Worked examples — Joint distributions — joint PMF - PDF, marginal, conditional

The scenario matrix

Every joint-distribution problem is one of these cases. The right column names the worked example that covers it.

We hit all eight cells below.

Example 1 — Discrete table (Cell C1)

Step 1 — check it's a valid PMF. Why this step? A table is only a distribution if all entries are $\ge 0$ and sum to $1$; otherwise everything downstream is meaningless. $0.05+0.10+0.05+0.15+0.30+0.15+0.05+0.05+0.10=\mathrm{1.00.}✓$

Step 2 — marginal of $X$ (sum each row). Why this step? $p_X(x)={\sum }_y{p}_{X,Y}(x,y)$ — "$X=x$ and $Y$ anything" is the Law of Total Probability over the $Y$-values. $p_X(0)=0.20,p_X(1)=0.60,p_X(2)=\mathrm{0.20.}$

Step 3 — marginal of $Y$ (sum each column). $p_Y(0)=0.25,p_Y(1)=0.45,p_Y(2)=\mathrm{0.30.}$

Step 4 — the conditional. Why this step? Freeze $Y=1$: look only at that column $(0.10,0.30,0.05)$, then divide by its total $p_Y(1)=0.45$ to re-normalize the slice (see Conditional Probability). $P(X=1\mid Y=1)=\frac{p_{X,Y}(1,1)}{p_Y(1)}=\frac{0.30}{0.45}=\frac{2}{3}\approx \mathrm{0.6667.}$

Step 5 — independence test. Why this step? Independence is a statement about the joint factoring, never about marginals alone. Test one cell. $p_{X,Y}(0,0)=0.05\text{vs}{p}_X(0)p_Y(0)=0.20\times 0.25=\mathrm{0.05.}$ One matching cell is not enough — check another: $p_{X,Y}(2,2)=0.10\text{vs}{p}_X(2)p_Y(2)=0.20\times 0.30=\mathrm{0.06.}\ne$ Since one cell fails, $X,Y$ are dependent.

Verify: rows sum to $0.20,0.60,0.20$ (total $1$); columns to $0.25,0.45,0.30$ (total $1$); conditional column $(0.10,0.30,0.05)/0.45$ sums to $1$. ✔

Example 2 — The independent case (Cell C2)

Step 1 — build the joint by multiplying. Why this step? Independence of Random Variables means $p_{X,Y}(x,y)=p_X(x)p_Y(y)$ — the joint is literally the outer product of the marginals.

Step 2 — check the total. $0.28+0.12+0.42+0.18=\mathrm{1.00.}✓$

Step 3 — the tell-tale of independence: conditional = marginal. Why this step? Under independence, knowing $Y$ tells you nothing about $X$, so the slice equals the marginal. $P(X=1\mid Y=0)=\frac{0.42}{0.28+0.42}=\frac{0.42}{0.70}=0.6=p_X(1).✓$

Verify: conditional $0.6$ equals the marginal $p_X(1)=0.6$ exactly — the fingerprint of independence. ✔

Example 3 — Continuous, rectangular support (Cell C3)

Step 1 — find $c$ from total probability $=1$. Why this step? Every density must integrate to $1$ over its support; that pins the constant. ${\int }_0^1{\int }_0^{1}c(x+y)dxdy=c{\int }_0^1(\frac{1}{2}+y)dy=c(\frac{1}{2}+\frac{1}{2})=c=1\Rightarrow c=1.$

Step 2 — marginal of $X$ (integrate out $y$, constant bounds $0$ to $1$). Why this step? $f_X(x)=\int f_{X,Y}dy$; the rectangle means $y$ always runs the full $0\to 1$. $f_X(x)={\int }_0^1(x+y)dy=x+\frac{1}{2},0<x<1.$

Step 3 — marginal of $Y$ (by symmetry). $f_Y(y)={\int }_0^1(x+y)dx=y+\frac{1}{2},0<y<1.$

Step 4 — conditional. Why this step? Divide the joint by the marginal of the conditioning variable. $f_{X\mid Y}(x\mid y)=\frac{x+y}{y+\frac{1}{2}},0<x<1.$

Step 5 — independent? $f_X(x)f_Y(y)=(x+\frac{1}{2})(y+\frac{1}{2})\ne x+y=f_{X,Y}(x,y).$ So dependent — even on a clean square, an additive density does not factor.

