Joint Distributions

Suppose you record both the height and weight of a randomly chosen adult. Each measurement alone is a familiar random variable with its own distribution. But neither number alone tells you whether tall people also tend to be heavy. To capture how two random quantities behave together, we need a single mathematical object. It must track their simultaneous behavior. This is not two separate one-dimensional pictures glued together. Instead, it is one genuinely two-dimensional (or higher) description. This is the role of a joint distribution. It is the natural extension of the Cumulative Distribution Function to several random variables at once. It is the foundation on which almost all of multivariate probability is built. Once we have joint distributions in hand, we can recover the individual behavior of each variable (its marginal distribution). We can also ask whether the variables are unrelated (independence). And we can compute the distribution of combinations like sums (convolution). All of this depends on having captured the joint information correctly in the first place. The central idea is deceptively simple: instead of asking P(X≤x)P(X \le x), we ask P(X≤x,Y≤y)P(X \le x, Y \le y) for every pair (x,y)(x,y). This function of two variables encodes everything there is to know about the pair (X,Y)(X,Y). When the variables are discrete, this specializes to a joint probability mass function. When they admit densities, it specializes to a joint probability density function. Both are used constantly. They appear in modeling correlated failures in engineering systems, joint stock returns in finance, pixel intensities across color channels, or errors in repeated measurements. This article gives the precise definitions of joint CDFs, joint PMFs, and joint PDFs for two (and more) random variables. It states and proves the basic consistency theorem linking them. It also works through concrete asymmetric examples. These examples show how to compute probabilities of events involving several variables simultaneously.

Key takeaways
  • A joint CDF FX,Y(x,y)=P(X≤x,Y≤y)F_{X,Y}(x,y) = P(X\le x, Y \le y) fully describes the simultaneous behavior of a pair of random variables.
  • Discrete pairs are described by a joint PMF p(x,y)=P(X=x,Y=y)p(x,y) = P(X=x,Y=y) summing to 11 over all values.
  • Continuous pairs are described by a joint PDF f(x,y)≥0f(x,y) \ge 0 with ∬f=1\iint f = 1, integrated to get probabilities of regions.
  • Rectangle probabilities from a joint CDF are computed by a signed sum of four corner values (the "inclusion–exclusion" identity).
  • Any monotonicity or normalization failure means a candidate function is not a valid joint CDF/PMF/PDF.
  • Joint distributions are the raw material from which marginals, independence, and sums are later built.
  • Key idea: knowing each variable's own distribution is not enough — the joint distribution is a genuinely new object encoding their relationship.

Where this lives today

Joint distributions are the working substrate of essentially all multivariate reasoning in applied probability and statistics. In finance, the joint distribution of asset returns drives portfolio risk models, value-at-risk calculations, and copula-based dependence modeling. In engineering reliability, joint distributions of component lifetimes determine system-level failure probabilities in series and parallel configurations. In machine learning and statistics, the joint distribution of features and labels is the object that generative models attempt to learn, and Bayesian inference is fundamentally a manipulation of joint distributions of parameters and data via Conditional Distributions. In physics, joint distributions of position and momentum (subject to quantum constraints) and joint distributions of particle velocities in statistical mechanics rely on exactly the density/PDF formalism developed here. In signal processing and communications, joint distributions of noise samples at different times determine detection and estimation performance. Every one of these applications begins with the same three objects defined in this article — joint CDF, joint PMF, joint PDF — before moving on to marginals, independence, and sums.

Ask the tutor: How do I compute the marginal PDF of XX from a joint PDF fX,Y(x,y)f_{X,Y}(x,y), and how does this connect to checking independence?

DefinitionJoint Distribution of Random Variables

Let XX and YY be random variables defined on the same Probability Space (Ω,F,P)(\Omega, \mathcal F, P).

Joint CDF. The joint cumulative distribution function of XX and YY is the function FX,Y:R2→[0,1]F_{X,Y}: \mathbb R^2 \to [0,1] given by

FX,Y(x,y)  =  P(X≤x, Y≤y),(x,y)∈R2,F_{X,Y}(x,y) \;=\; P(X \le x,\ Y \le y), \qquad (x,y)\in\mathbb R^2,

where {X≤x,Y≤y}\{X\le x, Y\le y\} denotes the event {ω∈Ω:X(ω)≤x and Y(ω)≤y}\{\omega\in\Omega: X(\omega)\le x \text{ and } Y(\omega)\le y\}.

