Joint, marginal, and conditional distributions: Worked examples

Definition Notation on this page (self-contained)

We use two symbols and they mean different things — read this once and never be confused again.

Uppercase $P (\cdot)$ = the probability of an event, a single number in $[0, 1]$ . Example: $P (X = 1 ∣ Y = 1)$ is "the probability the event $X = 1$ happens given $Y = 1$ ."
Lowercase $p$ / $f$ = a whole function (a distribution), not one number.
- $p_{X, Y} (x, y) = P (X = x and Y = y)$ — the joint pmf for discrete variables.
- $f_{X, Y} (x, y)$ — the joint pdf for continuous variables (a density, not a probability; it can exceed $1$ , but $\iint f = 1$ ).
The marginal operators we build every case with: $p_{X} (x) = \sum_{y} p_{X, Y} (x, y) (discrete), f_{X} (x) = \int f_{X, Y} (x, y) d y (continuous) .$ In words: $p_{X}$ ("marginal of $X$ ") is the joint with the other variable summed/integrated away.
The conditional operator: $p_{Y ∣ X} (y ∣ x) = \frac{p _{X, Y} ( x , y )}{p _{X} ( x )}$ (and likewise $f_{Y ∣ X}$ ) — the slice at $X = x$ rescaled so it sums/integrates to $1$ .

Rule of thumb: if you can plug a single pair of numbers and get one probability, it is $P (\dots)$ . If it is a rule that gives a value for every input, it is $p$ or $f$ .

The scenario matrix

Before any example, let us list every distinct kind of case this topic can throw at you. Read the table as a checklist — every row gets its own worked example below.

Cell	Case class	What is tricky about it	Example
A	Discrete 2×2, get marginal + conditional	must sum a whole row, not grab a cell	Ex 1
B	Discrete: both directions of the bar	$p (X ∣ Y) \neq = p (Y ∣ X)$	Ex 2
C	Degenerate slice: condition on a zero-probability event	division by $0$ — undefined	Ex 3
D	Continuous, independent density (conditional loses $x$ )	joint must factor as $g (x) h (y)$	Ex 4
E	Continuous, dependent density (limits depend on $x$ )	marginal support changes; not separable	Ex 5
F	Real-world word problem (medical test, Bayes)	base-rate / prosecutor's fallacy	Ex 6
G	Uncorrelated but dependent (limiting/degenerate check)	zero covariance $\neq =$ independence	Ex 7
H	Exam twist: 3-variable chain rule / marginalize the middle	law of total probability under the hood	Ex 8

Cells A–H below are each labelled. Together they touch every branch: discrete/continuous, both bar directions, independent/dependent, a zero-denominator degeneracy, a real-world base-rate trap, and a multivariable extension.

Ex 1 — Discrete marginal + conditional (Cell A)

Forecast: guess now — is $P (Y = 1 ∣ X = 1)$ bigger or smaller than the raw $p_{Y} (1)$ ? (Car owners commute far more, so we expect it bigger.)

Marginal $p_{X} (1) = 0.15 + 0.45 = 0.60$ . Why this step? To "forget" $Y$ we sum the whole $X = 1$ row — that erases the commute info, leaving pure car-ownership probability. (This is the $p_{X}$ operator from the notation box.)
Conditional $P (Y = 1 ∣ X = 1) = \frac{p ( 1 , 1 )}{p _{X} ( 1 )} = \frac{0.45}{0.60} = 0.75$ . Why this step? We restrict the world to car owners (total mass $0.60$ ), then ask what fraction of that world commutes far. Dividing re-normalizes the slice back to sum $1$ .

Verify: the two conditionals in the row must sum to $1$ : $\frac{0.15}{0.60} + \frac{0.45}{0.60} = 0.25 + 0.75 = 1$ . ✓ And $0.75 > p_{Y} (1) = 0.10 + 0.45 = 0.55$ , matching the forecast.

Ex 2 — Both directions of the bar (Cell B)

Forecast: will the two match? They almost never do — the denominators differ.

