Worked examples — Expectation, variance, and standard deviation

This page is a firing range. We take the parent topic's three tools — expectation $E[X]$ (the balance point), variance $\text{Var}(X)$ (average squared spread), and standard deviation $\sigma =\sqrt{\text{Var}(X)}$ (spread in original units) — and we throw every kind of input at them until nothing can surprise you.

Throughout this page we write $\mu$ (Greek letter "mu") as a shorthand for the mean $E[X]$ — it is just a one-symbol name for the balance point, nothing new. If any other symbol feels unfamiliar, it was built from zero in the parent note and in random variables & distributions. Here we only use them, on every case.

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The scenario matrix

Think of this table as a checklist. Every row is a "kind of situation" this topic can hand you. Every worked example below is stamped with the cell(s) it clears.

#	Case class	What makes it tricky	Cleared by
C1	Plain discrete	baseline: sum of value × probability	Ex 1
C2	Negative values in the support	signs cannot cancel in variance	Ex 2
C3	Continuous PDF	integral, not a sum	Ex 3
C4	Degenerate (constant) RV	zero spread — the edge case	Ex 4
C5	Scaling & shifting $aX+b$	$a^2$ for variance, $|a|$ for SD	Ex 5
C6	Sum of independent RVs	variances add, SDs do NOT	Ex 6
C7	Correlated sum (limiting signs)	the $+2\text{Cov}$ term, both signs	Ex 7
C8	Real-world word problem	translate English → $E,\text{Var}$	Ex 8
C9	Exam twist (given $E[X],E[X^2]$ only)	back out variance without the distribution	Ex 9

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Ex 1 — Plain discrete (C1)

Figure s01 (caption): a horizontal number line from 0 to 4 with two cyan mass spikes — a tall one at $x=3$ (height $0.6$) and a shorter one at $x=1$ (height $0.4$); a dotted white line marks the midpoint $2$, and an amber triangular pivot sits at $2.2$ showing the balance point pulled right of the midpoint by the heavier mass.

Step 1 — Expectation. $E[X]=3(0.6)+1(0.4)=1.8+0.4=2.2$. Why this step? Expectation is each value weighted by how likely it is — the balance point. Because heads is more likely, the balance sits nearer 3, matching a sensible forecast. In the figure, the amber pivot marker at $2.2$ is dragged right of the dotted midpoint $2$ by the taller $0.6$ mass at $3$.

Step 2 — Get $E[X^2]$. $E[X^2]=3^2(0.6)+1^2(0.4)=9(0.6)+1(0.4)=5.4+0.4=5.8$. Why this step? Variance's easy form is $E[X^2]-(E[X]{)}^2$; we need the average of the squared payout first.

Step 3 — Variance. $\text{Var}(X)=5.8-(2.2{)}^2=5.8-4.84=0.96$. Why this step? Subtracting $(E[X]{)}^2$ strips out the "center" and leaves pure spread.

Step 4 — SD. $\sigma =\sqrt{0.96}\approx 0.98$. Why this step? Square-root returns us to dollars, the original unit.

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Ex 2 — Negative values (C2)

Figure s02 (caption): a number line from $-5.5$ to $3.5$ with cyan dots at $X=2$ and $X=-4$; a dashed amber vertical line marks the mean $\mu =-1$; two cyan horizontal arrows run from the mean to each dot, labelled $+3$ (rightward to $2$) and $-3$ (leftward to $-4$) — equal in length, opposite in direction.

Step 1 — Expectation. $E[X]=2(0.5)+(-4)(0.5)=1-2=-1$. Why this step? A negative value is just a point to the left of zero on the number line; it enters the weighted sum normally.

Step 2 — Deviations, then square. Deviations from $\mu =-1$: for $X=2$ it is $2-(-1)=3$; for $X=-4$ it is $-4-(-1)=-3$. Why this step? This is exactly why we square: the two deviations $+3$ and $-3$ would cancel if we just averaged them, falsely reporting zero spread. In the figure the two cyan arrows have equal length $3$ but point in opposite directions — plain averaging of $+3$ and $-3$ gives $0$.

Step 3 — Variance. $\text{Var}(X)=0.5(3{)}^2+0.5(-3{)}^2=0.5(9)+0.5(9)=9$. Why this step? Squaring makes both deviations positive, so opposite signs reinforce instead of cancel — variance is always $\ge 0$.

Step 4 — SD. $\sigma =\sqrt{9}=3$. Why this step? Variance $9$ lives in squared dollars; taking the square root returns us to plain dollars so the spread is directly comparable to the payouts $2$ and $-4$.

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Ex 3 — Continuous PDF (C3)

Figure s03 (caption): the triangular density $f(x)=2x$ drawn as a cyan line rising from $(0,0)$ to $(1,2)$ with the area beneath shaded cyan; a dashed amber vertical line at $x=2/3$ marks the mean, and an amber arrow near the base points rightward captioned "more mass on the right."

