Mathematics · Stochastics

Expected Value

Q: What is the difference between expected value and mean?

The expected value $\mu{} = E(X)$ is a theoretical parameter of the probability model: it is computed from the probabilities before any experiment takes place. The mean x̄ (arithmetic average) is an empirical parameter of concrete data: it is computed from actually observed values after the experiment has run. The two are linked by the law of large numbers: as the number of trials grows, x̄ approaches the expected value μ. If you roll a die 10 times, x̄ might be 3.9; the model still says $\mu{} = 3.5$. Statistics uses this relation in reverse: x̄ serves as an estimator for an unknown μ. Linguistically: "expected value" for the model, "mean" for the sample.

The expected value is the long-run average of a random variable: values times probabilities, summed up.

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Formula

E (X) = μ = i \sum x_{i} \cdot p_{i}

E(X) in the unit of the random variable X · p_i dimensionless

Variables & units

Variables & units
Symbol	Meaning	Unit
E(X), μ	Expected value of the random variable X	same as X
$xᵢ$	Possible values of the random variable	same as X
$pᵢ$	Probability P(X = x_i)	dimensionless

LaTeX

E(X) = \mu = \sum_{i} x_{i} \cdot p_{i}

QUANTA STUDYquanta-study.de

Bar chart with five bars for the values 0 to 4; a triangle below the axis at 1.6 marks the expected value as the balance point. — The expected value is the balance point of the distribution: with the probabilities pᵢ as weights at the values xᵢ, it balances at μ = Σ xᵢ·pᵢ = 1.6.

Derivation & background

Christiaan Huygens introduced the concept in 1657 for games of chance. The expected value is the weighted mean of all possible values; by the law of large numbers the average of many repetitions stabilizes at E(X). A game is called fair if the expected value of the winnings is 0. For the binomial distribution simply $E(X) = n\cdot{}p$ .

Exam blueprint

Validity range

Applies to discrete random variables with finitely many values; all p_i sum to 1. For continuous random variables an integral replaces the sum.

Derivation steps

Weighted mean: each value counts according to its probability.

1In N repetitions, x_i occurs about N·p_i times; the average is Σx_i·N·p_i/N.
2Cancelling N gives E(X) = Σx_i·p_i; the law of large numbers guarantees the stabilization.

Rearrangements

Binomial distribution

E(X) = n \cdot p

Short formula instead of summing over all k.

Linearity

E(aX + b) = a \cdot E(X) + b

Linear transformations pass directly into the expected value.

Fair game

E(G) = 0

G = winnings minus stake; $E(G) = 0$ means fair.

Task variant

A wheel pays 10 € with $p = 0.2$ , otherwise nothing. Stake 3 €. Is the game worth it?

E(payout) = 10·0.2 = 2 €.
Winnings G: $E(G) = 2 - 3 = -1$ €.
On average you lose 1 € per game, the game is not worth it.

X takes the values 0, 1, 2 with $p =\frac{\frac{ 0.3}{0.5}}{0.2}$ . Compute E(X).

$E(X) = 0\cdot{}0.3 + 1\cdot{}0.5 + 2\cdot{}0.2 = 0.9$ .

Common mistakes

Simply taking the arithmetic mean of the values.

Correct only for a uniform distribution; otherwise weight with the p_i.

Interpreting E(X) as the "most likely value".

$E(X) = 3.5$ for a die is never rolled; it is the long-run average.

Forgetting the stake in a fair game.

Fair means E(winnings) = 0 including the stake, not E(payout) = 0.

Exam context

Games of chance and payout tasks, distribution parameters, fair stakes.

Formula cluster

Core stochastics

Location and spread together describe a distribution.

P(X=k) = (n über k)·pᵏ·(1−p)ⁿ⁻ᵏ

Binomial Distribution

σ = √Var(X)

Standard Deviation

Worked example

Fair die: $E(X) =\frac{ (1 + 2 + 3 + 4 + 5 + 6)\cdot{}1}{6} =\frac{ 21}{6} = 3.5$ . Binomial distribution with $n = 100$ and $p = 0.3$ : $E(X) = n\cdot{}p = 30$ successes on average.

Applications

Evaluating games of chance and insurance, fair stakes, mean success counts in exams, decision theory, risk analysis

How do you calculate with Expected Value?

Here is how to work through a typical Expected Value (E(X) = Σ xᵢ·pᵢ) task step by step:

1
Task
A wheel pays 10 € with $p = 0.2$ , otherwise nothing. Stake 3 €. Is the game worth it?
Solution path
1. E(payout) = 10·0.2 = 2 €.
2. Winnings G: $E(G) = 2 - 3 = -1$ €.
3. On average you lose 1 € per game, the game is not worth it.
2
Task
X takes the values 0, 1, 2 with $p =\frac{\frac{ 0.3}{0.5}}{0.2}$ . Compute E(X).
Solution path
$E(X) = 0\cdot{}0.3 + 1\cdot{}0.5 + 2\cdot{}0.2 = 0.9$ .

Frequently asked questions about Expected Value

How do I calculate the expected value of a random variable?+

First set up the probability distribution as a table: all possible values x_i of the random variable with their probabilities p_i. The expected value is then the sum of products: E(X) = x₁·p₁ + x₂·p₂ + ... Die example: E(X) = 1·(1/6) + 2·(1/6) + ... $+ 6\cdot{}(\frac{1}{6}) =\frac{ 21}{6} = 3.5$ . Prize game example: if a game pays 10 € with probability 0.2 and otherwise nothing, E(payout) = 10·0.2 + 0·0.8 = 2 €. Before computing, check that all p_i add to 1, otherwise the distribution is incomplete. The expected value need not be a possible value: 3.5 cannot be rolled, yet it is the correct long-run average.