Verify: ${\int }_0^1{f}_{X}dx={\int }_0^1(x+\frac{1}{2})dx=1$ ✔; ${\int }_0^1{f}_{X\mid Y}(x\mid y)dx=\frac{\frac{1}{2}+y}{y+\frac{1}{2}}=1$ for every $y$ ✔.

Example 4 — Continuous, triangular support (Cell C4)

Step 1 — read the bounds off the picture. Why this step? The region is $\{0<x<y<1\}$: for a fixed $x$, $y$ runs from $x$ up to $1$; for a fixed $y$, $x$ runs from $0$ to $y$. Constant bounds would give the wrong answer. Look at the amber slice in the figure.

Step 2 — confirm total $=1$. ${\int }_0^1{\int }_0^{y}8xydxdy={\int }_0^{1}8y\cdot \frac{y^2}{2}dy={\int }_0^{1}4y^{3}dy=1.✓$

Step 3 — marginal of $X$ (for fixed $x$, $y$ from $x$ to $1$). $f_X(x)={\int }_x^{1}8xydy=8x\cdot \frac{1-x^2}{2}=4x(1-x^2),0<x<1.$

Step 4 — marginal of $Y$ (for fixed $y$, $x$ from $0$ to $y$). $f_Y(y)={\int }_0^{y}8xydx=8y\cdot \frac{y^2}{2}=4y^3,0<y<1.$

Step 5 — conditional of $Y$ given $X=x$. Why this step? We want $P(Y>0.5\mid X=0.25)$, so condition on $X$. Divide joint by $f_X$; the support becomes $y\in (x,1)$. $f_{Y\mid X}(y\mid x)=\frac{8xy}{4x(1-x^2)}=\frac{2y}{1-x^2},x<y<1.$

Step 6 — the requested probability at $x=0.25$. $f_{Y\mid X}(y\mid 0.25)=\frac{2y}{1-0.0625}=\frac{2y}{0.9375}.$ $P(Y>0.5\mid X=0.25)={\int }_{0.5}^1\frac{2y}{0.9375}dy=\frac{1}{0.9375}[y^2{]}_{0.5}^1=\frac{1-0.25}{0.9375}=\frac{0.75}{0.9375}=\mathrm{0.8.}$

Verify: ${\int }_x^1{f}_{Y\mid X}(y\mid x)dy=\frac{1-x^2}{1-x^2}=1$ ✔; the numeric answer $0.8$ is re-checked below.

Example 5 — Degenerate / zero conditioning (Cell C5)

Step 1 — evaluate the marginal at the conditioning value. Why this step? The conditional $f_{X\mid Y}(x\mid y)=f_{X,Y}(x,y)/f_Y(y)$ is only defined where $f_Y(y)>0$. $f_Y(y)=4y^3\Rightarrow f_Y(0)=0,f_Y(1.5)=0(\text{outside support}).$

Step 2 — declare it undefined. Both give a $0/0$ or "$0$ over $0$-support" situation: $f_{X\mid Y}(x\mid 0)=\frac{0}{0}\text{— undefined},f_{X\mid Y}(x\mid 1.5)=\frac{0}{0}\text{— undefined.}$

Verify: $f_Y(0)=4\cdot 0^3=0$ and the support requires $y<1$, so $y=1.5$ has zero density — both conditionals are undefined. ✔ (No numeric answer; the "answer" is undefined.)

Example 6 — Limiting behaviour (Cell C6)

Step 1 — write the conditional of $X$ given $Y=y$. Why this step? Condition on $Y$ this time: divide joint by $f_Y(y)=4y^3$, support $0<x<y$. $f_{X\mid Y}(x\mid y)=\frac{8xy}{4y^3}=\frac{2x}{y^2},0<x<y.$

Step 2 — limit $y\to 1^{-}$. $f_{X\mid Y}(x\mid y)\to 2x,0<x<1.$ A wide, gently increasing slope over the full interval — the picture's cyan curve.

Step 3 — limit $y\to 0^{+}$. Why this step? Look at the shape: on the shrinking interval $0<x<y$ the density is $2x/y^2$. Its peak at $x=y$ is $2/y\to \infty$, but its width is $y\to 0$. The mass stays $1$ but concentrates into a spike at $x=0$. ${\int }_0^y\frac{2x}{y^2}dx=\frac{1}{y^2}[x^2{]}_0^y=1\text{for every }y>0.$ So as $y\to 0^{+}$ the slice becomes an ever-taller, ever-narrower spike (a "point mass forming"), even though at $y=0$ itself the object is undefined (Example 5). The amber curves in the figure show this squeeze.

