Expected value, variance and covariance: where a distribution lies, how wide it is and what goes with what

The expected value as a centre of mass, linearity, variance and standard deviation, Chebyshev's inequality, covariance and correlation — and why zero covariance does not yet mean independence.

14 minutes Matěj Žalmánek 2026-10-15

Series AI basics Part 16 of 52

Full series syllabus 52 of 83 topics published

What you will take away

What the expected value is, why it is the centre of mass of a distribution and why expected values always add up.
What variance and standard deviation measure, when variances add up and what Chebyshev's inequality guarantees.
What covariance and correlation are, how they show up in data and why zero covariance does not mean independence.

Useful to know first: Probability density and probability mass functions: why a continuous variable has zero probability at a point

The distribution of a random variable tells you everything — but “everything” is often too much. To compare two distributions quickly, a few numbers are enough: where the distribution lies, how wide it is and whether two variables grow together or pull against each other. These numbers are called the expected value, the variance and the covariance, and you will meet them in every machine learning model — from the loss function to the Adam optimizer.

This part builds on Random variable and Probability density function.

1. Expected value

Definition — expected value of a discrete variable

Let X be a discrete random variable, H = {x₁, x₂, …} ⊆ ℝ an at most countable set of pairwise distinct values with P(X ∈ H) = 1 and p : ℝ → [0, 1] the probability mass function of X (see the parts Random variable and Probability density function). We say that X has a finite expected value if and only if ∑_i |x_i| · p(x_i) < ∞ (the series converges absolutely). Then the expected value of X is the number

E[X] = ∑_i x_i · p(x_i).

Absolute convergence guarantees that the sum does not depend on the order of the values x_i. It is also written μ or μ_X. The expected value is thus the average of the values weighted by their probabilities.

In plain words

A weighted average of the possible values, with probabilities as the weights. For a single die, 1 · 1/6 + 2 · 1/6 + … + 6 · 1/6 = 1 + 2 + … + 6/6 = 3.5 (colours as in the formula below). The expected value does not have to be a value that can actually come up — a die never shows 3.5. It is “the average you end up with in the long run”.

E[X] = ∑_i x_i · p(x_i)

E[X]: the expected value — a single number meant to stand in for the whole distribution: where it lies “on average”
∑_i: over all possible values — none is left out, each one contributes to the average
x_i: the value — the thing being averaged
p(x_i): the weight of the value — more frequent values should pull the average more; the weights add up to 1, so there is nothing to divide by at the end

In N trials, x_i comes up roughly N · p(x_i) times; the ordinary average of the results is then ∑ x_i · Np(x_i)/N — exactly this formula.

Physically, the expected value is the centre of mass. Put the probabilities on a lever as weights: a pivot keeps it level only when it sits at the expected value. The picture shows the larger of the two numbers on a pair of dice — the maximum. Large numbers are more common (six is the maximum in 11 of 36 cases, one in just a single case), so the centre of mass lies to the right of the middle: E[max] = 161/36 ≈ 4.47.

The expected value is the centre of mass: the lever balances only at 4.47

Example — the expected value of the maximum of two dice and the balance of the lever

The maximum is at most k exactly when both dice are at most k: that is k² pairs. So it is exactly k in k² − (k − 1)² = 2k − 1 pairs.

p(1), …, p(6) = 1/36, 3/36, 5/36, 7/36, 9/36, 11/36

E[max] = 1 · 1/36 + 2 · 3/36 + 3 · 5/36 + 4 · 7/36 + 5 · 9/36 + 6 · 11/36 = 1 + 6 + 15 + 28 + 45 + 66/36 = 161/36 ≈ 4.47

(colours as in the formula above)

The lever: the moment about a point c is ∑ (k − c) · p(k) = E[max] − c, and it is zero exactly for c = E[max].

Definition — expected value of a continuous variable

Let X be a random variable with density f : ℝ → [0, ∞). We say that X has a finite expected value if and only if ∫_−∞^∞ |x| · f(x) dx < ∞ (the integral converges absolutely). Then the expected value of X is the number

E[X] = ∫_−∞^∞ x · f(x) dx.

It is the same weighted average as in the discrete case, except that the sum over values is replaced by an integral and the probability p(x_i) by the “probability on a stretch” f(x) dx.

In plain words

The same thing, except that instead of summing over points you integrate over the area under the density. For the e-mail wait from the previous part (density 0.5 · e^−0.5x), E[X] = 2 minutes — exactly the average we built into the example.

