Probability course
Probability course
Module 5 of 8

Expected value

Use weighted outcomes and indicators to avoid case explosions.

18 min2 worked examples3 concept checks

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Learning objectives
  • Calculate a weighted average of possible payoffs.
  • Use linearity even when terms are dependent.
  • Break counting problems into indicator variables.

Concepts

Expectation is a weighted average

For a discrete variable, E[X]=∑xxP(X=x)E[X]=\sum_x xP(X=x) when the expectation exists. It describes a model average, not a promise for the next outcome and not necessarily a value X can take. Keep losses negative and distinguish a gross payout from a net payoff after fees.

Linearity is the workhorse

E[aX+bY+c]=aE[X]+bE[Y]+cE[aX+bY+c]=aE[X]+bE[Y]+c for variables with finite expectations. Independence is not required. This often avoids finding the entire distribution of a complicated total. In contrast, E[XY]=E[X]E[Y]E[XY]=E[X]E[Y] is not generally valid without additional assumptions such as independence.

Count with indicators

An indicator I is 1 when an event occurs and 0 otherwise, so E[I]=P(event). If a total counts how many events occur, write it as a sum of indicators and add their probabilities. You can also average conditional means: split the model into exhaustive cases and weight each conditional expectation by its case probability.

Examples

Worked example 1

A signed payoff

A game gives a net payoff of +8 units with probability 1/4 and −2 units otherwise. What is its expected net payoff?
  1. List all net outcomes

    The payoffs already include any costs, so use +8 and −2 directly. Their probabilities are 1/4 and 3/4.
  2. Weight and add

    E[X]=8(1/4)+(−2)(3/4)=2−1.5=0.5E[X]=8(1/4)+(-2)(3/4)=2-1.5=0.5.
  3. Interpret the answer

    The model average is +0.5 units per game. No individual game pays 0.5, and a positive mean does not remove risk.
Worked example 2

Correctly returned badges

Ten named badges are randomly permuted among their ten owners, with every permutation equally likely. How many owners get their own badge, on average?
  1. Use one indicator per owner

    Let I_i be 1 if owner i receives the correct badge. The total correct returns is X=I_1+⋯+I_10.
  2. Use symmetry for one owner

    Each owner is equally likely to receive any of ten badges, so E[I_i]=P(I_i=1)=1/10.
  3. Add without requiring independence

    E[X]=10×(1/10)=1. Correct-return events are dependent, but linearity still applies.
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Expectation is a balance point

Place probability weight at each possible payoff. The mean balances those weights and stays between the outcomes; it need not be a payoff you can receive.

A probability-weighted payoff balance75.0% of the weight is at -2; 25.0% is at 8. The balance point is 0.50. Circle areas represent probability weights.75.0%25.0%Mean 0.50-28Payoff in units
Expected payoff0.50 units

75.0% × -2 + 25.0% × 8. A positive expectation describes a model average, not a guaranteed gain.

How each outcome contributes to the mean
PayoffProbabilityWeighted contribution
-275.0%-1.50
825.0%2.00

Concept checks

Choose an answer, then check it. You can retry; your latest checked answer determines completion.

0 of 3 concept checks passed
Concept check 1X and Y are dependent and have finite means. Is E[X+Y]=E[X]+E[Y] valid?
Concept check 2An indicator equals 1 with probability 0.12 and 0 otherwise. What is its expected value?
Concept check 3A process is in state A 30% of the time, with conditional mean output 10; otherwise its conditional mean is 4. What is the overall mean?

Key results

  • Keep payoff signs and fees consistent.
  • Add expectations before seeking a full distribution.
  • Write counts as sums of indicators.
  • Average conditional means with the correct weights.

Practice problems

Solve related problems or try a timed probability drill.

Next in the course

Measure what the mean hides

Track spread, scale, and the way variables move together.

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