Marginal $p_{Y} (1) = 0.10 + 0.45 = 0.55$ . Why this step? Now we condition on $Y$ , so the denominator is the $Y = 1$ column total.
$P (X = 1 ∣ Y = 1) = \frac{p ( 1 , 1 )}{p _{Y} ( 1 )} = \frac{0.45}{0.55} \approx 0.818$ . Why this step? Same joint cell $0.45$ , different normalizer — that is exactly why the bar is not symmetric.

Verify: cross-check with Bayes' theorem: $P (X = 1 ∣ Y = 1) = \frac{P ( Y = 1 ∣ X = 1 ) p _{X} ( 1 )}{p _{Y} ( 1 )} = \frac{0.75 \times 0.60}{0.55} \approx 0.818$ . ✓ Same number two ways.

Ex 3 — Degenerate slice: zero-probability condition (Cell C)

Forecast: feel the trouble coming — dividing by zero.

Write the definition: $P (X = 1 ∣ Y = 2) = \frac{p ( 1 , 2 )}{p _{Y} ( 2 )} = \frac{0}{0}$ . Why this step? The conditional formula requires the denominator $p_{Y} (y) > 0$ . Here it is $0$ .
Conclusion: the conditional is undefined. Why this step? "Restrict to the world where $Y = 2$ " is meaningless — that world has probability $0$ , there is nothing to re-normalize. The definition in the parent note explicitly demanded $p_{X} (x) > 0$ ; the same guard applies to $p_{Y} (y) > 0$ .

Verify: any assignment $c \in [0, 1]$ satisfies $0 = c \cdot 0$ , so no unique value exists → genuinely undefined, not " $0$ " and not " $1$ ". This is the degenerate corner of the matrix: always check the denominator before dividing.

Ex 4 — Continuous, independent density (Cell D)

Forecast: the joint is a product $4 \cdot x \cdot y$ — that is $g (x) h (y)$ shape. Guess whether the conditional will still contain $x$ . (Prediction: it should drop $x$ , the signature of independence.)

Check it is a pdf: $\int_{0}^{1} \int_{0}^{1} 4 x y d x d y = 4 (\frac{1}{2}) (\frac{1}{2}) = 1$ . ✓ Why this step? A density must integrate to $1$ before any question is legal.
Marginal of $X$ : integrate $y$ away. $f_{X} (x) = \int_{0}^{1} 4 x y d y = 4 x \cdot \frac{1}{2} = 2 x, 0 \leq x \leq 1.$ Why this step? Marginal = "collapse the $Y$ -direction," i.e. integrate out $y$ (the $f_{X}$ operator).
Conditional: $f_{Y ∣ X} (y ∣ x) = \frac{4 x y}{2 x} = 2 y$ . Why this step? Slice at fixed $x$ , then divide by the marginal so the slice integrates to $1$ in $y$ . The $x$ cancels.

Verify: conditional integrates to $1$ ? $\int_{0}^{1} 2 y d y = 1$ . ✓ Independent? $f_{Y ∣ X} (y ∣ x) = 2 y$ does not depend on $x$ ⇒ knowing $X$ tells you nothing about $Y$ ⇒ independent. Equivalently the joint factors: $f_{X} f_{Y} = (2 x) (2 y) = 4 x y = f_{X, Y}$ . ✓ This is the true Cell D: a genuinely independent continuous case where conditioning loses $x$ .

Look at the figure: the plot below shows the three conditional slices $f_{Y ∣ X} (y ∣ x)$ at $x = 0, \frac{1}{2}, 1$ . They lie exactly on top of each other — the single line $2 y$ . That coincidence is independence: the shape of $Y$ never changes when you move $x$ . Contrast this with Ex 5, where the slices will visibly differ.

Figure — Joint, marginal, and conditional distributions

Ex 5 — Continuous, dependent (support depends on $x$ ) (Cell E)

Forecast: the tricky part is the limits: given $x$ , $y$ ranges only over $[x, 1]$ , not $[0, 1]$ . Guess whether the conditional keeps $x$ this time. (Prediction: yes — so dependent.)