What does the expected value mean intuitively?+

It is the long-run average: if you repeat the random experiment very often and average the results, this mean stabilizes at the expected value; the law of large numbers guarantees it. In 6000 die rolls you expect a total near $6000\cdot{}3.5 = 21000$ . Two distinctions matter: E(X) is not a prediction for a single experiment, an individual roll can deviate arbitrarily. And E(X) is not the most likely value (that would be the mode); for skewed distributions the two lie far apart, as in lotteries, where the most frequent prize is 0 but the expected value can be positive. As a physical picture: E(X) is the centre of mass of the distribution, the point where the probability mass balances.

When is a game fair and how do I check it?+

A game is called fair if the expected value of the winnings is zero, where winnings = payout minus stake. So compute the expected payout and subtract the stake. Example: a wheel of fortune pays 10 € with $p = 0.2$ , the stake is 3 €. E(payout) = 2 €, so E(winnings) = 2 − 3 = −1 €: on average the player loses 1 € per round, the game is unfair in the operator's favour. A stake of exactly 2 € would be fair. Typical exam task: "Determine the stake so that the game is fair", the answer is always stake = expected payout. Real gambling (lottery, roulette, slot machines) has a systematically negative expected value; that is what the operator lives on.

Why does

E(X) = n\cdot{}p

simply hold for the binomial distribution?+

Because of the additivity of the expected value. A binomially distributed X is the sum of n indicator variables X₁, ..., X_n, where X_i = 1 for a success in the i-th trial and 0 otherwise. Each has E(X_i) = 1·p + 0·(1−p) = p. Since the expected value of a sum is always the sum of the expected values (this even holds without independence!), $E(X) = n\cdot{}p$ follows immediately, without evaluating the sum formula with binomial coefficients. Example: 100 attempts at rolling a six, $p =\frac{ 1}{6}$ , give on average $\frac{100}{6} \approx{} 16.7$ successes. The formula is intuitive too: a fraction p of n trials succeeds. The same decomposition yields the variance $n\cdot{}p\cdot{}(1-p)$ , though there independence is required.

What is the difference between expected value and mean?+

The expected value $\mu{} = E(X)$ is a theoretical parameter of the probability model: it is computed from the probabilities before any experiment takes place. The mean x̄ (arithmetic average) is an empirical parameter of concrete data: it is computed from actually observed values after the experiment has run. The two are linked by the law of large numbers: as the number of trials grows, x̄ approaches the expected value μ. If you roll a die 10 times, x̄ might be 3.9; the model still says $\mu{} = 3.5$ . Statistics uses this relation in reverse: x̄ serves as an estimator for an unknown μ. Linguistically: "expected value" for the model, "mean" for the sample.

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Which formula describes Expected Value?

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How do you rearrange Expected Value for Binomial distribution?

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, Introductory Statistics 2e, §4.2: Mean or Expected Value and Standard Deviation

Related formulas

f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}

Normal Distribution (Gaussian Bell Curve)

P(X = k) = \frac{\lambda^{k}}{k!} \cdot e^{-\lambda}

Poisson Distribution

More formulas: Probability and statistics

P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}

Bayes' Theorem

P(A \mid B) = \frac{P(A \cap B)}{P(B)} \quad (P(B) > 0)

Conditional Probability

b = \frac{\sum (x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum (x_{i}-\bar{x})^{2}}, \quad a = \bar{y} - b\bar{x}

Linear Regression (Line of Best Fit)

\binom{n}{k} = \frac{n!}{k!\,(n-k)!}

Binomial Coefficient (n choose k)

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Expected Value

The expected value is the long-run average of a random variable: values times probabilities, summed up.

BasicExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

E(X), μ

Expected value of the random variable X

same as X

xᵢ

Possible values of the random variable

same as X

pᵢ

Probability P(X = x_i)

dimensionless

Derivation & background

Exam blueprint

Validity range

Applies to discrete random variables with finitely many values; all p_i sum to 1. For continuous random variables an integral replaces the sum.

Derivation steps

Weighted mean: each value counts according to its probability.

1In N repetitions, x_i occurs about N·p_i times; the average is Σx_i·N·p_i/N.
2Cancelling N gives E(X) = Σx_i·p_i; the law of large numbers guarantees the stabilization.

How do you calculate with Expected Value?

Here is how to work through a typical Expected Value (E(X) = Σ xᵢ·pᵢ) task step by step:

Task

A wheel pays 10 € with $p = 0.2$ , otherwise nothing. Stake 3 €. Is the game worth it?

Solution path

E(payout) = 10·0.2 = 2 €.
Winnings G: $E(G) = 2 - 3 = -1$ €.
On average you lose 1 € per game, the game is not worth it.

Task

X takes the values 0, 1, 2 with $p =\frac{\frac{ 0.3}{0.5}}{0.2}$ . Compute E(X).

Solution path

$E(X) = 0\cdot{}0.3 + 1\cdot{}0.5 + 2\cdot{}0.2 = 0.9$ .

Frequently asked questions about Expected Value

How do I calculate the expected value of a random variable?+

What does the expected value mean intuitively?+

When is a game fair and how do I check it?+

Why does

E(X) = n\cdot{}p

simply hold for the binomial distribution?+

What is the difference between expected value and mean?+

Retain Expected Value for exams

Create a curated FSRS exam set for E(X) = Σ xᵢ·pᵢ: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential