Expected value, variance, standard deviation — properties: Worked examples

This page is a drill through every case the properties of expected value, variance and standard deviation can throw at you. Before we solve anything, we lay out a map of all the case classes so you can see that nothing is left out. Then every worked example is tagged with the map cell it covers.

If a symbol here is unfamiliar, it was built in the parent note — but we re-anchor each one as we use it, from zero.

The scenario matrix

Read this as: "what could possibly change from problem to problem?" Each row is a distinct kind of situation. Our examples below are chosen so that every row gets hit at least once.

#	Case class	What is tricky about it	Covered by
A	Positive scale, shift ( $a X + b$ , $a > 0$ )	shift must vanish, scale squares	Ex 1
B	Negative scale ( $a < 0$ )	SD uses $∣ a ∣$ ; sign traps	Ex 2
C	Zero scale / degenerate constant ( $a = 0$ )	variance collapses to $0$	Ex 3
D	Sum of independent variables	covariance term dies	Ex 4
E	Sum of dependent variables (incl. $X + X$ )	must keep $2 Cov$	Ex 5
F	Limiting behaviour (Bernoulli spread vs $p$ )	where is spread biggest/zero?	Ex 6
G	Real-world word problem (units, money)	translate words → symbols	Ex 7
H	Exam twist — average of $n$ i.i.d. (shrinking spread)	connects to big theorems	Ex 8

We will lean on three tools. Here they are in one place, each with the question it answers:

Here $E [X]$ means the average value (balance point) of $X$ ; $Var (X)$ is the average squared distance from that balance point; $SD (X) = Var (X)$ brings us back to ordinary units; and $Cov (X, Y) = E [X Y] - E [X] E [Y]$ measures how two variables move together.

Our running character is the fair die: $X \in {1, 2, 3, 4, 5, 6}$ , each with probability $\frac{1}{6}$ . From the parent note we already know its two summaries:

$E [X] = 3.5, Var (X) = \frac{35}{12} \approx 2.9167, SD (X) \approx 1.708.$

The picture below is the whole idea of stretch-and-slide on a mass distribution — keep it in your mind for the first three examples.

Figure — Expected value, variance, standard deviation — properties

Example 1 — Cell A: positive scale + shift

Worked example Convert die pips to a score

A game turns each die roll into a score $Y = 10 X + 5$ . Find $E [Y]$ , $Var (Y)$ , $SD (Y)$ .

Forecast: guess before reading — does the " $+ 5$ " change the spread? Does the " $10$ " multiply the variance by $10$ or by something else?

Step 1. $E [Y] = 10 E [X] + 5 = 10 (3.5) + 5 = 40$ . Why this step? Mean is linear — a stretch scales the centre, a slide moves it. Both survive.

Step 2. $Var (Y) = 1 0^{2} Var (X) = 100 \cdot \frac{35}{12} = \frac{3500}{12} \approx 291.67$ . Why this step? The $+ 5$ slides the whole mass rigidly — distances-to-centre are unchanged, so it vanishes from variance. The $10$ stretches every distance by $10$ ; squared distances by $1 0^{2}$ .

Step 3. $SD (Y) = ∣ 10 ∣ \cdot 1.708 = 17.08$ . Why this step? SD is a distance, so it scales by the raw factor (absolute value), not its square.

Verify: $291.67 \approx 17.08$ ✓, matching Step 3. Units check: if pips scale by $10$ , spread in "score units" scales by $10$ — consistent.

Example 2 — Cell B: negative scale (the sign trap)

Worked example Flip the die

Let $Z = - 2 X + 1$ . Find $E [Z]$ , $Var (Z)$ , $SD (Z)$ .

Forecast: the $- 2$ flips the distribution left-to-right. Does variance become negative? Does SD become $- 2 \cdot 1.708$ ?

Step 1. $E [Z] = - 2 (3.5) + 1 = - 6$ . Why this step? Linearity of the mean holds for any sign of $a$ ; the flip drags the centre to the negative side.

Step 2. $Var (Z) = (- 2)^{2} Var (X) = 4 \cdot \frac{35}{12} = \frac{140}{12} = \frac{35}{3} \approx 11.667$ . Why this step? $a^{2} = (- 2)^{2} = 4$ — squaring erases the sign. Variance can never be negative; a mirror flip does not change how scattered the points are.

Step 3. $SD (Z) = ∣ - 2 ∣ \cdot 1.708 = 2 \cdot 1.708 = 3.416$ . Why this step? SD uses $∣ a ∣$ . Writing $- 2 \cdot 1.708$ would give a negative "distance", which is impossible.