Joint PMF (discrete case). If XX and YY take values in a countable set, the joint probability mass function is

pX,Y(x,y)  =  P(X=x, Y=y).p_{X,Y}(x,y) \;=\; P(X=x,\ Y=y).

It satisfies pX,Y(x,y)≥0p_{X,Y}(x,y)\ge 0 for all (x,y)(x,y) and

∑x∑ypX,Y(x,y)=1,\sum_{x}\sum_{y} p_{X,Y}(x,y) = 1,

the sum being over all values x,yx,y that X,YX,Y can take.

Joint PDF (continuous case). XX and YY are jointly continuous if there exists a function fX,Y:R2→[0,∞)f_{X,Y}:\mathbb R^2\to[0,\infty), the joint probability density function, such that for every (Borel) set A⊆R2A \subseteq \mathbb R^2,

P((X,Y)∈A)=∬AfX,Y(x,y) dx dy,P\big((X,Y)\in A\big) = \iint_A f_{X,Y}(x,y)\, dx\, dy,

and in particular

FX,Y(x,y)=∫−∞x∫−∞yfX,Y(u,v) dv du,∬R2fX,Y(x,y) dx dy=1.F_{X,Y}(x,y) = \int_{-\infty}^{x}\int_{-\infty}^{y} f_{X,Y}(u,v)\, dv\, du, \qquad \iint_{\mathbb R^2} f_{X,Y}(x,y)\,dx\,dy = 1.

General nn variables. For random variables X1,…,XnX_1,\dots,X_n, the joint CDF is

FX1,…,Xn(x1,…,xn)=P(X1≤x1,…,Xn≤xn),F_{X_1,\dots,X_n}(x_1,\dots,x_n) = P(X_1\le x_1,\dots, X_n \le x_n),

with analogous joint PMF (discrete) or joint PDF (continuous, requiring ∫⋯∫Rnf=1\int\cdots\int_{\mathbb R^n} f = 1) definitions.

Intuition

Think of FX,Y(x,y)F_{X,Y}(x,y) as the probability mass that a "sensor" sweeping down-and-left from (x,y)(x,y) has accumulated: it is the total probability assigned to the entire lower-left quadrant with corner (x,y)(x,y). As (x,y)(x,y) moves to (+∞,+∞)(+\infty,+\infty), this sweeps up all the probability in the plane, so FX,Y(x,y)→1F_{X,Y}(x,y)\to 1. As (x,y)→(−∞,−∞)(x,y)\to(-\infty,-\infty) in either coordinate, the quadrant becomes empty and FX,Y→0F_{X,Y}\to 0. For discrete pairs, imagine placing a little tower of height pX,Y(x,y)p_{X,Y}(x,y) above every possible pair of values (x,y)(x,y) in the plane; the joint PMF is this whole "skyline" of towers, and their total volume (sum of heights) is exactly 11. For continuous pairs, replace towers with a continuous surface z=fX,Y(x,y)z=f_{X,Y}(x,y) hovering over the plane; probability of any region AA is the volume trapped between this surface and AA, and the total volume under the whole surface is 11. The picture below shows a discrete joint PMF as a grid of probability masses, and a continuous joint PDF as a density surface represented by shaded contour bands — both integrating (or summing) to unit total mass.

Figure
Left: a discrete joint PMF as a grid of point masses. Right: a continuous joint PDF as density contours, both normalized to total mass 1.
x y Joint PMF p(x,y) .10 .25 .08 .17 .30 .10 Σ p(x,y) = 1 Joint PDF f(x,y) x y ∬ f(x,y) dx dy = 1

The rectangle identity: extracting probabilities from a joint CDF

Before stating the central theorem, note the essential computational fact used throughout multivariate probability: the joint CDF alone determines the probability of any rectangle, via a four-term inclusion–exclusion formula. This is the two-dimensional analogue of the single-variable fact P(a<X≤b)=FX(b)−FX(a)P(a < X \le b) = F_X(b) - F_X(a).

