9.3 Linearity of expectation

Sometimes the full distribution of a total is difficult to find, but the average contribution of each part is easy. Linearity of expectation lets you add those contributions. It is especially useful for counts of many events that overlap or depend on one another.

This is intermediate probability. Review expectation in Module 9.2, then see MIT OpenCourseWare's lecture notes on indicators and linearity.

A basic idea: add the parts

For random variables $X_1,\ldots,X_n$ with finite expectations,

$$E[X_1+\cdots+X_n]=E[X_1]+\cdots+E[X_n].$$

The variables do not need to be independent. This is a statement about averages, not a claim that their probabilities multiply. If a total is a count, let $I_i$ be $1$ when event $i$ happens and $0$ otherwise. Then $E[I_i]=P(I_i=1)$ and

$$E[\text{number of events}]=\sum_i P(\text{event }i).$$

A quick tour of properties

Constants also pass through sums: $E[aX+bY+c]=aE[X]+bE[Y]+c$. Linearity works for any finite number of terms with finite expectations. It does not say $E[XY]=E[X]E[Y]$ without further conditions.

Application 1: matching cards

Three distinct cards are placed at random, one per envelope, into three matching labelled envelopes. Let $I_i=1$ if card $i$ lands in its matching envelope. How many matches are expected?

1. Each card has probability $1/3$ of matching its own envelope.
2. The total matches are $M=I_1+I_2+I_3$.
3. $E[M]=1/3+1/3+1/3=1$ match.

The match events are dependent: matching two cards forces the third. Linearity still works.

Application 2: two dependent costs

Rain has probability $0.3$. Transport costs $10$ on a dry day and $12$ on a rainy day; on rainy days you also pay $4$ for shelter. Let $X$ be the transport cost and $Y$ the shelter cost. Both depend on the weather, so they are dependent. Still,

$$E[X+Y]=E[X]+E[Y]=[10(0.7)+12(0.3)]+[4(0.3)]=10.6+1.2=11.8.$$

Check directly: the total is $10$ or $16$, and $10(0.7)+16(0.3)=11.8$.