Example — the expected waiting time for an e-mail

We compute the integral of the product by integration by parts, ∫ u v′ = [u v] − ∫ u′ v, with u = x and v′ = λ e^−λx, i.e. v = −e^−λx:

E[X] = ∫₀^∞ x · λ e^−λx dx = [−x e^−λx]₀^∞ + ∫₀^∞ e^−λx dx

= (0 − 0) + [−e^−λx/λ]₀^∞ = 0 + 1/λ = 1/λ

λ = 0.5 → E[X] = 1/0.5 = 2 minutes

The term x e^−λx goes to zero as x → ∞, because the exponential falls faster than x grows.

2. Linearity of expectation

Theorem — linearity of expectation

Let (Ω, 𝓕, P) be a probability space, X, Y : Ω → ℝ random variables with finite expected values and a, b ∈ ℝ. Then the variables aX + b and X + Y (defined outcome by outcome: (X + Y)(ω) = X(ω) + Y(ω)) have finite expected values and

E[aX + b] = a · E[X] + b, E[X + Y] = E[X] + E[Y],

and this holds even when X and Y are not independent.

Proof (for a finite sample space Ω with 𝓕 = 2^Ω, such as two dice). Write P(ω) = P({ω}). The expected value can be summed over outcomes instead of over values: E[X] = ∑_{ω ∈ Ω} X(ω) P(ω) (the outcomes with the same value X(ω) = x together give x · p(x)). Then

E[X + Y] = ∑_ω (X(ω) + Y(ω)) P(ω) = ∑_ω X(ω) P(ω) + ∑_ω Y(ω) P(ω) = E[X] + E[Y].

Nowhere did we need to know how X and Y are related — both are evaluated at the same outcome ω. The equality for aX + b follows in the same way from ∑ P(ω) = 1. The general case is proved by the same argument with integrals. ∎

In plain words

The average of a sum is the sum of the averages. The sum of two dice has expected value 3.5 + 3.5 = 7 — no table needed. And if you roll just once and double the result, you also get 2 · 3.5 = 7. Linearity is the most useful property in all of probability: break a complicated variable into simple pieces and add up their averages.

Example — how many sixes come up on two dice

For an event A ∈ 𝓕, the indicator of A is the random variable 𝟙_A : Ω → {0, 1}, 𝟙_A(ω) = 1 for ω ∈ A and 𝟙_A(ω) = 0 for ω ∉ A. We split the number of sixes N into two indicators: I₁ = 𝟙_A₁ for A₁ = “a six on the first die” and I₂ = 𝟙_A₂ for A₂ = “a six on the second die”. Then N = I₁ + I₂.

E[I₁] = 1 · 1/6 + 0 · 5/6 = 1/6, E[I₂] = 1/6 (colours as in the formula for the expected value above)

E[N] = E[I₁] + E[I₂] = 1/6 + 1/6 = 1/3

Checking by enumeration: two sixes in 1 of 36 cases, one in 10 of 36 → 2 · 1 + 1 · 10/36 = 12/36 = 1/3

The expected value of an indicator is the probability of its event — thanks to that, linearity lets you compute the average count of anything.

3. Variance and standard deviation

Definition — variance

Let X be a random variable with finite expected value μ = E[X]. If the random variable (X − μ)² has a finite expected value (which happens if and only if E[X²] < ∞), the variance of X is the number

Var(X) = E[(X − μ)²]

and we say that X has a finite variance. For a discrete X, E[(X − μ)²] = ∑_i (x_i − μ)² · p(x_i); for X with density f it is ∫_−∞^∞ (x − μ)² · f(x) dx. The variance is thus the expected value of the squared deviation from the expected value; always Var(X) ≥ 0.

In plain words

The average squared distance from the average. When values land close to the expected value, the variance is small; when they fly far away, it is large. Squaring makes sure that deviations up and down do not cancel out, and it punishes large deviations more than small ones.

Var(X) = E[(X − μ)²]

Var(X): the variance — a single number for the width of the distribution
X − μ: the deviation from the centre — we measure the width from the expected value, so shifting the whole distribution does not change the variance
²: the square — without it, deviations up and down would cancel out and the result would always be zero (E[X − μ] = μ − μ = 0); moreover, it counts one large deviation more than two half-size ones
E: the average over the whole distribution — frequent deviations weigh more than rare ones, just as with the expected value

Why the square and not the absolute value? The square can be differentiated and expanded, and thanks to that the rules below hold — the variances of independent variables add up.