TheoremRectangle Probability and Consistency of the Joint CDF

Let X,YX,Y be random variables on a common probability space (Ω,F,P)(\Omega,\mathcal F, P) with joint CDF FX,YF_{X,Y}.

(a) Rectangle formula. For all real numbers a1<a2a_1 < a_2 and b1<b2b_1< b_2,

P(a1<X≤a2, b1<Y≤b2)  =  FX,Y(a2,b2)−FX,Y(a1,b2)−FX,Y(a2,b1)+FX,Y(a1,b1).P(a_1 < X \le a_2,\ b_1 < Y \le b_2) \;=\; F_{X,Y}(a_2,b_2) - F_{X,Y}(a_1,b_2) - F_{X,Y}(a_2,b_1) + F_{X,Y}(a_1,b_1).

(b) Basic properties. FX,YF_{X,Y} is: 1. Monotone nondecreasing in each argument: if x1≤x2x_1\le x_2 then FX,Y(x1,y)≤FX,Y(x2,y)F_{X,Y}(x_1,y)\le F_{X,Y}(x_2,y) for every yy, and similarly in yy. 2. Right-continuous in each argument: lim⁡x↓x0FX,Y(x,y)=FX,Y(x0,y)\lim_{x\downarrow x_0} F_{X,Y}(x,y) = F_{X,Y}(x_0,y) for every yy, and similarly in yy. 3. Has correct limits:

lim⁡x→−∞FX,Y(x,y)=0=lim⁡y→−∞FX,Y(x,y),lim⁡x,y→∞FX,Y(x,y)=1.\lim_{x\to-\infty} F_{X,Y}(x,y) = 0 = \lim_{y\to -\infty}F_{X,Y}(x,y), \qquad \lim_{x,y\to\infty} F_{X,Y}(x,y) = 1.

4. Marginal consistency: lim⁡y→∞FX,Y(x,y)=FX(x)\lim_{y\to\infty} F_{X,Y}(x,y) = F_X(x) and lim⁡x→∞FX,Y(x,y)=FY(y)\lim_{x\to\infty} F_{X,Y}(x,y) = F_Y(y). 5. The right side of the rectangle formula in part (a) is always nonnegative (this is a genuinely new constraint, not implied by ordinary one-variable monotonicity — see the counterexample below).

Proof

We prove each part using elementary set operations and the axioms of probability (countable additivity and monotonicity of PP).

(a) Rectangle formula. Fix a1<a2a_1<a_2, b1<b2b_1<b_2. Define the events

A={X≤a2,Y≤b2},B={X≤a1,Y≤b2},C={X≤a2,Y≤b1},D={X≤a1,Y≤b1}.A = \{X\le a_2, Y\le b_2\}, \quad B=\{X\le a_1, Y\le b_2\}, \quad C=\{X\le a_2, Y\le b_1\}, \quad D = \{X\le a_1, Y\le b_1\}.

By definition P(A)=FX,Y(a2,b2)P(A)=F_{X,Y}(a_2,b_2), P(B)=FX,Y(a1,b2)P(B)=F_{X,Y}(a_1,b_2), P(C)=FX,Y(a2,b1)P(C)=F_{X,Y}(a_2,b_1), P(D)=FX,Y(a1,b1)P(D)=F_{X,Y}(a_1,b_1). Consider the event R={a1<X≤a2, b1<Y≤b2}R=\{a_1<X\le a_2,\ b_1<Y\le b_2\}, whose probability we want. We claim

A=R∪B∪C,with B∩C=D and R∩(B∪C)=∅.A = R \cup B \cup C, \qquad \text{with } B\cap C = D \text{ and } R \cap (B\cup C) = \varnothing.

Indeed, ω∈A\omega \in A means X(ω)≤a2X(\omega)\le a_2 and Y(ω)≤b2Y(\omega)\le b_2. Split on whether X(ω)≤a1X(\omega)\le a_1 or X(ω)>a1X(\omega)>a_1, and whether Y(ω)≤b1Y(\omega)\le b_1 or Y(ω)>b1Y(\omega)> b_1:

  • If X≤a1X\le a_1: then ω∈B\omega\in B (regardless of YY, since Y≤b2Y\le b_2 already holds).
  • If X>a1X> a_1 and Y≤b1Y\le b_1: then ω∈C\omega \in C (since X≤a2X\le a_2 holds).
  • If X>a1X>a_1 and Y>b1Y> b_1: combined with X≤a2, Y≤b2X\le a_2,\,Y\le b_2 this is exactly ω∈R\omega \in R.

