Worked examples — Random variables (discrete and continuous)

This page is a drill ground. The parent note built the machinery — PMF, PDF, expected value. Here we push that machinery through every kind of input it can meet, so no exam question or real dataset ever surprises you.

Before we start, one promise: every symbol you see was defined in the parent note. If you want a refresher on any of them, the relevant Expectation and Variance and Gaussian Distribution notes are one click away.

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The scenario matrix

Think of a random-variable problem as a machine with inputs. The inputs can be nice (a clean die) or nasty (a degenerate constant, a zero-probability point, an infinite tail). The matrix below lists every case class — every fundamentally different kind of input — that this topic can throw at you.

#	Case class	What makes it special	Example that hits it
C1	Finite discrete, uneven weights	PMF values differ; must weight the sum	Ex 1 (loaded die)
C2	Countably infinite discrete	Sum runs to $\infty$; needs a series	Ex 2 (geometric wait)
C3	Degenerate RV	All probability on ONE value	Ex 3 (constant "random" variable)
C4	Continuous, bounded support	PDF nonzero only on $[a,b]$	Ex 4 (uniform interval)
C5	Zero-probability point	$P(X=x)=0$ for continuous — the classic trap	Ex 5 (exact-value question)
C6	Continuous, infinite tail	Support is all of $\mathbb{R}$; tails must decay	Ex 6 (Gaussian window)
C7	Mixed / limiting behaviour	Binomial with large $n$ → looks Gaussian	Ex 7 (limit crossover)
C8	Real-world word problem	Translate English → RV first	Ex 8 (spam filter)
C9	Exam twist: transformed RV	Compute $E[g(X)]$, not $E[X]$	Ex 9 (variance as $E[X^2]-E[X{]}^2$)

Each example below is tagged with the cell it covers. Together they fill the whole table.

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Ex 1 — Loaded die (C1: finite, uneven weights)

Forecast: guess before computing — will $E[X]$ be above or below the fair-die average of $3.5$? (The six is favoured, so bet above.)

Step 1 — Turn "twice as often" into numbers. Let each ordinary face have probability $q$. The six has $2q$. $q+q+q+q+q+2q=7q=1\Rightarrow q=\frac{1}{7}.$ Why this step? The parent note's normalization rule (${\sum }_{x}P(X=x)=1$) is the only equation that pins down the unknown $q$. Everything must sum to 1 because something happens.

So the PMF is $P(X=1)=\cdots =P(X=5)=\frac{1}{7}$ and $P(X=6)=\frac{2}{7}$.

Step 2 — Apply the discrete expected-value formula. $E[X]={\sum }_{x}xP(X=x)=(1+2+3+4+5)\cdot \frac{1}{7}+6\cdot \frac{2}{7}.$ Why this step? $E[X]={\sum }_{x}xP(X=x)$ weights each value by how often it appears in the long run — the six, appearing twice as often, pulls harder.

Step 3 — Arithmetic. $E[X]=\frac{15}{7}+\frac{12}{7}=\frac{27}{7}\approx \mathrm{3.857.}$

Verify: $3.857>3.5$ ✓ matches our forecast. Sanity check on the PMF: $5\cdot \frac{1}{7}+\frac{2}{7}=\frac{7}{7}=1$ ✓.

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Ex 2 — Waiting for the first success (C2: infinite discrete)

Forecast: with heads only 1-in-3, do you expect the first heads early or late? (Late — guess $E[X]$ around 3.)

Step 1 — Build the PMF. To need exactly $k$ flips you get $k-1$ tails then a head: $P(X=k)=(1-p{)}^{k-1}p,k=1,2,3,\dots$ Why this step? Flips are independent, so probabilities multiply (parent note's binomial logic). This is the geometric distribution — a countably infinite discrete RV.

Step 2 — Check it normalizes over an infinite sum. ${\sum }_{k=1}^{\infty }(1-p{)}^{k-1}p=p\cdot \frac{1}{1-(1-p)}=\frac{p}{p}=1.$ Why this step? For infinite support we must confirm the geometric series converges to 1 — this is case C2's whole point: "sum to 1" now means "the series adds to 1".

Step 3 — Expected value. Using the known geometric-series result $E[X]=\frac{1}{p}$: $E[X]=\frac{1}{1/3}=3.$ Why this step? We need a series, not a finite sum. The identity ${\sum }_{k\ge 1}k(1-p{)}^{k-1}p=\frac{1}{p}$ is the tool that answers "how many trials on average?".

Verify: $E[X]=3$ matches the forecast (rare event → wait $\approx 1/p$ trials). Unit sanity: $E[X]$ is a count of flips, dimensionless, positive ✓.