Theorem — computational formula for variance

Let X be a random variable with finite variance. Then

Var(X) = E[X²] − (E[X])².

Proof. Write μ = E[X]. We expand and use linearity, μ being a constant: E[(X − μ)²] = E[X² − 2μX + μ²] = E[X²] − 2μ · E[X] + μ² = E[X²] − 2μ² + μ² = E[X²] − μ². ∎

In plain words

“The mean of the squares minus the square of the mean.” For a single die, E[X²] = 1 + 4 + 9 + 16 + 25 + 36/6 = 91/6, so Var(X) = 91/6 − 3.5² = 35/12 ≈ 2.92. It follows from linearity: just expand (X − μ)².

Definition — standard deviation

Let X be a random variable with finite variance Var(X) ≥ 0. The standard deviation of X is the non-negative number

σ = σ_X = √Var(X).

Thus σ ≥ 0 and σ² = Var(X); σ has the same units as X.

In plain words

Variance comes in odd units — for a wait measured in minutes, it is in minutes squared. The square root brings it back: σ is “the typical distance from the average” in the same units as the data. For a single die σ ≈ 1.71; for the e-mail wait the variance is 4 and σ = 2 minutes.

From data instead of from a distribution

Often we do not know the distribution, all we have are the measured numbers x₁, …, x_n. Then each of them gets the weight 1/n — as if it were a distribution in which every measurement is equally likely:

def prumer_a_rozptyl(data):
    n = len(data)
    mu = sum(data) / n                          # E[X] with weights 1/n
    var = sum((x - mu) ** 2 for x in data) / n  # E[(X − μ)²] with weights 1/n
    return mu, var, var ** 0.5                  # μ, variance, σ

Example — the mean and variance of eight numbers

data 2, 4, 4, 4, 5, 5, 7, 9; μ = 40/8 = 5

deviations x_i − μ: −3, −1, −1, −1, 0, 0, 2, 4; squares 9, 1, 1, 1, 0, 0, 4, 16

Var = 9 + 1 + 1 + 1 + 0 + 0 + 4 + 16/8 = 32/8 = 4, σ = √4 = 2

(colours as in the breakdown of the variance above: the average E with weights 1/8 is division by eight)

When the data are only a sample from a larger whole and we estimate the variance of the whole, one often divides by n − 1 instead of n (the so-called sample variance, here 32/7 ≈ 4.57): we measure the deviations from the sample mean, which fits the data better than the true μ, and dividing by n would underestimate the variance on average.

4. Variance of a sum: why rolling twice pays off

Theorem — variance of a linear function and of a sum

Let X, Y be random variables on the same probability space with finite variance and a, b ∈ ℝ. Then

Var(aX + b) = a² · Var(X);
if X and Y are independent (defined in the section Covariance and correlation), then Var(X + Y) = Var(X) + Var(Y).

Proof of the first part. By linearity E[aX + b] = aμ + b, so the deviation from the centre is (aX + b) − (aμ + b) = a(X − μ) — the b cancels. Then Var(aX + b) = E[a²(X − μ)²] = a² Var(X). We prove the second part in the section on covariance. ∎

In plain words

Shifting by b does not change the width; stretching a times makes it a² times larger. And the variances of independent variables add up — unlike the standard deviations.

Let us compare two games with the same expected value of 7. In the first, you roll two dice and add them up: Var = 35/12 + 35/12 = 35/6 ≈ 5.83, σ ≈ 2.42. In the second, you roll once and double the result: Var = 2² · 35/12 = 35/3 ≈ 11.67, σ ≈ 3.42. The average is the same, but the second game is much wilder — with two dice, bad luck on one is often offset by good luck on the other.

Same average, different variance: the green segment is μ ± σ

The same principle is behind diversifying investments and behind the fact that an average of several measurements fluctuates less than a single measurement. Exactly how much less is the subject of the next part, on the law of large numbers.

5. Chebyshev's inequality

Theorem — Chebyshev's inequality

Let X be a random variable with finite variance, μ = E[X] and σ = √Var(X) > 0. Then for every k ∈ ℝ, k > 0,

P(|X − μ| ≥ kσ) ≤ 1/k²,

where {|X − μ| ≥ kσ} = {ω ∈ Ω : |X(ω) − μ| ≥ kσ}.