These three cases are mutually exclusive and exhaust AA, confirming A=R⊔B′⊔C′A = R\sqcup B' \sqcup C' is not quite a clean disjoint union of B,CB,C themselves (since B,CB, C overlap), so we instead use inclusion–exclusion directly on B∪CB\cup C. By definition, B∪C={X≤a1 or Y≤b1}∩{X≤a2,Y≤b2}B\cup C = \{X\le a_1 \text{ or } Y \le b_1\} \cap \{X\le a_2, Y\le b_2\} (since B,C⊆AB,C\subseteq A), and this union together with RR partitions AA: indeed ω∈A∖(B∪C)\omega\in A\setminus(B\cup C) iff X(ω)>a1X(\omega) > a_1 and Y(ω)>b1Y(\omega)>b_1 (together with X≤a2,Y≤b2X\le a_2, Y\le b_2), i.e. iff ω∈R\omega \in R. Also R∩(B∪C)=∅R\cap(B\cup C)=\varnothing by construction. Hence, using finite additivity of PP,

P(A)=P(R)+P(B∪C).P(A) = P(R) + P(B\cup C).

By inclusion–exclusion for two events,

P(B∪C)=P(B)+P(C)−P(B∩C).P(B\cup C) = P(B) + P(C) - P(B\cap C).

Finally B∩C={X≤a1,Y≤b2}∩{X≤a2,Y≤b1}={X≤a1,Y≤b1}=DB\cap C = \{X\le a_1, Y\le b_2\}\cap\{X\le a_2,Y\le b_1\} = \{X\le a_1, Y\le b_1\} = D, since a1<a2a_1<a_2 forces X≤a1⇒X≤a2X\le a_1\Rightarrow X\le a_2, and b1<b2b_1<b_2 forces Y≤b1⇒Y≤b2Y\le b_1 \Rightarrow Y\le b_2. Substituting,

P(A)=P(R)+P(B)+P(C)−P(D),P(A) = P(R) + P(B)+P(C) - P(D),

so

P(R)=P(A)−P(B)−P(C)+P(D)=FX,Y(a2,b2)−FX,Y(a1,b2)−FX,Y(a2,b1)+FX,Y(a1,b1),P(R) = P(A) - P(B) - P(C) + P(D) = F_{X,Y}(a_2,b_2) - F_{X,Y}(a_1,b_2) - F_{X,Y}(a_2,b_1) + F_{X,Y}(a_1,b_1),

which is the claimed identity.

(b1) Monotonicity. If x1≤x2x_1\le x_2 then {X≤x1,Y≤y}⊆{X≤x2,Y≤y}\{X\le x_1,Y\le y\}\subseteq \{X\le x_2, Y\le y\} as sets (any ω\omega with X(ω)≤x1X(\omega)\le x_1 automatically has X(ω)≤x2X(\omega)\le x_2). Monotonicity of probability measures (E⊆F⇒P(E)≤P(F)E\subseteq F \Rightarrow P(E)\le P(F), itself immediate from countable additivity applied to F=E⊔(F∖E)F = E \sqcup (F\setminus E)) gives FX,Y(x1,y)≤FX,Y(x2,y)F_{X,Y}(x_1,y)\le F_{X,Y}(x_2,y). The argument in yy is symmetric.