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Ex 3 — The constant "random" variable (C3: degenerate)

Forecast: if it's always 5, what's the average? What's the variance?

Step 1 — Expected value. $E[X]=5\cdot 1=5.$ Why this step? The formula still applies; there's just one term. A constant is a perfectly legal (degenerate) random variable — the "randomness" is zero.

Step 2 — Variance. $\mathrm{Var}(X)=E[(X-5{)}^2]=(5-5{)}^2\cdot 1=0.$ Why this step? Variance measures spread. No spread ⇒ variance $0$ — this is the boundary case every distribution collapses to when uncertainty vanishes (see Expectation and Variance).

Verify: $E[X]=5$, $\mathrm{Var}=0$. This is the limiting case of any distribution as its spread shrinks to nothing — a useful mental anchor.

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Ex 4 — Uniform on an interval (C4: continuous, bounded)

Forecast: the interval is $8$ wide; the sub-interval $[0,3]$ is $3$ wide. Guess $P\approx 3/8$.

Step 1 — Density. $f(x)=\frac{1}{b-a}=\frac{1}{6-(-2)}=\frac{1}{8}\text{on }[-2,6],0\text{ elsewhere.}$ Why this step? Uniform means constant density; the height is forced to $\frac{1}{b-a}$ so the total area is 1.

Step 2 — Probability as area. $P(0\le X\le 3)={\int }_0^3\frac{1}{8}dx=\frac{1}{8}\cdot 3=\frac{3}{8}=\mathrm{0.375.}$ Why this step? For continuous RVs probability = area under $f$ (look at the shaded blue strip in the figure), because $P(X=x)=0$ at any single point.

Step 3 — Expected value (midpoint). $E[X]=\frac{a+b}{2}=\frac{-2+6}{2}=2.$ Why this step? Symmetry: equal density on both sides ⇒ the balance point is the centre.

Verify: $P=0.375$ matches forecast. $E[X]=2$ lies inside $[-2,6]$ and at its centre ✓.

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Ex 5 — The zero-probability trap (C5)

Forecast: careful — this is the classic mistake. Is $P(X=2)$ equal to $f(2)=\frac{1}{8}$?

Step 1 — Exact-value probability. $P(X=2)={\int }_2^{2}f(x)dx=0.$ Why this step? An integral over an interval of zero width is zero — geometrically, a line segment has no area.

Step 2 — Boundary probability. $P(X\ge 6)={\int }_6^{6}f+{\int }_6^{\infty }0dx=0.$ Why this step? At and beyond $x=6$ the density is $0$, so there is no area to accumulate. This also shows $P(X\le b)=1$ — the whole distribution lives inside $[-2,6]$.

Verify: Both answers are $0$. Consistency check: $P(0\le X\le 3)$ was $0.375$ whether or not we include the endpoints, precisely because endpoints contribute $0$ ✓.

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Ex 6 — Gaussian window (C6: infinite tail)

Forecast: the famous "68–95–99.7" rule — guess about 68%.

Step 1 — Write the PDF. $f(x)=\frac{1}{\sqrt{2\pi \cdot 4}}\exp \left(-\frac{(x-0{)}^2}{2\cdot 4}\right).$ Why this step? This is the Gaussian form from the parent note; the tails stretch to $\pm \infty$ but decay fast enough that the total area is still exactly 1 (case C6: infinite support, decaying tail).

Step 2 — Standardize to a $z$-score. Let $z=\frac{x-\mu }{\sigma }$. Then $[-2,2]$ maps to $z\in [-1,1]$. $P(-2\le X\le 2)=P(-1\le Z\le 1).$ Why this step? Every Gaussian is a stretched/shifted copy of the standard $\mathcal{N}(0,1)$. Rescaling by $\sigma$ turns "within one $\sigma$" into "within $z=1$", which we can read off one universal table.

Step 3 — Read the area. $P(-1\le Z\le 1)\approx \mathrm{0.6827.}$ Why this step? This is the fixed area under the standard bell between $\pm 1$ (shaded orange in the figure). It never changes — that's why the 68% rule works for any Gaussian.

Verify: $0.6827\approx 68\%$ ✓ matches forecast and the empirical rule. See Central Limit Theorem for why Gaussians appear so often.

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Ex 7 — Binomial marching toward Gaussian (C7: limiting behaviour)

Forecast: the mean should be $50$. Within $\pm 5$ of $50$ — is that roughly one $\sigma$? Guess "a bit more than 68%".