Verify: area under $f_{X\mid Y}(\cdot \mid y)$ is $1$ for all $y>0$; limit at $y=1$ gives ${\int }_0^{1}2xdx=1$. ✔

Example 7 — Real-world word problem (Cell C7)

Step 1 — check validity & get $p_Y(1)$. Why this step? We condition on $Y=1$, so we need the weight of that column. Total $=0.20+0.05+0.25+0.20+0.10+0.20=1.00$ ✔. $p_Y(1)=0.05+0.20+0.20=\mathrm{0.45.}$

Step 2 — conditional distribution of $X$ given $Y=1$. Why this step? Slice the $Y=1$ column and re-normalize by $0.45$. $P(X=0\mid Y=1)=\frac{0.05}{0.45}=\frac{1}{9},P(X=1\mid Y=1)=\frac{0.20}{0.45}=\frac{4}{9},P(X=2\mid Y=1)=\frac{0.20}{0.45}=\frac{4}{9}.$

Step 3 — answer the headline question. $P(X=2\mid Y=1)=\frac{0.20}{0.45}=\frac{4}{9}\approx \mathrm{0.4444.}$ Compare baseline $p_X(2)=0.10+0.20=0.30$. Buying a pastry raises the chance of 2 coffees ($0.44>0.30$) — they go together.

Step 4 — conditional expectation (see Expectation and Variance). Why this step? Expected value under the conditional distribution weights each coffee-count by its conditional probability. $E[X\mid Y=1]=0\cdot \frac{1}{9}+1\cdot \frac{4}{9}+2\cdot \frac{4}{9}=\frac{0+4+8}{9}=\frac{12}{9}=\frac{4}{3}\approx \mathrm{1.3333.}$

Verify: conditional probs $\frac{1}{9}+\frac{4}{9}+\frac{4}{9}=1$ ✔; $P(X=2\mid Y=1)=4/9$ and $E[X\mid Y=1]=4/3$ re-checked below.

Example 8 — Exam twist: rebuild the joint (Cell C8)

Step 1 — chain rule to get the joint. Why this step? $f_{X,Y}(x,y)=f_{Y\mid X}(y\mid x)f_X(x)$ — multiply the slice by the weight of the thing you conditioned on. $f_{X,Y}(x,y)=\frac{1}{x}\cdot 2x=2,0<y<x<1.$ A flat density of height $2$ on the triangle $0<y<x<1$.

Step 2 — sanity: does it integrate to $1$? ${\int }_0^1{\int }_0^{x}2dydx={\int }_0^{1}2xdx=1.✓$

Step 3 — marginal of $Y$ (integrate out $x$; for fixed $y$, $x$ runs $y\to 1$). $f_Y(y)={\int }_y^{1}2dx=2(1-y),0<y<1.$

Step 4 — reverse conditional (Bayes direction). Why this step? Now we want $X$ given $Y=y$: divide the joint by $f_Y(y)$. $f_{X\mid Y}(x\mid y)=\frac{2}{2(1-y)}=\frac{1}{1-y},y<x<1.$ That's uniform on $(y,1)$ — the mirror of the forward "uniform on $(0,x)$".

Verify: joint integrates to $1$ ✔; ${\int }_0^1{f}_Y(y)dy={\int }_0^{1}2(1-y)dy=1$ ✔; ${\int }_y^1{f}_{X\mid Y}(x\mid y)dx=\frac{1-y}{1-y}=1$ ✔.

Recall

Connections

• Joint distributions — joint PMF - PDF, marginal, conditional — the parent whose machinery every example here exercises.
• Conditional Probability — the slice-and-renormalize rule behind every conditional above.
• Independence of Random Variables — tested in Examples 1–3.
• Law of Total Probability — the justification for summing/integrating to get marginals.
• Bayes' Theorem — the reversal direction in Example 8.
• Expectation and Variance — conditional expectation in Example 7.
• Covariance and Correlation — natural next step once you can compute joint moments.
• Bivariate Normal Distribution — the continuous joint you'll meet next.