FE Home→Engineering Economics→Uncertainty and Expected Value Example

Uncertainty and Expected Value Example

Medium DifficultyFE Engineering Economics

Find the expected value (mean) and standard deviation of a discrete random outcome (e.g. NPV under scenarios) to quantify risk.

Concept

Under uncertainty, outcomes can be modeled with probabilities. The expected value E (X) is the probability-weighted average of outcomes. The variance measures spread; standard deviation σ = Var (X) is often used to describe risk. For a discrete random variable: E (X) = \sum x_{k} P (X = x_{k}) and Var (X) = E (X^{2}) - [E (X)]^{2} . E (X) = k \sum x_{k} \cdot P (X = x_{k}) Var (X) = E (X^{2}) - [E (X)]^{2}, σ = Var (X) Notation: E (X) = expected value (mean), x_{k} = outcome of scenario k, P (X = x_{k}) = probability of outcome x_{k}, Var (X) = variance, σ = standard deviation (risk).

A project has three possible net present value (NPV) outcomes and associated probabilities: NPV = -$10k with probability 0.3, NPV = $20k with probability 0.5, NPV = $50k with probability 0.2. Find: Expected NPV Standard deviation of NPV (risk measure)

NPV = -10 (thousands) with P = 0.3 NPV = 20 with P = 0.5 NPV = 50 with P = 0.2

E (X) = k \sum x_{k} \cdot P (X = x_{k}) E (N P V) = (0.3) (- 10) + (0.5) (20) + (0.2) (50) = - 3 + 10 + 10 = 17

E (N P V^{2}) = (0.3) (100) + (0.5) (400) + (0.2) (2500) = 30 + 200 + 500 = 730 Var (N P V) = 730 - (17)^{2} = 730 - 289 = 441, σ = 441 = 21

(1) Expected NPV: E (N P V) = $17, 000 (2) Standard deviation: σ = $21, 000 (measure of risk or spread of outcomes).

E (X) = \sum x_{k} P (x_{k}); Var (X) = E (X^{2}) - [E (X)]^{2} Notation: E (X) = expected value, x_{k} = outcome k, P (x_{k}) = probability of x_{k}, Var (X) = variance, σ = standard deviation. Higher σ means more variability (risk). Decision-makers often compare expected value vs. risk across alternatives.