Step 1 — Mean and variance of Binomial. $E[X]=np=100\cdot 0.5=50,\mathrm{Var}(X)=np(1-p)=100\cdot 0.5\cdot 0.5=25.$ So $\sigma =\sqrt{25}=5$. Why this step? $E[X]=np$ (each flip adds $p$) and $\mathrm{Var}=np(1-p)$ come from summing $n$ independent Bernoullis (see Bernoulli and Categorical Distributions).

Step 2 — Invoke the Gaussian limit. For large $n$, Binomial $\approx \mathcal{N}(50,25)$. The window $[45,55]$ is exactly $\mu \pm \sigma$. Why this step? By the Central Limit Theorem, a sum of many independent trials becomes bell-shaped — this is case C7, where a discrete RV is well-approximated by a continuous one.

Step 3 — Use the one-sigma area. $P(45\le X\le 55)\approx P(-1\le Z\le 1)\approx \mathrm{0.6827.}$

Verify: $E[X]=50$, $\sigma =5$; the window is one standard deviation, giving $\approx 68\%$ ✓, matching the forecast "a bit more than 68%" (exact binomial is $\approx 0.729$; the Gaussian estimate is close).

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Ex 8 — Spam filter word problem (C8: real-world translation)

Forecast: expected flagged $=np=2$. "At least 2" — probably high, guess $\gtrsim 60\%$.

Step 1 — Identify the RV. Let $X$ = number flagged. Independent flags with fixed $p$ ⇒ $X\sim \text{Binomial}(n=10,p=0.2)$. Why this step? The English "each independently, count the successes" is the definition of a Binomial — always translate words to an RV first.

Step 2 — Use the complement. $P(X\ge 2)=1-P(X=0)-P(X=1).$ $P(X=0)=(0.8{)}^{10},P(X=1)=(\genfrac{}{}{0ex}{}{10}{1})(0.2)(0.8{)}^9.$ Why this step? "At least 2" over many terms is easier as $1-$(the two small cases). $\left(\genfrac{}{}{0ex}{}{10}{1}\right)=10$ counts which single email is flagged.

Step 3 — Compute. $P(X=0)=0.1074,P(X=1)=10\cdot 0.2\cdot 0.1342=\mathrm{0.2684.}$ $P(X\ge 2)=1-0.1074-0.2684=\mathrm{0.6242.}$ And $E[X]=np=10\cdot 0.2=2$.

Verify: $P(X\ge 2)\approx 0.624$ matches "$\gtrsim 60\%$"; $E[X]=2$ ✓. Connects to Maximum Likelihood Estimation (estimating $p$ from flagged counts).

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Ex 9 — Exam twist: variance via $E[X^2]$ (C9: transformed RV)

Forecast: a fair die has variance $\approx 2.92$. The loading concentrates weight at 6, slightly changing spread — guess "a little under 3".

Step 1 — Compute $E[X^2]$ (a transformed RV). $E[X^2]={\sum }_x{x}^{2}P(X=x)=(1+4+9+16+25)\cdot \frac{1}{7}+36\cdot \frac{2}{7}.$ Why this step? Variance needs $E[X^2]$, and $X^2$ is a function of $X$. The rule $E[g(X)]={\sum }_{x}g(x)P(X=x)$ — the law of the unconscious statistician — lets us reuse the same PMF without finding the distribution of $X^2$.

Step 2 — Arithmetic. $E[X^2]=\frac{55}{7}+\frac{72}{7}=\frac{127}{7}\approx \mathrm{18.143.}$

Step 3 — Subtract the square of the mean. From Ex 1, $E[X]=\frac{27}{7}$, so $(E[X]{)}^2=\frac{729}{49}\approx 14.878$. $\mathrm{Var}(X)=18.143-14.878\approx \mathrm{3.265.}$ Why this step? The shortcut $E[X^2]-(E[X]{)}^2$ avoids computing $E[(X-\mu {)}^2]$ term by term — algebraically identical but faster.

Verify: $\mathrm{Var}\approx 3.265$, close to but not equal to the fair-die $2.917$; loading raised the spread by pulling weight to the extreme value 6. Variance is positive ✓.

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Recall Quick self-test

Why is $P(X=2)=0$ for a continuous RV but not for a discrete one? ::: A continuous RV spreads probability over an interval as area; a single point has zero width ⇒ zero area. A discrete RV places lumps of mass on specific points. For $U[-2,6]$, what is $f(x)$ on the interval? ::: $\frac{1}{8}$. Binomial $(n=100,p=0.5)$: mean and $\sigma$? ::: mean $50$, $\sigma =5$. What tool turns "within one $\sigma$" into a table lookup? ::: The $z$-score $z=(x-\mu )/\sigma$.

See also: Joint and Conditional Distributions, Sampling Methods, Information Theory.