In plain words

There cannot be much probability far from the average — and this holds for every distribution. At most a quarter of the values land more than two standard deviations from the average, at most a ninth more than three. For the sum of two dice, 2σ ≈ 4.83, and only the sums 2 and 12 are that far from seven: in fact 2/36 ≈ 0.056, well below the guaranteed bound of 1/2² = 0.25 (colours as in the formula below).

P(|X − μ| ≥ kσ) ≤ 1/k²

P: the probability of a “runaway” value — this is what we want to bound from above without knowing the shape of the distribution
|X − μ|: the distance from the expected value — an absolute value, because we care about runaways in both directions
kσ: a threshold measured in standard deviations — that way one inequality holds for all distributions and all units; “far” means far relative to the typical distance
1/k²: the bound — the square comes from the variance (the average square of the deviation): every value at least kσ from the centre contributes at least k²σ² times its probability to the variance; if together they had more than 1/k², the variance would come out larger than σ²

For k ≤ 1 the bound is ≥ 1 and says nothing; it is only useful for k > 1.

Theorem — Markov's inequality and the proof of Chebyshev's

Markov's inequality: Let Y be a random variable with finite expected value that is non-negative (Y(ω) ≥ 0 for every ω ∈ Ω), and a ∈ ℝ, a > 0. Then

P(Y ≥ a) ≤ E[Y]/a.

Proof. Always Y ≥ a · I, where I = 𝟙_{{Y ≥ a}} is the indicator of the event {Y ≥ a} (1 when Y ≥ a, otherwise 0) (for Y ≥ a this reads Y ≥ a, otherwise Y ≥ 0). The expected value preserves the inequality: E[Y] ≥ a · E[I] = a · P(Y ≥ a).

Chebyshev. Apply Markov to Y = (X − μ)² ≥ 0 and a = k²σ² > 0:

P(|X − μ| ≥ kσ) = P((X − μ)² ≥ k²σ²) ≤ E[(X − μ)²]/k²σ² = σ²/k²σ² = 1/k². ∎

How loose the bound is can be seen on the sum of two dice (σ = √35/6 ≈ 2.42) for several values of k (colours as in the formula above):

`k`	threshold `k`σ	sums that far from 7	true `P`	bound 1/`k`²
1	2.42	2, 3, 4, 10, 11, 12	12/36 ≈ 0.333	1
1.5	3.62	2, 3, 11, 12	6/36 ≈ 0.167	0.444
2	4.83	2, 12	2/36 ≈ 0.056	0.25
3	7.25	none	0	0.111

For the continuous e-mail wait (μ = σ = 2 minutes), the threshold for k = 2 is 4 minutes; downwards the wait cannot get that far, upwards it means waiting at least 6 minutes: P(X ≥ 6) = e⁻³ ≈ 0.050, again well below 0.25. The bound is crude because it has to hold even for the worst possible distribution.

6. Covariance and correlation

Definition — covariance

Let X, Y : Ω → ℝ be random variables on the same probability space (Ω, 𝓕, P) with finite variance and μ_X = E[X], μ_Y = E[Y]. The covariance of X and Y is the number

Cov(X, Y) = E[(X − μ_X)(Y − μ_Y)].

This expected value is finite, because |(X − μ_X)(Y − μ_Y)| ≤ (X − μ_X)² + (Y − μ_Y)²/2. The covariance is thus the expected value of the product of the deviations of both variables from their expected values.

In plain words

When X is above its average, is Y usually above its average too? Then the product of the deviations is mostly positive and so is the covariance. When they pull against each other, it is negative. In the picture, each product is a rectangle from the point where the averages cross to the data point: green ones are positive, orange ones negative. The covariance is their average signed area.

Definition — correlation coefficient

Let X, Y be random variables on the same probability space with finite variance and standard deviations σ_X > 0, σ_Y > 0. The correlation coefficient (correlation) of X and Y is the number

ρ(X, Y) = Cov(X, Y)/σ_X σ_Y.

Always −1 ≤ ρ(X, Y) ≤ 1 (this follows from the Cauchy–Schwarz inequality |Cov(X, Y)| ≤ σ_X σ_Y).

In plain words

Covariance depends on the units — in centimetres it comes out 10,000× larger than in metres. Correlation divides it by both standard deviations and gets a unitless number: 1 means the points lie on a rising line, −1 on a falling one, and 0 means no linear relationship.

