Expectation and Variance

Long-run center, spread, standard deviation, and risk for random variables.

Probability distributions can be large, but two summaries are especially useful: expected value and variance. Expected value describes the long-run center. Variance describes spread around that center. Together, they help compare games, risks, measurements, and uncertain decisions.

Learning objectives

After this lesson you should be able to:

  • compute expected value as a weighted average;
  • explain why expected value may not be an actual outcome;
  • compute variance and standard deviation from a small table;
  • compare distributions with the same mean but different spread;
  • use linearity of expectation in simple examples;
  • check calculations with fcmath, Python, and SymPy.

Expected value as a weighted average

For a discrete random variable, expected value is:

E[X] = sum of value * probability

If X takes values 0 and 10 with probabilities 0.7 and 0.3, then

E[X] = 0(0.7) + 10(0.3) = 3

This means that over many repetitions, the average value approaches 3.

Expected value is not always a possible outcome

A fair die has values 1,2,3,4,5,6, each with probability 1/6.

E[X] = (1+2+3+4+5+6)/6 = 21/6 = 3.5

But a die can never roll 3.5. Expected value is a long-run average, not a promise about one trial.

Variance and standard deviation

Variance measures how far values tend to be from the mean. For a discrete random variable:

Var(X) = sum of probability * (value - mean)^2

Standard deviation is the square root of variance:

SD(X) = sqrt(Var(X))

Variance uses squared distances, so it is sensitive to values far from the mean.

Step-by-step variance table

Suppose X is 0 with probability 0.5 and 2 with probability 0.5.

First compute the mean:

E[X] = 0(0.5) + 2(0.5) = 1

Now build a table:

value probability value - mean squared distance weighted squared distance
0 0.5 -1 1 0.5
2 0.5 1 1 0.5

Add the last column:

Var(X) = 0.5 + 0.5 = 1
SD(X) = sqrt(1) = 1

Linearity of expectation

Expected value has a very useful property:

E[X + Y] = E[X] + E[Y]

This is true even when X and Y are not independent. For example, if one fair die has expected value 3.5, then two fair dice have total expected value:

3.5 + 3.5 = 7

Risk interpretation

Two games can have the same expected value but different risk.

  • Game A pays $5 every time.
  • Game B pays $0 half the time and $10 half the time.

Both have expected value $5, but Game B has more spread. Variance helps describe that uncertainty.

Worked example 1: fair game

A game costs $2. It pays $5 with probability 0.4 and $0 with probability 0.6. Is the game favorable in expected value?

Expected payout:

5(0.4) + 0(0.6) = 2

Expected net gain:

expected payout - cost = 2 - 2 = 0

The game is fair in expected value. It is not guaranteed to break even on one play.

Worked example 2: compare spread

Distribution A:

X = 4 with probability 1

Distribution B:

Y = 0 with probability 1/2
Y = 8 with probability 1/2

Both have expected value 4. But X never varies, while Y jumps between 0 and 8. So Y has larger variance and more risk.

Common pitfalls

  • Forgetting to multiply each value by its probability.
  • Treating expected value as a guaranteed outcome.
  • Computing variance before computing the mean.
  • Forgetting to square deviations in variance.
  • Comparing risk using only the mean.

Practice exercises

  1. Compute the expected value of values 0, 10 with probabilities 0.7, 0.3.
  2. What is the expected value of a fair die?
  3. Why can expected value be impossible as a single outcome?
  4. Compute the variance for values 0, 2 with probabilities 0.5, 0.5.
  5. Which is riskier: a guaranteed $5 or $0/$10 with equal probabilities?
  1. 0(0.7)+10(0.3)=3.
  2. 3.5.
  3. It is a long-run weighted average, not necessarily a value the random variable can take.
  4. The mean is 1. Variance is 0.5(0-1)^2 + 0.5(2-1)^2 = 1.
  5. The $0/$10 game is riskier because it has more spread.

Subtopic guided practice and checkpoints

Expectation and variance calculations are safest when you use a table and fill one column at a time.

Lab 1: compute expected value row by row

Guess first. Which contributes more to expected value: a large value with small probability or a smaller value with large probability?

Guided exercise.

For values 1, 4, 10 with probabilities 0.2, 0.5, 0.3:

E[X] = 1(0.2) + 4(0.5) + 10(0.3)
     = 0.2 + 2.0 + 3.0
     = 5.2

Each row contributes value * probability.

Checkpoint. Compute expected value for values -2, 4 with probabilities 0.25, 0.75.

Lab 2: interpret an impossible expected value

Guess first. If the expected number of heads in one fair coin flip is 0.5, can one flip produce 0.5 heads?

Guided exercise.

One flip produces either 0 heads or 1 head. The expected value is

0(0.5) + 1(0.5) = 0.5

This is a long-run average over many flips.

Checkpoint. Explain why the expected value 3.5 for a fair die is useful even though no die face is 3.5.

Lab 3: compute variance with deviations

Guess first. Why do we subtract the mean before squaring?

Guided exercise.

For values 0, 4 with probabilities 0.5, 0.5, the mean is 2. Deviations are:

0 - 2 = -2
4 - 2 = 2

Square and weight:

Var(X) = 0.5(-2)^2 + 0.5(2)^2
       = 2 + 2
       = 4

Checkpoint. Compute the variance for values 1, 3 with probabilities 0.5, 0.5.

Lab 4: compare risk

Guess first. Which has more variance: always getting 5, or getting 0 or 10 with equal probability?

Guided exercise.

Always getting 5 has no spread, so variance is 0. Getting 0 or 10 has the same mean but values far from the mean, so it has positive variance.

Checkpoint. Two investments both have expected return 6%. One always returns 6%; the other returns -4% or 16% with equal probability. Which has larger variance?

Lab 5: use linearity of expectation

Guess first. If one fair die has expected value 3.5, do three fair dice have expected total 10.5?

Guided exercise.

By linearity of expectation:

E[die1 + die2 + die3] = 3.5 + 3.5 + 3.5 = 10.5

You do not need to list all 216 outcomes.

Checkpoint. If one raffle ticket has expected value $1.20, what is the expected value of 5 identical tickets?

  1. (-2)(0.25)+4(0.75)=-0.5+3=2.5.
  2. It predicts the long-run average over many rolls and helps compare games or repeated experiments.
  3. The mean is 2. Variance is 0.5(1-2)^2 + 0.5(3-2)^2 = 1.
  4. The second investment has larger variance.
  5. 5 * 1.20 = $6.00.

Expectation and variance guessing game

Use this checkpoint to practice center, spread, and fair-game reasoning.

Using this lesson with fcmath and SymPy

Use fcmath to compute expected value and variance.

from fcmath.probability import expected_value, standard_deviation, variance

values = [0, 2]
probabilities = [0.5, 0.5]
expected_value(values, probabilities), variance(values, probabilities), standard_deviation(values, probabilities)
(1.0, 1.0, 1.0)

Compare two games.

safe_values = [5]
safe_probabilities = [1]
risky_values = [0, 10]
risky_probabilities = [0.5, 0.5]

variance(safe_values, safe_probabilities), variance(risky_values, risky_probabilities)
(0.0, 25.0)

Use SymPy for exact variance arithmetic.

import sympy as sp

values = [0, 4]
probabilities = [sp.Rational(1, 2), sp.Rational(1, 2)]
mean = sum(value * probability for value, probability in zip(values, probabilities))
var = sum(probability * (value - mean) ** 2 for value, probability in zip(values, probabilities))
mean, var
(2, 4)
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