(b2) Right-continuity. Fix x0,yx_0,y. Let xn↓x0x_n \downarrow x_0 be any sequence decreasing to x0x_0. The events En={X≤xn,Y≤y}E_n = \{X\le x_n, Y\le y\} form a decreasing sequence of events (En+1⊆EnE_{n+1}\subseteq E_n since xn+1≤xnx_{n+1}\le x_n, by the monotonicity argument above) with ⋂nEn={X≤x0,Y≤y}\bigcap_n E_n = \{X\le x_0, Y\le y\} — this intersection equality holds because X(ω)≤xnX(\omega)\le x_n for all nn iff X(ω)≤inf⁡nxn=x0X(\omega) \le \inf_n x_n = x_0 (using that the reals are Archimedean and xn↓x0x_n\downarrow x_0). By continuity of probability measures from above (a standard consequence of countable additivity: for decreasing En↓EE_n \downarrow E, P(En)→P(E)P(E_n)\to P(E)), we get FX,Y(xn,y)=P(En)→P(⋂nEn)=FX,Y(x0,y)F_{X,Y}(x_n,y) = P(E_n) \to P\left(\bigcap_n E_n\right) = F_{X,Y}(x_0,y). Since the sequence xn↓x0x_n\downarrow x_0 was arbitrary, lim⁡x↓x0FX,Y(x,y)=FX,Y(x0,y)\lim_{x\downarrow x_0}F_{X,Y}(x,y) = F_{X,Y}(x_0,y). The argument in yy is symmetric.

(b3) Limits. Let xn→−∞x_n\to -\infty. Then En={X≤xn,Y≤y}E_n = \{X\le x_n, Y\le y\} decreases to ⋂nEn\bigcap_n E_n. If ω∈⋂nEn\omega\in\bigcap_n E_n then X(ω)≤xnX(\omega)\le x_n for every nn, forcing X(ω)=−∞X(\omega) = -\infty, impossible since XX is real-valued; hence ⋂nEn=∅\bigcap_n E_n = \varnothing. By continuity of probability from above, FX,Y(xn,y)→P(∅)=0F_{X,Y}(x_n,y)\to P(\varnothing)=0. The case y→−∞y\to-\infty is symmetric. For the joint limit to 11: let xn,yn→∞x_n,y_n\to\infty; then En={X≤xn,Y≤yn}E_n=\{X\le x_n,Y\le y_n\} increases to ⋂nEncc=Ω\bigcap_n E_n^c{}^c = \Omega (every ω\omega eventually satisfies both inequalities since X(ω),Y(ω)X(\omega),Y(\omega) are finite real numbers), and continuity of probability from below gives FX,Y(xn,yn)→P(Ω)=1F_{X,Y}(x_n,y_n)\to P(\Omega)=1.

(b4) Marginal consistency. Let yn→∞y_n\to\infty. Then En={X≤x,Y≤yn}E_n = \{X\le x, Y\le y_n\} increases to {X≤x}\{X\le x\} (since for any ω\omega, eventually Y(ω)≤ynY(\omega)\le y_n, while the constraint X(ω)≤xX(\omega)\le x is unaffected). Continuity of probability from below gives FX,Y(x,yn)→P(X≤x)=FX(x)F_{X,Y}(x,y_n) \to P(X\le x) = F_X(x). The other marginal limit is symmetric.

(b5) Nonnegativity. The quantity in (a) equals P(R)P(R) for an actual event RR, and probabilities are nonnegative by the axioms of probability. ■\blacksquare

∎
CorollaryDiscrete and continuous specializations

If X,YX,Y are discrete with joint PMF pX,Yp_{X,Y}, then for any set A⊆R2A\subseteq \mathbb R^2,

P((X,Y)∈A)=∑(x,y)∈ApX,Y(x,y),P((X,Y)\in A) = \sum_{(x,y)\in A} p_{X,Y}(x,y),

and in particular FX,Y(x,y)=∑u≤x∑v≤ypX,Y(u,v)F_{X,Y}(x,y) = \sum_{u\le x}\sum_{v\le y} p_{X,Y}(u,v). If X,YX,Y are jointly continuous with density fX,Yf_{X,Y}, then wherever fX,Yf_{X,Y} is continuous,

∂2∂x ∂yFX,Y(x,y)=fX,Y(x,y),\frac{\partial^2}{\partial x\,\partial y} F_{X,Y}(x,y) = f_{X,Y}(x,y),

by the Fundamental Theorem of Calculus applied twice (differentiating the double integral in the definition, first in yy then in xx, using continuity of the integrand to justify differentiating under the integral sign).