Check pdf: area of triangle $= \frac{1}{2}$ , times height $2$ = $1$ . ✓ Why this step? Constant density $\times$ region area must equal $1$ .
Marginal of $X$ : for a fixed $x$ , $y$ runs from $x$ to $1$ . $f_{X} (x) = \int_{x}^{1} 2 d y = 2 (1 - x), 0 \leq x \leq 1.$ Why this step? The integration limits depend on $x$ because the support is a triangle — this is exactly what makes $X, Y$ dependent.
Conditional: $f_{Y ∣ X} (y ∣ x) = \frac{2}{2 ( 1 - x )} = \frac{1}{1 - x}$ for $x \leq y \leq 1$ . Why this step? Given $x$ , $Y$ is uniform on $[x, 1]$ — an interval whose length shrinks as $x$ grows. The conditional depends on $x$ ⇒ dependent.

Verify: $\int_{x}^{1} \frac{1}{1 - x} d y = \frac{1 - x}{1 - x} = 1$ . ✓ And marginal check $\int_{0}^{1} 2 (1 - x) d x = 2 (1 - \frac{1}{2}) = 1$ . ✓

Look at the figure: the lavender triangle is the support. The coral vertical strip is the slice at $x = 0.4$ : notice $y$ is trapped in $[0.4, 1]$ , a shorter interval than the full $[0, 1]$ . Because that interval shrinks as $x$ grows, the conditional density $\frac{1}{1 - x}$ changes with $x$ — the opposite of Ex 4's identical slices. Watching the strip get shorter is watching dependence happen.

Ex 6 — Real-world word problem: medical test (Cell F)

Forecast: most people guess $\approx 90%$ . Watch the base rate destroy that intuition.

Total probability of a positive (marginalize over disease status): $P (+) = P (+ ∣ D) P (D) + P (+ ∣ \neg D) P (\neg D) = 0.99 (0.01) + 0.05 (0.99) .$ Why this step? $+$ can come from a sick person or a healthy false positive; we sum both routes — the law of total probability. Everything here is an event probability, so we use uppercase $P$ . Numerically $0.0099 + 0.0495 = 0.0594$ .
Bayes' theorem: $P (D ∣ +) = \frac{P ( + ∣ D ) P ( D )}{P ( + )} = \frac{0.0099}{0.0594} \approx 0.1667.$ Why this step? We want $D ∣ +$ but were given $+ ∣ D$ ; Bayes flips the bar using the marginal denominator $P (+)$ from step 1.

Verify: the two posteriors must sum to $1$ : $P (\neg D ∣ +) = \frac{0.0495}{0.0594} \approx 0.8333$ , and $0.1667 + 0.8333 = 1$ . ✓ Only ~ $16.7%$ — the tiny base rate $0.01$ dominates. This is the base-rate cell of the matrix.

Ex 7 — Uncorrelated but dependent (Cell G)

Forecast: guess whether zero covariance forces independence. (The parent warned: no!)

Covariance. By symmetry $E [X] = \frac{1}{3} (- 1 + 0 + 1) = 0$ . $Cov (X, Y) = E [X Y] - E [X] E [Y] = E [X^{3}] - 0.$ Why this step? $X Y = X \cdot X^{2} = X^{3}$ ; and $E [X] = 0$ kills the second term. See Covariance and correlation and Expectation and variance.
$E [X^{3}] = \frac{1}{3} ((- 1)^{3} + 0 + 1^{3}) = \frac{1}{3} (- 1 + 0 + 1) = 0$ , so $Cov = 0$ : uncorrelated. Why this step? Odd powers of a symmetric variable average to $0$ .
Dependence via factorization. Take $x = 0$ : $P (Y = 0 ∣ X = 0) = 1$ but $P (Y = 0) = P (X = 0) = \frac{1}{3}$ . Since $1 \neq = \frac{1}{3}$ , $p_{Y ∣ X} \neq = p_{Y}$ ⇒ dependent. Why this step? The only real test of independence is $p_{X, Y} = p_{X} p_{Y}$ — correlation is a weaker summary. See Independence and conditional independence.