Covariance as an average of rectangles: positive, zero, negative

The first die and the sum grow together: Cov = 35/12 ≈ 2.92 and ρ ≈ 0.71. The first and second die are independent, and the rectangles cancel out exactly: Cov = 0. And the higher the first die, the smaller the difference “second minus first”: Cov ≈ −2.92, ρ ≈ −0.71.

Theorem — computational formula for covariance

Let X, Y be random variables on the same probability space with finite variance. Then

Cov(X, Y) = E[XY] − E[X] · E[Y].

Proof. Just as with the variance, we expand: E[XY − μ_YX − μ_XY + μ_Xμ_Y] = E[XY] − μ_Yμ_X − μ_Xμ_Y + μ_Xμ_Y. For Y = X this is the computational formula for variance: Cov(X, X) = Var(X). ∎

Example — the first die and the sum

X = first die, Y = second, S = X + Y; E[XS] = E[X²] + E[XY] = 91/6 + 3.5 · 3.5 = 91/6 + 49/4 = 329/12

Cov(X, S) = 329/12 − 3.5 · 7 = 329/12 − 294/12 = 35/12 ≈ 2.92

ρ = 35/12/√35/12 · √35/6 = √35/12/√35/6 = √1/2 ≈ 0.71

We were allowed to use E[XY] = E[X] · E[Y] because the dice are independent (theorem below).

Definition — independent random variables

Let X, Y : Ω → ℝ be random variables on the same probability space (Ω, 𝓕, P) with distribution functions F_X, F_Y. X and Y are independent if and only if

∀x, y ∈ ℝ : P(X ≤ x, Y ≤ y) = F_X(x) · F_Y(y),

where {X ≤ x, Y ≤ y} = {X ≤ x} ∩ {Y ≤ y}; the events {X ≤ x} and {Y ≤ y} are thus independent for every pair x, y (for independent events see the part Conditional probability and Bayes’ theorem). For discrete X and Y this is equivalent to the condition

∀x, y ∈ ℝ : P(X = x, Y = y) = P(X = x) · P(Y = y).

In plain words

Learning the value of one variable changes nothing about the chances of the other. For two dice: P(first 6, second 6) = 1/36 = 1/6 · 1/6. For the sum and the difference (first − second) this does not hold: a sum of 12 together with a difference of 1 cannot come up, the probability is 0, but P(sum 12) · P(difference 1) = 1/36 · 5/36 ≠ 0.

Theorem — variance of a sum in general

Let X, Y be random variables on the same probability space with finite variance. Then

Var(X + Y) = Var(X) + Var(Y) + 2 · Cov(X, Y).

If moreover X and Y are independent, then E[XY] = E[X] · E[Y], hence Cov(X, Y) = 0 and Var(X + Y) = Var(X) + Var(Y).

Proof. The deviation of the sum from its centre is (X − μ_X) + (Y − μ_Y); its square is (X − μ_X)² + (Y − μ_Y)² + 2(X − μ_X)(Y − μ_Y), and taking the expected value by linearity gives the first equality. For independent discrete variables E[XY] = ∑_x,y xy · P(X = x) P(Y = y) = (∑_x x P(X = x)) · (∑_y y P(Y = y)) = E[X] · E[Y], so Cov = 0; for variables with a density one proceeds in the same way with integrals. ∎

Var(X + Y) = Var(X) + Var(Y) + 2 · Cov(X, Y)

Var(X + Y): how much the sum fluctuates — say a whole portfolio, or the sum of two measurement errors
Var(X) + Var(Y): the fluctuation of each part on its own — if they were unrelated, this would be the whole result
Cov(X, Y): the interplay — when both variables go up together, the deviations add up and the sum fluctuates more (positive); when they pull against each other, the deviations cancel (negative)
2: the cross term 2ab from (a + b)² — the pairs (X, Y) and (Y, X) contribute equally

The sum of two dice: 35/12 + 35/12 + 2 · 0 = 35/6. One die × 2 is “the sum of a die with itself”: 35/12 + 35/12 + 2 · 35/12 = 35/3.

In plain words

Variables that grow together increase the variance of the sum; variables that pull against each other reduce it — which is exactly what hedging does in investing. For independent variables the covariance term vanishes, leaving the rule from the previous section.

Careful, the converse does not hold: zero covariance does not mean independence. The difference of the dice D = first − second and their sum S have covariance Var(first) − Var(second) = 0. Yet they are dependent: if the sum is 12, the difference must be 0. Covariance measures only a linear relationship.

