Decision Models Reference
Expected Value and Decision Models
Expected value is the long-run average outcome of a random process. It helps compare games, choices, and risk-reward tradeoffs.
Solved Examples
Jump directly to a worked example by expected-value task.
Fact Table
| Idea | Meaning | Watch For |
|---|---|---|
| Random variable | Numerical outcome assigned to each event. | List every possible value and probability. |
| Expected value | Probability-weighted average. | It need not be a possible one-time outcome. |
| Fair game | Expected net gain is $0$. | Include the cost to play. |
| Decision model | Compares expected outcomes of alternatives. | Expected value measures long-run average, not a guarantee. |
Content Formulas
Expected Value
$$E(X)=\sum x\,P(X=x)$$
Net Gain
$$\text{net gain}=\text{payout}-\text{cost}$$
Decision Rule
$$\text{compare expected net values under the same conditions}$$
Classic Examples
Find Expected Value from a Distribution
A game pays $0$ with probability $\frac12$, $2$ with probability $\frac14$, and $6$ with probability $\frac14$. Find the expected payout.
Solution Steps
- Multiply each payout by its probability.
- Add the weighted payouts.
- State the long-run average payout.
$$0\left(\frac12\right)+2\left(\frac14\right)+6\left(\frac14\right)$$$$0+\frac12+\frac32=2$$$$\boxed{E(X)=2}$$
Test a Fair Game
A game costs $2$ to play and has expected payout $2$. Is it fair?
Solution Steps
- Find expected net gain by subtracting the cost.
- Compare the result to zero.
- State the fairness conclusion.
$$E(\text{net})=2-2$$$$E(\text{net})=0$$$$\boxed{\text{The game is fair in the long run.}}$$
Compare Two Decisions
Game A has expected net gain $1$. Game B has expected net gain $\frac12$. Which is better by expected value?
Solution Steps
- Write both expected net gains.
- Compare the values.
- State the higher expected-value choice.
$$E(A)=1,\qquad E(B)=\frac12$$$$1>\frac12$$$$\boxed{\text{Choose Game A by long-run expected value.}}$$
Model an Insurance Decision
A $100$ loss has probability $0.01$. What is the expected loss before insurance?
Solution Steps
- Assign the loss as a negative outcome and no loss as zero.
- Multiply the loss by its probability.
- State the expected loss.
$$E(X)=(-100)(0.01)+0(0.99)$$$$E(X)=-1$$$$\boxed{\text{Expected loss }=\$1\text{ per event.}}$$