Worked examples

Example 1 (discrete, asymmetric joint PMF). A repair shop receives, on a given day, XX = number of urgent tickets (0,1,20,1,2) and YY = number of routine tickets (0,10,1), with joint PMF given by the table

p(x,y)p(x,y)y=0y=0y=1y=1
x=0x=00.100.100.150.15
x=1x=10.200.200.250.25
x=2x=20.050.050.250.25

Check normalization: 0.10+0.15+0.20+0.25+0.05+0.25=1.000.10+0.15+0.20+0.25+0.05+0.25 = 1.00. Good — this is a valid joint PMF, since every entry is also ≥0\ge 0. Question 1: Find P(X≥1,Y=1)P(X \ge 1, Y=1). This is the sum of table entries with x∈{1,2}x\in\{1,2\}, y=1y=1:

P(X≥1,Y=1)=p(1,1)+p(2,1)=0.25+0.25=0.50.P(X\ge 1, Y=1) = p(1,1) + p(2,1) = 0.25 + 0.25 = 0.50.

Question 2: Find the joint CDF value FX,Y(1,0)F_{X,Y}(1, 0). By definition this is P(X≤1,Y≤0)=p(0,0)+p(1,0)=0.10+0.20=0.30P(X\le 1, Y\le 0) = p(0,0)+p(1,0) = 0.10+0.20 = 0.30. Question 3 (rectangle formula check): Compute P(0<X≤2, 0≤Y≤1)P(0 < X \le 2,\ 0 \le Y \le 1) two ways. Direct sum over the table with x∈{1,2}x\in\{1,2\} (since 0<X≤20<X\le2 excludes x=0x=0) and y∈{0,1}y \in\{0,1\}:

p(1,0)+p(1,1)+p(2,0)+p(2,1)=0.20+0.25+0.05+0.25=0.75.p(1,0)+p(1,1)+p(2,0)+p(2,1) = 0.20+0.25+0.05+0.25 = 0.75.

Via the rectangle formula with a1=0,a2=2,b1=−1,b2=1a_1=0,a_2=2,b_1=-1,b_2=1 (choosing b1=−1b_1=-1 so that b1<0≤Yb_1 < 0 \le Y is captured as Y≤b1Y \le b_1 excluding all mass, consistent with "Y≥0Y\ge 0" always holding here):

FX,Y(2,1)−FX,Y(0,1)−FX,Y(2,−1)+FX,Y(0,−1).F_{X,Y}(2,1) - F_{X,Y}(0,1) - F_{X,Y}(2,-1) + F_{X,Y}(0,-1).

Here FX,Y(2,1)=1F_{X,Y}(2,1)=1 (all mass), FX,Y(0,1)=p(0,0)+p(0,1)=0.25F_{X,Y}(0,1) = p(0,0)+p(0,1) = 0.25, FX,Y(2,−1)=0F_{X,Y}(2,-1)=0 and FX,Y(0,−1)=0F_{X,Y}(0,-1)=0 (no mass with Y<0Y<0). This gives 1−0.25−0+0=0.751 - 0.25 - 0 + 0 = 0.75, matching the direct computation. ✓.

Example 2 (continuous, asymmetric joint PDF). Let X,YX,Y have joint density

fX,Y(x,y)={c (x+2y),0<x<1, 0<y<2,0,otherwise.f_{X,Y}(x,y) = \begin{cases} c\,(x + 2y), & 0 < x < 1,\ 0<y<2, \\ 0, & \text{otherwise.} \end{cases}

Step 1: Find cc. Normalization requires

∫01∫02c(x+2y) dy dx=1.\int_0^1\int_0^2 c(x+2y)\,dy\,dx = 1.

Compute the inner integral in yy:

∫02(x+2y) dy=[xy+y2]02=2x+4.\int_0^2 (x+2y)\,dy = \left[xy + y^2\right]_0^2 = 2x + 4.

Then

∫01(2x+4) dx=[x2+4x]01=1+4=5.\int_0^1 (2x+4)\,dx = \left[x^2 + 4x\right]_0^1 = 1 + 4 = 5.

So c⋅5=1c\cdot 5 = 1, giving c=15c = \tfrac15. Step 2: Verify nonnegativity and support. On 0<x<1, 0<y<20<x<1,\ 0<y<2 we have x+2y>0x+2y>0, so fX,Y≥0f_{X,Y}\ge 0 everywhere, confirming this is a valid density. Step 3: Compute P(X<12,Y>1)P(X < \tfrac12, Y > 1).

