Recipe

Compute E(X) using the outcome-sum definition

Use when: you know every outcome's probability and need E(X), especially when several outcomes share the same X-value

Goal: the value of E(X), computed outcome by outcome and checked against the table method

Taught in Expected value, revisited

1. 1

   Write down every individual outcome s and its probability P(s), not yet grouped by shared X-value.

   Example

   Flipping a coin twice gives four outcomes, HH, HT, TH, TT, each with probability \frac14.

   Why: the outcome-sum formula adds one term per outcome, so every outcome needs its own probability listed first.

2. 2

   For each outcome s, compute X(s), the random variable's value on that specific outcome.

   Example

   With X= number of heads: X(HH)=2, X(HT)=1, X(TH)=1, X(TT)=0.

   Why: X(s) is a function of the outcome itself, evaluated before any grouping by shared value happens.

3. 3

   Compute X(s)\cdot P(s) for every outcome and add all of these products together.

   Example

   2(\tfrac14)+1(\tfrac14)+1(\tfrac14)+0(\tfrac14) = \tfrac24+\tfrac14+\tfrac14+\tfrac04=1.

   ⚡ Shortcut

   If X(s)=c, the same constant, for every outcome, skip the sum entirely: E(X)=c. Every term becomes c\cdot P(s), and factoring c out leaves c\sum_{s\in S}P(s)=c\cdot1, since a sample space's probabilities always add to 1.

   Why: this is the outcome-sum formula E(X)=\sum_{s\in S}X(s)P(s) applied directly, term by term.

4. 4

   Group the outcomes by shared X-value, multiply each distinct value by the total probability of that group, and add. Confirm the two totals agree.

   Example

   Grouping the same outcomes by value gives P(X=0)=\tfrac14, P(X=1)=\tfrac12, P(X=2)=\tfrac14, so the table way gives 0(\tfrac14)+1(\tfrac12)+2(\tfrac14)=1, matching Step 3.

   Why: the two methods always agree, so a mismatch signals an arithmetic slip in one of them, not a real disagreement.