Verify: $35/3 \approx 3.416$ ✓. Sanity: reflecting a cloud of darts about a point leaves the cloud equally wide.

Example 3 — Cell C: zero scale / degenerate constant

Example 4 — Cell D: sum of independent variables

Worked example Two independent dice

Roll two independent fair dice $X_{1}, X_{2}$ and let $S = X_{1} + X_{2}$ . Find $E [S]$ , $Var (S)$ , $SD (S)$ .

Forecast: the mean surely doubles to $7$ . Does the variance also just double?

Step 1. $E [S] = E [X_{1}] + E [X_{2}] = 3.5 + 3.5 = 7$ . Why this step? Linearity of expectation holds always, independent or not.

Step 2. Because the dice are independent, $Cov (X_{1}, X_{2}) = 0$ , so $Var (S) = Var (X_{1}) + Var (X_{2}) = \frac{35}{12} + \frac{35}{12} = \frac{70}{12} = \frac{35}{6} \approx 5.833.$ Why this step? One die tells you nothing about the other, so there is no "moving together" term to add. See Covariance and Correlation for why independence forces $Cov = 0$ .

Step 3. $SD (S) = 35/6 \approx 2.415$ . Why this step? Root returns to pip units.

Verify: $2 \times 2.9167 = 5.833$ ✓. Note $SD (S) \approx 2.415$ is less than $2 \times 1.708 = 3.416$ : independent spreads add in quadrature (like Pythagoras), not linearly. This is the seed of the Law of Large Numbers.

Example 5 — Cell E: dependent sum, including $X + X$

Worked example The self-sum trap

For one die $X$ , compare $Var (X + X)$ with the "independent guess" $2 Var (X)$ .

Forecast: $X + X$ and $2 X$ are the same thing. Does the sum rule still give $2 Var (X)$ ?

Step 1. $X + X = 2 X$ , so by the scale rule $Var (2 X) = 2^{2} Var (X) = 4 \cdot \frac{35}{12} = \frac{35}{3} \approx 11.667$ . Why this step? This is stretch-by- $2$ , cell B/A machinery — no independence assumed.

Step 2. Now via the sum rule with $Y = X$ : $Cov (X, X) = Var (X)$ (a variable is perfectly correlated with itself). So $Var (X + X) = Var (X) + Var (X) + 2 Var (X) = 4 Var (X) .$ Why this step? $Cov (X, X) = E [X \cdot X] - E [X] E [X] = E [X^{2}] - μ^{2} = Var (X)$ — that IS the definition of variance.

Step 3. Both routes agree: $4 \cdot \frac{35}{12} = \frac{35}{3} \approx 11.667$ , not $2 \cdot \frac{35}{12} \approx 5.833$ . Why this step? The naive "just add variances" answer is wrong by exactly the $2 Cov$ term.

Verify: $35/3 \approx 11.667$ from Step 1 equals $4 \times 2.9167$ ✓. The dependent answer is double the independent answer of Ex 4 — dependence matters enormously.

Example 6 — Cell F: limiting behaviour (where is spread maximal?)

Worked example Bernoulli spread as

$p$ varies An indicator $X$ is $1$ with probability $p$ , else $0$ . Its variance is $Var (X) = p (1 - p)$ . For which $p$ is spread largest? What happens at the limits $p \to 0$ and $p \to 1$ ?

Forecast: a coin is "most uncertain" at $p = \frac{1}{2}$ . Do you expect variance to peak there and vanish at the ends?

Step 1. Write $v (p) = p (1 - p) = p - p^{2}$ . Why this step? We turn "spread vs $p$ " into a plain function of one variable so we can find its top.

Step 2. The top is where the slope is zero: $v^{'} (p) = 1 - 2 p = 0 \Rightarrow p = \frac{1}{2}$ , giving $v (\frac{1}{2}) = \frac{1}{4}$ . Why the derivative? $v^{'} (p)$ answers "is spread still rising as I nudge $p$ ?" It crosses zero exactly at the peak.

Step 3. Limits: $v (0) = 0$ and $v (1) = 0$ . Why this step? At $p = 0$ (never happens) or $p = 1$ (always happens) the outcome is a constant — that is the degenerate zero-variance case from Ex 3.

Verify: $v (0.5) = 0.25$ ; $v (0.1) = 0.09$ ; $v (0) = v (1) = 0$ ✓. The curve is an upside-down parabola peaking at the fair coin — see Bernoulli and Binomial Distributions.

