Learning Module 4
Probability Trees
Key Outcomes Summary & Practice Problems
What you must be able to do
Curriculum Year: 2026
Calculate expected values, variances, and standard deviations and demonstrate their application to investment problems.
Formulate an investment problem as a probability tree and explain the use of conditional expectations in investment applications.
Calculate and interpret an updated probability in an investment setting using Bayes' formula.
1 · Expected Value, Variance & Standard Deviation
A random variable is a quantity whose future value is uncertain. For a discrete random variable X with n possible outcomes, each weighted by its probability, we calculate three fundamental descriptors.
Probability-weighted average of all possible outcomes:
E(X) = P(X₁)X₁ + P(X₂)X₂ + … + P(Xₙ)Xₙ = Σᵢ P(Xᵢ)Xᵢ
E(X) is our forward-looking forecast, not a historical average.
It equals the population mean μ — what we expect on average over many trials.
Probability-weighted average of squared deviations from E(X):
σ²(X) = Σᵢ P(Xᵢ)[Xᵢ − E(X)]²
σ²(X) ≥ 0 always. If σ²(X) = 0 → outcome is certain (no risk).
Units: squared units of X (e.g., return² in %²).
Increasing variance = increasing dispersion and uncertainty.
Positive square root of variance — same units as X:
σ(X) = √σ²(X) = √[Σᵢ P(Xᵢ)(Xᵢ − E(X))²]
Easier to interpret than variance because it is in original units.
If X is return in %, σ(X) is also in %. σ is the main risk measure.
Key distinction — Expected Value vs. Sample Mean: E(X) is a forward-looking, probability-weighted forecast (population concept). The sample mean X̄ is a backward-looking equally-weighted average of observed data (historical). In investment analysis, we use E(X) to forecast future outcomes and σ²(X) to quantify the risk around that forecast.
Worked example — BankCorp EPS: Probability distribution: P(USD 2.60)=0.15, P(USD 2.45)=0.45, P(USD 2.20)=0.24, P(USD 2.00)=0.16.E(EPS) = 0.15(2.60) + 0.45(2.45) + 0.24(2.20) + 0.16(2.00) = USD 2.3405σ²(EPS) = 0.15(2.60−2.34)² + 0.45(2.45−2.34)² + 0.24(2.20−2.34)² + 0.16(2.00−2.34)² = 0.0388σ(EPS) = √0.0388 ≈ USD 0.197
Define mutually exclusive and exhaustive scenarios S₁, S₂, … that drive the outcome. Assign prior probability P(Sᵢ) to each.
For each scenario, list possible outcomes Xᵢ with their conditional probabilities P(Xᵢ | Sⱼ). These branch off each scenario node.
Compute E(X | Sⱼ) = Σᵢ P(Xᵢ|Sⱼ)×Xᵢ for each scenario — the expected value conditioned on that scenario occurring.
Apply the total probability rule: E(X) = Σⱼ E(X|Sⱼ)×P(Sⱼ). This gives the unconditional expected value today.
Unconditional expected value from conditional expectations:
E(X) = E(X|S)P(S) + E(X|Sᶜ)P(Sᶜ) (two scenarios)
General form:
E(X) = Σⱼ E(X|Sⱼ) × P(Sⱼ) where S₁,…,Sₙ are mutually exclusive & exhaustive
Joint probability of any terminal node: P(Xᵢ) = P(Xᵢ|Sⱼ) × P(Sⱼ)
Sum of all joint probabilities = 1 ✓
Conditional expected value E(X|S): the expected value of X given scenario S has occurred. It revises the unconditional estimate once we condition on new information.
Conditional variance σ²(X|S): measures dispersion of outcomes around E(X|S) within a specific scenario. Like E(X|S), it uses only outcomes and probabilities relevant to that scenario.
Consistency requirement: unconditional and conditional probabilities must be consistent. If not, investment strategies could exploit the inconsistency for riskless profit — an arbitrage opportunity.
Practical application: analysts decompose uncertain outcomes (EPS, revenue, operating costs) into scenarios with assigned probabilities, forming conditional forecasts for each scenario, then roll back to an unconditional forecast.
3 · Bayes' Formula — Updating Probabilities
Bayes' formula is a rational method for revising (updating) prior probabilities when new information arrives. It reverses the conditional probability — using the occurrence of new information to infer the probability of the scenario that generated it.
P(Event | Information) = [P(Information | Event) / P(Information)] × P(Event)
Where:
P(Event) = prior probability (before new information)
P(Information|Event)= likelihood of observing the info IF the event is true
P(Information) = unconditional probability of the new information
P(Event|Information)= posterior probability (updated after new information)
P(Information) computed via the total probability rule:
P(Info) = Σⱼ P(Info|Sⱼ) × P(Sⱼ)
Term | Meaning | DriveMed Example |
|---|---|---|
Prior P(Event) | Probability before new info | P(EPS exceeded) = 0.45 |
Likelihood P(Info|Event) | How likely is the info IF the event occurred? | P(expands|EPS exceeded) = 0.75 |
P(Information) | Unconditional prob of the new info (total prob rule) | P(DriveMed expands) = 0.41 |
Posterior P(Event|Info) | Updated probability after receiving new info | P(EPS exceeded|expands) = 0.823 |
DriveMed worked example: Prior: P(EPS exceeded) = 0.45. New info: DriveMed announces factory expansion.
Step 1 — P(expands) via total probability rule:P(expands) = 0.75(0.45) + 0.20(0.30) + 0.05(0.25) = 0.41
Step 2 — Apply Bayes' formula:P(exceeded|expands) = (0.75/0.41) × 0.45 = 0.823
Interpretation: prior belief of 45% revised upward to 82.3% upon learning DriveMed is expanding — strong signal that EPS beat expectations. The three posterior probabilities sum to 1: 0.823 + 0.146 + 0.031 = 1.000 ✓
Inverse probability: Bayes' formula reverses the conditioning. We know P(Info|Event) from our models, and we want P(Event|Info) — the posterior. Bayes' formula connects the two.
Ratio interpretation: the likelihood ratio P(Info|Event)/P(Info) adjusts the prior. If this ratio >1, the new info increases the event's probability. If <1, it decreases it.
Diffuse (equal) priors: when prior probabilities are equal, P(Event|Info) = P(Info|Event). The posterior is fully determined by the likelihoods alone.
Bayesian updating is continuous: each posterior becomes the new prior when the next piece of information arrives. This is how analysts iteratively refine their views as new data comes in.
Must sum to 1: updated probabilities across all mutually exclusive and exhaustive events must always sum to exactly 1 — a useful check on calculations.