Theorem — covariance is bilinear

Let X, Y, Z be random variables on the same probability space with finite variance and a, b ∈ ℝ. Then

Cov(X, Y) = Cov(Y, X) (symmetry);
Cov(X, X) = Var(X);
Cov(aX + bY, Z) = a · Cov(X, Z) + b · Cov(Y, Z) (linearity in the first argument; with symmetry also in the second).

It follows from the linearity of expectation, just as for the variance.

Example — why the difference and the sum have zero covariance

Cov(X − Y, X + Y) = Cov(X, X) + Cov(X, Y) − Cov(Y, X) − Cov(Y, Y)

= Var(X) − Var(Y) = 35/12 − 35/12 = 0

The cross terms cancel thanks to symmetry, leaving the difference of the variances — zero, because both dice are the same. This changes nothing about the dependence.

7. Expected value and variance in artificial intelligence

Loss function. The mean squared error (MSE) is the expected value of the squared error, E[(Y − ŷ)²]. The best constant prediction is exactly the expected value of Y, and the error that remains is the variance.
Input standardization. Before data go into a network, they are converted to z = x − μ/σ — expected value 0, standard deviation 1. Training then runs faster and more stably; batch normalization does the same inside the network.
Covariance matrix and PCA. The covariances of all pairs of features form a matrix, and its eigenvectors (see the part on linear algebra) point in the directions in which the data vary the most.
The Adam optimizer (from the part on gradient descent) keeps running estimates of the expected value of the gradient and of its square — and divides the step by the square root of the latter. Where the gradient fluctuates a lot, it takes more cautious steps.
Bias and variance of a model. A model's error can be split into a systematic error (bias) and fluctuation between training sets (variance). Simple models have high bias, complex ones high variance.

Why the best constant prediction is the expected value

Theorem — decomposition of the mean squared error

Let Y be a random variable with finite variance and c ∈ ℝ any number. Then

E[(Y − c)²] = Var(Y) + (E[Y] − c)².

The function c ↦ E[(Y − c)²] therefore has a unique minimum at the point c = E[Y], and its value there is Var(Y).

Proof. Var(Y − c) = Var(Y) (a shift does not change the width), and the computational formula for the variance of Y − c gives E[(Y − c)²] = Var(Y − c) + (E[Y − c])² = Var(Y) + (E[Y] − c)². ∎

E[(Y − c)²] = Var(Y) + (E[Y] − c)²

E[(Y − c)²]: the mean squared error (MSE) of the prediction c — the loss the model minimises
Var(Y): the irreducible error — c does not appear in it at all, so no constant prediction can reduce it; it is the noise of the data themselves
(E[Y] − c)²: the bias of the prediction, squared — the only thing the choice of c affects; it is zero exactly for c = E[Y]

Hence the claim from the first bullet: the MSE is smallest for c = E[Y], and the variance is what remains. The same decomposition in a more general form is the bias–variance decomposition of a model.

Example — predicting a die roll and standardization

Prediction c = 3: E[(Y − 3)²] = 4 + 1 + 0 + 1 + 4 + 9/6 = 19/6 ≈ 3.17

By the decomposition: 35/12 + (3.5 − 3)² = 35/12 + 1/4 = 38/12 = 19/6 ✓; for c = 3.5 only 35/12 + 0 ≈ 2.92 (colours as in the formula above)

Standardizing a six: z = 6 − 3.5/1.71 ≈ 1.46 — a six lies 1.46 standard deviations above the average

8. Summary

The expected value is a weighted average — the centre of mass of the distribution. It is linear, even for dependent variables.
Variance is the average squared deviation from the average; the standard deviation is its square root, in the same units as the data.
The variances of independent variables add up; that is why the sum of two dice fluctuates less than one die × 2.
Chebyshev's inequality: at most 1/k² of the probability lies farther than kσ from the average.
Covariance and correlation measure the linear relationship between two variables; zero covariance does not yet mean independence.

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Expected value, variance and covariance: where a distribution lies, how wide it is and what goes with what

1. Expected value

2. Linearity of expectation

3. Variance and standard deviation

From data instead of from a distribution

4. Variance of a sum: why rolling twice pays off

5. Chebyshev's inequality

6. Covariance and correlation

7. Expected value and variance in artificial intelligence

Why the best constant prediction is the expected value

8. Summary

An Agent on your computer, driven from your phone