1 and 1 with probability 1/2 each. For Y=X, X+Y=2X takes −2 and 2 equally, so variance is 4. For Y=−X, the sum is always 0, so variance is 0. Both Y variables have the same marginal distribution as X. Independence would instead give sum probabilities 1/4,1/2,1/4 at −2,0,2 and variance 2. None of the three cases can be selected from the marginals alone