P(X<12,Y>1)=∫01/2∫1215(x+2y) dy dx.P\Big(X<\tfrac12, Y>1\Big) = \int_0^{1/2}\int_1^{2} \tfrac15(x+2y)\,dy\,dx.

Inner integral:

∫12(x+2y) dy=[xy+y2]12=(2x+4)−(x+1)=x+3.\int_1^2 (x+2y)\,dy = \left[xy+y^2\right]_1^2 = (2x+4) - (x+1) = x+3.

Then

∫01/215(x+3) dx=15[x22+3x]01/2=15(18+32)=15⋅138=1340=0.325.\int_0^{1/2} \tfrac15(x+3)\,dx = \tfrac15\left[\tfrac{x^2}{2}+3x\right]_0^{1/2} = \tfrac15\left(\tfrac18 + \tfrac32\right) = \tfrac15\cdot\tfrac{13}{8} = \frac{13}{40} = 0.325.

Step 4: Find the joint CDF at (x,y)=(0.5,1)(x,y) = (0.5, 1) directly, cross-checking part of Step 3.

FX,Y(0.5,1)=∫00.5∫0115(u+2v) dv du.F_{X,Y}(0.5,1) = \int_0^{0.5}\int_0^1 \tfrac15(u+2v)\,dv\,du.

Inner integral: ∫01(u+2v) dv=u+1\int_0^1(u+2v)\,dv = u + 1. Then ∫00.515(u+1) du=15[u22+u]00.5=15(0.125+0.5)=15(0.625)=0.125\int_0^{0.5}\tfrac15(u+1)\,du = \tfrac15\left[\tfrac{u^2}2+u\right]_0^{0.5} = \tfrac15(0.125+0.5)=\tfrac15(0.625)=0.125. As a rectangle-formula sanity check, note

P(X<0.5,Y>1)=FX(0.5)−FX,Y(0.5,1),P(X<0.5, Y>1) = F_X(0.5) - F_{X,Y}(0.5,1),

where FX(0.5)=P(X<0.5)=∫00.5∫0215(x+2y) dy dxF_X(0.5) = P(X<0.5) = \int_0^{0.5}\int_0^2 \tfrac15(x+2y)\,dy\,dx. The inner integral is 2x+42x+4 (computed above with the general bound), so FX(0.5)=∫00.515(2x+4) dx=15[x2+4x]00.5=15(0.25+2)=15(2.25)=0.45F_X(0.5) = \int_0^{0.5}\tfrac15(2x+4)\,dx = \tfrac15[x^2+4x]_0^{0.5} = \tfrac15(0.25+2)=\tfrac15(2.25)=0.45. Then FX(0.5)−FX,Y(0.5,1)=0.45−0.125=0.325F_X(0.5) - F_{X,Y}(0.5,1) = 0.45 - 0.125 = 0.325, matching Step 3 exactly. ✓

CounterexampleWhy nonnegativity of the rectangle formula cannot be dropped

Property (b5) of the theorem — that F(a2,b2)−F(a1,b2)−F(a2,b1)+F(a1,b1)≥0F(a_2,b_2)-F(a_1,b_2)-F(a_2,b_1)+F(a_1,b_1)\ge 0 for all rectangles — is an extra requirement beyond ordinary coordinate-wise monotonicity (b1), right-continuity (b2), and correct limits (b3). To see that (b1)–(b3) do not imply (b5), consider the function

G(x,y)={1if x≥0 and y≥0 and x+y≥1,0otherwise.G(x,y) = \begin{cases} 1 & \text{if } x\ge 0 \text{ and } y \ge 0 \text{ and } x+y\ge 1, \\ 0 & \text{otherwise.} \end{cases}

Check coordinate-wise monotonicity: fixing y≥0y\ge 0, increasing xx can only move a point from the "0" region into the "1" region (since increasing xx only increases x+yx+y), never the reverse, so GG is nondecreasing in xx for fixed yy; symmetric argument for yy. One can also check GG is right-continuous in each variable separately (it is piecewise constant with the jump occurring in a right-continuous manner along x+y=1x+y=1), and G→0G\to 0 as either coordinate →−∞\to -\infty (in fact whenever x<0x<0 or y<0y<0, G=0G=0), G→1G\to 1 as both →∞\to \infty. Now test the rectangle formula on a1=0,a2=1,b1=0,b2=1a_1=0,a_2=1,b_1=0,b_2=1:

G(1,1)−G(0,1)−G(1,0)+G(0,0).G(1,1) - G(0,1) - G(1,0) + G(0,0).

