pebble-grouping

Sun Jul 05 2026

Why is $E(X+Y) = E(X) + E(Y)$ with no independence assumption? My instinct said you need the distribution of $X+Y$ first, and building that distribution means working out how $X$ and $Y$ interact. This note is the regrouping move that dissolved the itch, and the same move later turned out to prove LOTUS too.

Setup

Two fair coins. $X = 1$ if the first coin lands heads, else 0. $Y$ the same for the second coin. The route I trusted: find the distribution of $X+Y$ (values $\{0,1,2\}$ with probabilities $\{1/4, 1/2, 1/4\}$, where the middle $1/2$ comes from merging HT and TH), then $E(X+Y) = 1$. That merge step felt essential. It is where the two coins combine, so I assumed any proof of linearity had to pass through it.

Now the second route. The sample space has four pebbles, HH, HT, TH, TT, each with weight $1/4$. Sum directly over pebbles, never forming any distribution:

$E(X+Y) = \sum_s \big(X(s) + Y(s)\big)\,P(\{s\}) = 2 \cdot \tfrac14 + 1 \cdot \tfrac14 + 1 \cdot \tfrac14 + 0 \cdot \tfrac14 = 1.$

Each term splits as $X(s)P(\{s\}) + Y(s)P(\{s\})$, so the whole sum splits into two pebble sums, one for $X$ and one for $Y$.

Where I got stuck

I refused to call those two sums $E(X)$ and $E(Y)$. The sum $\sum_s X(s)P(\{s\})$ runs over the four-pebble Cartesian-product world, but $X$ was born on a two-outcome world with its own PMF. The underlying experiment changed, so why would a sum over the wrong sample space compute $E(X)$? I could feel the informal rescue (each $P(\{s\})$ shrank, and the number of pebbles grew in proportion, so it balances out) but I could not formalize it.

The key move: group by value

Take the $X$-part of the sum and, instead of adding left to right, collect the pebbles by the value $X$ takes:

$\sum_s X(s)P(\{s\}) = 1 \cdot \big(\tfrac14 + \tfrac14\big) + 0 \cdot \big(\tfrac14 + \tfrac14\big) = 1 \cdot \tfrac12 + 0 \cdot \tfrac12.$

Inside each group $X$ is constant, so it factors out. And each group's total weight is $P(X = x)$ by definition, on whatever sample space $X$ happens to live. That is the marginal probability, computed by pooling pebbles. So the pebble sum is $\sum_x x\,P(X=x) = E(X)$, always. My "shrank but grew in proportion" intuition was this grouping, stated loosely. One correction to my first attempt at formalizing it: I leaned on "each $x$ repeats $|Y|$ times", which only works for this product-shaped example. The grouping argument needs no product structure and no equal weights. Any pebble world, same move.

The picture to keep: expectation has two accounting systems. Pebble-by-pebble sums over outcomes, value-by-value sums over the PMF. They are the same sum with the terms regrouped, and regrouping a finite sum changes nothing. So you compute in whichever system is convenient, and equalities proven in one system transfer to the other for free.

Payoff 1: linearity is unconditional

The split $(X(s)+Y(s))P(\{s\}) = X(s)P(\{s\}) + Y(s)P(\{s\})$ happens per pebble, before any distribution is formed. No joint structure exists yet, so independence never gets consulted. The merge I thought was essential (HT and TH pooling into $P(X+Y=1) = 1/2$) is just the grouping move applied to the value of $X+Y$, and it happens after linearity is already settled. That is the whole reason $E(X+Y) = E(X) + E(Y)$ holds for dependent variables too.

Payoff 2: the same move is the LOTUS proof

Group the pebbles by the value of $g(X)$ instead of the value of $X$ and you get LOTUS: $E(g(X)) = \sum_x g(x)P(X=x)$. I did this by hand before knowing the theorem's name. Computing $E(W)$ for $W = (X-3.5)^2$ on a die, I instinctively paired the six terms into three collision braces ($x=1$ with $x=6$, and so on), and each brace held a constant value of $W$ with pooled weight $\tfrac16 + \tfrac16 = \tfrac13 = P(W=w)$. The braces are super-pebbles. Same grouping, different sorting key. The full LOTUS story, including what it does not license, lives in lotus-and-variance.

One move, two theorems: split before grouping gives linearity, group by $g(X)$ gives LOTUS.