Example 7 — Cell G: real-world word problem (money & units)

Worked example Daily profit

A stall's daily demand $D$ (units sold) has $E [D] = 40$ and $SD (D) = 6$ . Each unit gives $3 profit but there is a fixed $20 daily rent. Profit is $P = 3 D - 20$ . Find the mean profit and its standard deviation.

Forecast: does the $20 rent affect how variable profit is day to day?

Step 1. Translate words to symbols: $P = 3 D - 20$ , so $a = 3$ (dollars per unit), $b = - 20$ (fixed rent). Why this step? Every "per-unit price" is a scale $a$ ; every "fixed cost" is a shift $b$ .

Step 2. $E [P] = 3 E [D] - 20 = 3 (40) - 20 = 100$ dollars. Why this step? Linearity of the mean.

Step 3. $Var (D) = SD (D)^{2} = 6^{2} = 36$ , so $Var (P) = 3^{2} \cdot 36 = 324$ , and $SD (P) = ∣ 3 ∣ \cdot 6 = 18$ dollars. Why this step? Fixed rent slides profit down but does not change day-to-day scatter, so $b = - 20$ drops out. The $3/unit stretches the scatter by $3$ .

Verify: $324 = 18$ ✓. Units: SD of $P$ is in dollars ( $=3\ \tfrac{\$ }{\text{unit}}\times 6\ \text{units} $) ✓. Mean \$ 100, typical swing $18.

Example 8 — Cell H: exam twist (average of $n$ i.i.d.)

Worked example Why averaging tames randomness

Take $n$ independent fair dice and form the sample mean $\overset{ˉ}{X}_{n} = \frac{X _{1} + \dots + X _{n}}{n}$ . Find $E [\overset{ˉ}{X}_{n}]$ and $Var (\overset{ˉ}{X}_{n})$ . What happens as $n \to \infty$ ?

Forecast: the average is still "around $3.5$ ". Does its spread stay the same, grow, or shrink as $n$ grows?

Step 1. $E [\overset{ˉ}{X}_{n}] = \frac{1}{n} \sum_{i = 1}^{n} E [X_{i}] = \frac{1}{n} \cdot n \cdot 3.5 = 3.5$ . Why this step? Linearity: the mean of an average of same-mean variables is that common mean, for any $n$ .

Step 2. Independence lets variances add: $Var (\sum_{i} X_{i}) = n Var (X)$ . Then the $\frac{1}{n}$ is a scale, squared: $Var (\overset{ˉ}{X}_{n}) = \frac{1}{n ^{2}} \cdot n Var (X) = \frac{Var ( X )}{n} = \frac{35/12}{n} .$ Why this step? Sum rule (cell D) then scale rule (cell A) with $a = \frac{1}{n}$ , so $a^{2} = \frac{1}{n ^{2}}$ .

Step 3. As $n \to \infty$ , $Var (\overset{ˉ}{X}_{n}) = \frac{35/12}{n} \to 0$ , and $SD (\overset{ˉ}{X}_{n}) = \frac{1.708}{n}$ shrinks like $1/ n$ . Why this step? Dividing by $n$ drives the spread to zero — the average concentrates on $3.5$ .

Verify (numeric): for $n = 100$ , $Var (\overset{ˉ}{X}_{100}) = \frac{35/12}{100} \approx 0.029167$ and $SD \approx 0.1708$ ✓. The mean is pinned at $3.5$ but the swing is $10 \times$ smaller than one die. This is the mechanism behind the Law of Large Numbers and the $1/ n$ scale in the Central Limit Theorem.

Recall

Recall One-line answers to the whole matrix

Shift $b$ in $Var$ ? ::: Always vanishes — spread is shift-invariant. Scale $a$ in $Var$ ? ::: Multiplies by $a^{2}$ ; in SD by $∣ a ∣$ . $Var$ of a constant? ::: Exactly $0$ (degenerate). $Var (X + Y)$ for independent $X, Y$ ? ::: $Var (X) + Var (Y)$ . $Var (X + X)$ ? ::: $4 Var (X)$ , not $2 Var (X)$ — $Cov (X, X) = Var (X)$ . Where is Bernoulli spread largest? ::: At $p = \frac{1}{2}$ , value $\frac{1}{4}$ ; zero at $p = 0, 1$ . $Var$ of the average of $n$ i.i.d.? ::: $Var (X) / n \to 0$ ; SD shrinks like $1/ n$ .