Verify: $Cov = 0$ (step 2) yet $P (Y = 0 ∣ X = 0) = 1 \neq = P (Y = 0) = 1/3$ . Both hold ⇒ uncorrelated-but-dependent confirmed. ✓

Ex 8 — Exam twist: 3-variable chain rule + marginalize the middle (Cell H)

Forecast: the trap is thinking you need the full 8-cell joint. You do not — marginalize $A$ , then $B$ .

Marginal of $B$ (sum out $A$ via total probability): $P (B = 1) = P (B = 1 ∣ A = 1) P (A = 1) + P (B = 1 ∣ A = 0) P (A = 0) = 0.7 (0.6) + 0.2 (0.4) = 0.42 + 0.08 = 0.50.$ Why this step? $C$ depends on $B$ , not directly on $A$ (that is what the chain structure of a graphical model encodes). So we only need $P (B)$ , obtained by marginalizing $A$ out.
Marginal of $C$ (sum out $B$ ): $P (C = 1) = P (C = 1 ∣ B = 1) P (B = 1) + P (C = 1 ∣ B = 0) P (B = 0) = 0.9 (0.5) + 0.3 (0.5) = 0.45 + 0.15 = 0.60.$ Why this step? Same law of total probability, one link further down the chain — this is how the chain rule composes conditionals into a marginal without ever forming the whole joint.

Verify: probabilities in range and $P (C = 1) = 0.60$ ; sanity: $P (C = 0) = (1 - 0.9) (0.5) + (1 - 0.3) (0.5) = 0.05 + 0.35 = 0.40$ , and $0.60 + 0.40 = 1$ . ✓

Coverage check

Recall Did we hit every cell?

A discrete marginal+conditional (Ex 1) ::: ✓ B both bar directions differ (Ex 2) ::: ✓ C zero-probability condition → undefined (Ex 3) ::: ✓ D continuous independent density, conditional loses $x$ (Ex 4) ::: ✓ E continuous with $x$ -dependent support (Ex 5) ::: ✓ F real-world Bayes / base-rate (Ex 6) ::: ✓ G uncorrelated but dependent (Ex 7) ::: ✓ H multivariable chain + marginalize middle (Ex 8) ::: ✓

Active recall

How do you handle conditioning on an event with probability

0

The conditional is undefined — the denominator is

0

, there is no world to re-normalize.

In Ex 4 the joint is

4 x y

; why are

X, Y

independent?

The conditional

\frac{4 x y}{2 x} = 2 y

drops

x

; equivalently the joint factors

f_{X} f_{Y} = (2 x) (2 y) = 4 x y

In Ex 5, what is the conditional distribution of

Y

given

X = x

Uniform on

[x, 1]

, density

\frac{1}{1 - x}

— its support shrinks with

x

, so

X, Y

are dependent.

Ex 6: why is

P (D ∣ +)

only ~17% despite 99% sensitivity?

The base rate

P (D) = 0.01

is tiny, so false positives (

0.05 \times 0.99

) outnumber true positives in the total.

Ex 7: what is the only reliable test of independence?

Factorization

p_{X, Y} = p_{X} p_{Y}

— zero covariance is necessary but not sufficient.

What is the difference between

P (\cdot)

and

p

f

on this page?

P

is one event's probability (a number);

p

f

is a whole distribution function (pmf / pdf).

Connections

Bayes' theorem — Ex 2 and Ex 6 flip the bar.
Marginalization and the law of total probability — the denominators in Ex 6 and Ex 8.
Independence and conditional independence — Ex 4, 5, 7 test factorization.
Covariance and correlation — Ex 7's uncorrelated-yet-dependent case.
Probabilistic graphical models — Ex 8's chain factorization.
Naive Bayes classifier — uses conditional independence like Ex 4/8.
Expectation and variance — the moments in Ex 7.

$p (x, y)$	$Y = 0$	$Y = 1$
$X = 0$	$0.30$	$0.10$
$X = 1$	$0.15$	$0.45$