Step 1 — Sanity: is it a valid PDF? ${\int }_0^{1}2xdx=[x^2{]}_0^1=1$. ✓ Why this step? Total probability must be $1$; if the slivers don't add to $1$, everything downstream is wrong.

Step 2 — Expectation (integral, not sum). $E[X]={\int }_0^{1}x\cdot 2xdx={\int }_0^{1}2x^{2}dx={\left[\frac{2x^3}{3}\right]}_0^1=\frac{2}{3}$. Why the integral? For a continuous variable there is no list of outcomes to sum; each sliver of width $dx$ around a point $x$ carries probability $f(x)dx$, we weight its position $x$ by that probability, and the integral adds all slivers. As forecast, the balance point $2/3>0.5$ because mass leans right — matching the amber mean line in the figure sitting past the midpoint.

Step 3 — Second moment. $E[X^2]={\int }_0^1{x}^2\cdot 2xdx={\int }_0^{1}2x^{3}dx={\left[\frac{x^4}{2}\right]}_0^1=\frac{1}{2}$. Why this step? Variance's easy form $E[X^2]-{\mu }^2$ needs the average of the squared variable, so we weight $x^2$ by the same density $f(x)=2x$ and integrate again — the continuous twin of "sum the squares" from the dice examples.

Step 4 — Variance & SD. $\text{Var}(X)=\frac{1}{2}-{\left(\frac{2}{3}\right)}^2=\frac{1}{2}-\frac{4}{9}=\frac{9-8}{18}=\frac{1}{18}\approx 0.0556$; $\sigma =\sqrt{1/18}\approx 0.236$. Why this step? Subtracting ${\mu }^2$ removes the "center" contribution and leaves pure spread; the final square-root converts squared-units variance back to the plain scale of $x$, so $\sigma \approx 0.236$ is a distance on the same $[0,1]$ axis.

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Ex 4 — Degenerate / constant RV (C4)

Figure s04 (caption): a number line from 3 to 11 carrying a single tall amber spike of height $1$ at $x=7$ (probability $1$) and nothing anywhere else; a cyan annotation points at the spike noting that with all mass on one point there is no distance to average, so $\text{Var}=0$ and $\sigma =0$.

Step 1 — Expectation of a constant. $E[X]=7\cdot 1=7$. Why this step? One outcome with probability $1$: the balance point is that single point.

Step 2 — Variance. $\text{Var}(X)=E[(X-7{)}^2]=(7-7{)}^2\cdot 1=0$. Why this step? Every "observation" sits exactly on the mean, so every squared deviation is $0$. This is the degenerate edge case: zero variance means zero randomness.

Step 3 — SD. $\sigma =\sqrt{0}=0$. Why this step? The square-root of a zero variance is zero, and it must be: with all mass piled on the single amber spike in the figure there is literally no distance to average, so the spread in original units is exactly $0$.

Recall What does

$\sigma =0$ tell you? The value is fixed ::: it never deviates from its mean, so it isn't really random.

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Ex 5 — Scaling and shifting $aX+b$ (C5)

Figure s05 (caption): two stacked number lines. The upper line shows the die faces $1$–$6$ as cyan dots with a dashed amber mean at $3.5$; the lower line shows $Y=2X-1$, whose dots are spaced twice as far apart (a stretch) and whose dashed amber mean sits at $6$ (a slide). A white note reads "ticks spread ×2 (stretch); whole line slides (shift)."

Step 1 — Mean uses full linearity. $E[Y]=2E[X]-1=2(3.5)-1=6$. Why this step? Expectation is linear: constants slide straight through, and the shift $-1$ does move the center.

Step 2 — Variance ignores the shift. $\text{Var}(Y)=2^2\text{Var}(X)=4\cdot \frac{35}{12}=\frac{140}{12}=\frac{35}{3}\approx 11.67$. Why $a^2$ and why drop the $-1$? Adding $-1$ slides the whole distribution but keeps distances between points identical, so it changes nothing about spread. Multiplying by $2$ stretches every distance by $2$; squared distances stretch by $2^2=4$. In the figure the lower line's dots are literally twice as far apart while its mean has merely slid.

Step 3 — SD. $\text{SD}(Y)=|2|\cdot \text{SD}(X)=2\sqrt{35/12}\approx 2(1.708)=3.416$. Why $|a|$ not $a^2$? SD is a square root of variance, so it picks up $\sqrt{a^2}=|a|$. The absolute value matters if $a$ is negative — spread can't be negative.

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Ex 6 — Sum of independent RVs (C6)

Figure s06 (caption): a bar chart with two bars — a cyan bar at the true SD $\sqrt{35/6}\approx 2.415$ and a taller amber bar at the wrong "SD + SD" $=3.416$ — visually showing that adding standard deviations overshoots the real spread of the sum.

Step 1 — Mean adds always. $E[S]=3.5+3.5=7$. Why this step? Linearity of expectation needs no independence — means always add.

Step 2 — Variance adds (independence). $\text{Var}(S)=\frac{35}{12}+\frac{35}{12}=\frac{70}{12}=\frac{35}{6}\approx 5.83$. Why this step? Independent dice don't influence each other, so the covariance term is $0$ and only the variances add.

