AcademyProbability
Academy
Expectation
Level 1 - Math II (Physics) topic page in Probability.
Principle
Expectation is a probability-weighted average. It describes the long-run centre of a random quantity, not necessarily a value that can occur in one trial.
Notation
\(X,Y\)
random variables
\(E[X]\)
expectation, or mean, of X
\(p_X(x)\)
PMF of a discrete random variable
\(f_X(x)\)
PDF of a continuous random variable
\(g(X)\)
a function of the random variable X
\(a,b\)
constants
Method
Step 1: Multiply each value by its probability or density weight
For a discrete variable, add value times probability. For a continuous variable, integrate value times density.
Discrete expectation
\[E[X]=\sum_{x\in D_X}x\,p_X(x)\]
Continuous expectation
\[E[X]=\int_{-\infty}^{\infty}x f_X(x)\,dx\]
Step 2: Average functions by applying the function first
If the quantity of interest is \(g(X)\), weight \(g(x)\), not \(x\).
Discrete function
\[E[g(X)]=\sum_{x\in D_X}g(x)p_X(x)\]
Continuous function
\[E[g(X)]=\int_{-\infty}^{\infty}g(x)f_X(x)\,dx\]
Step 3: Use expectation rules to simplify
Linearity lets constants pass through expectation and lets sums be averaged term by term.
Rules
Discrete mean
\[E[X]=\sum_{x\in D_X}x p_X(x)\]
Continuous mean
\[E[X]=\int_{-\infty}^{\infty}x f_X(x)\,dx\]
Function of a variable
\[E[g(X)]=\sum_{x\in D_X}g(x)p_X(x)\]
Linearity
\[E[aX+b]=aE[X]+b\]
Additivity
\[E[X+Y]=E[X]+E[Y]\]
Positivity
\[X\ge0\quad\Rightarrow\quad E[X]\ge0\]
Independent product
\[X,Y\ \text{independent}\quad\Rightarrow\quad E[XY]=E[X]E[Y]\]
Examples
Question
Find the expected score on a fair die.
Answer
Each value has probability
\[1/6\]
\[E[X]=1\cdot\frac{1}{6}+2\cdot\frac{1}{6}+3\cdot\frac{1}{6}+4\cdot\frac{1}{6}+5\cdot\frac{1}{6}+6\cdot\frac{1}{6}=\frac{21}{6}=3.5.\]
Checks
- Expectation has the same unit as the random variable.
- An expected value need not be a possible observed value.
- Probabilities or densities must define a valid distribution before computing expectation.
- Additivity does not require independence.
- The product rule \(E[XY]=E[X]E[Y]\) does require independence in this course setting.