Step 3 — SD does NOT add. $\text{SD}(S)=\sqrt{35/6}\approx 2.415$, not $1.71+1.71=3.42$. Why this step? Variances live in squared units, so they add; SD is a square root of the summed variance. Adding SDs directly double-counts spread. In the figure the amber "added SDs" bar clearly overshoots the true cyan bar.

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Ex 7 — Correlated sum, both signs (C7)

Figure s07 (caption): three bars for $\text{Var}(X+Y)$ against a dotted white baseline at $18$ — a short cyan bar at $6$ (Cov $=-6$), a white bar at $18$ (independent, Cov $=0$), and a tall amber bar at $30$ (Cov $=+6$) — showing the $+2\text{Cov}$ term pushing spread below or above the independent baseline.

Step 1 — General formula. $\text{Var}(X+Y)=\text{Var}(X)+\text{Var}(Y)+2\text{Cov}(X,Y)$. Why this step? Recall $\mu$ is our name for a mean. When variables move together, expanding $(X+Y-{\mu }_{X+Y}{)}^2$ — the squared deviation of the sum from its mean ${\mu }_{X+Y}={\mu }_X+{\mu }_Y$ — leaves a cross-term $2(X-{\mu }_X)(Y-{\mu }_Y)$ whose average is $2\text{Cov}(X,Y)$.

Step 2 — Positive covariance. $\text{Var}=9+9+2(6)=30$. Why this step? Positively correlated variables tend to be high together and low together, so their swings amplify — spread exceeds $18$ (the tall amber bar).

Step 3 — Negative covariance. $\text{Var}=9+9+2(-6)=6$. Why this step? Negatively correlated variables cancel each other's swings, so spread shrinks below $18$ (the short cyan bar). At perfect $\text{Cov}=-9$ it would hit $0$.

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Ex 8 — Real-world word problem (C8)

Figure s08 (caption): a number line of order counts with three cyan mass spikes at $100$ (height $0.2$), $150$ (height $0.5$, the tallest), and $250$ (height $0.3$); an amber triangular pivot at $170$ marks the mean, sitting left of the midrange because the heaviest $0.5$ mass at $150$ pulls it down.

Step 1 — Translate to a distribution. Support $\{100,150,250\}$ with probabilities $\{0.2,0.5,0.3\}$. Why this step? The English sentence is a discrete RV in disguise; naming it lets us apply the machinery.

Step 2 — Expected orders. $E[X]=100(0.2)+150(0.5)+250(0.3)=20+75+75=170$. Why this step? Expectation weights each possible order count by how often it occurs — the balance point of the three mass spikes in the figure. Note it need not equal any single outcome; $170$ sits between $150$ and $250$ because the heavier $0.5$ mass at $150$ pulls the pivot leftward of the midrange.

Step 3 — Second moment & variance. $E[X^2]=100^2(0.2)+150^2(0.5)+250^2(0.3)=2000+11250+18750=32000$. So $\text{Var}(X)=32000-170^2=32000-28900=3100$. Why this step? To measure spread we need $E[X^2]-{\mu }^2$, so we first average the squared order counts (weighting each $x^2$ by its probability), then subtract ${\mu }^2=170^2$ to strip out the center and leave pure spread in orders${}^2$.

Step 4 — Scale to revenue. E[R]=4(170)=\680$.$\text{Var}(R)=4^2(3100)=49600$,so$\text{SD}(R)=\sqrt{49600}\approx$222.7$.\ast Why$4^2$?\ast Multiplyingeveryordercountby$$4$scalesvarianceby$4^2$(Ex{5}^{\prime }srule),thenSDpicksupthe$\lvert 4\rvert$.

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Ex 9 — Exam twist: only moments given (C9)

Step 1 — Use the computational form. $\text{Var}(X)=E[X^2]-(E[X]{)}^2=40-5^2=40-25=15$. Why this step? The identity $\text{Var}=E[X^2]-(E[X]{)}^2$ was built precisely so that variance depends only on the two moments $E[X]$ and $E[X^2]$ — never on the individual outcomes. That is exactly why an exam can hand you just these two numbers and still expect an answer.

Step 2 — SD. $\sigma =\sqrt{15}\approx 3.873$. Why this step? Variance $15$ is in squared units; the square-root brings it back to the original scale of $X$ so the spread is directly comparable to the mean $\mu =5$. Notice $\sigma <\mu$ here, which is typical for a moderately concentrated variable.

Step 3 — Transform. $\text{Var}(3X+2)=3^2\text{Var}(X)=9(15)=135$; the $+2$ shift is invisible to spread. Why this step? By the scaling rule (Ex 5), multiplying by $3$ stretches every squared distance by $3^2=9$, while adding $2$ merely slides the whole distribution and leaves all inter-point distances — hence the spread — unchanged. We never needed the distribution to apply this.

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This spread-vs-noise theme returns as the "variance" axis of the bias–variance tradeoff.