Here G(1,1)=1G(1,1)=1 (since 1+1=2≥11+1=2\ge1), G(0,1)=1G(0,1) = 1 (since 0+1=1≥10+1=1\ge 1), G(1,0)=1G(1,0)=1 (since 1+0=1≥11+0=1\ge1), and G(0,0)=0G(0,0)=0 (since 0+0=0<10+0=0<1). So the rectangle sum is

1−1−1+0=−1<0.1 - 1 - 1 + 0 = -1 < 0.

A genuine joint CDF can never produce a negative value here, because that value must equal P(0<X≤1,0<Y≤1)≥0P(0<X\le 1, 0<Y\le 1) \ge 0 by the very proof of the rectangle formula (it literally is the probability of an event). Since GG violates nonnegativity of this particular rectangle combination, GG cannot be the joint CDF of any pair of random variables, even though it satisfies monotonicity, right-continuity, and the correct limits at infinity. This shows property (b5) is not automatic — it is a genuinely independent constraint that must be checked (or built into the construction, e.g. by starting from an actual PDF/PMF) whenever verifying that a candidate two-variable function is a legitimate joint CDF.

Scope, edge cases & common traps

SituationWhat to watch for
Mixed discrete/continuous pairNeither pure PMF nor pure PDF description applies; need the joint CDF or a mixed density/measure treatment.
Only marginals FX,FYF_X, F_Y knownThe joint CDF is not determined by the marginals alone (many joint distributions share the same marginals).
Candidate two-variable function proposed as a joint CDFMust check all of monotonicity, right-continuity, correct limits, and rectangle-nonnegativity — see counterexample.
Computing P(X≤x,Y≤y)P(X\le x, Y\le y) from a PDFIntegrate over the correct (often non-rectangular) region; do not assume independence of the bounds.
Support of the joint densityAlways state the region of (x,y)(x,y) where fX,Y>0f_{X,Y}>0; integrals outside this region contribute 00 automatically but must be tracked when setting limits.

The mathematician behind it

Origins · 1 minAndrei Nikolaevich Kolmogorov and how this was discovered
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Andrei Nikolaevich Kolmogorov
1903–1987 · Soviet Union, foundational work in measure-theoretic probability

Kolmogorov's 1933 monograph Grundbegriffe der Wahrscheinlichkeitsrechnung placed probability theory on a rigorous measure-theoretic footing, defining random variables as measurable functions on a probability space and joint distributions as the induced measures on product spaces Rn\mathbb R^n. This framework — built on countable additivity and the axioms he formalized — is precisely the machinery used above to prove continuity-from-above/below properties of joint CDFs. Kolmogorov also proved the extension theorem guaranteeing that consistent finite-dimensional joint distributions determine a well-defined stochastic process, a cornerstone result for the theory of sums and limiting behavior of sequences of random variables.

History
- 1900 — Hilbert's sixth problem calls for an axiomatic treatment of probability, motivating decades of foundational work.
- 1933 — Kolmogorov publishes the measure-theoretic axioms of probability, formally defining joint distributions as measures on product spaces.
- 1930s–1940s — Multivariate distribution theory (bivariate normal, multinomial) becomes standard tooling in statistics, following earlier bivariate correlation work by Galton and Pearson in the 1880s–1900s.
- 1950s — Joint density and convolution techniques become central to the emerging theory of stochastic processes.
- Present — Joint distributions underpin multivariate statistics, machine learning (joint likelihoods), and simulation methods across science and engineering.
What is the difference between a joint distribution and two separate (marginal) distributions?

Two marginal distributions describe XX and YY in isolation and say nothing about their relationship. A joint distribution describes the pair (X,Y)(X,Y) simultaneously, encoding how likely combinations of values are to occur together. Crucially, many different joint distributions can share the exact same pair of marginals — for instance, XX and YY could be independent, or could be perfectly correlated, and yet still have identical individual (marginal) distributions. Only the joint distribution can distinguish these cases, which is exactly why it is introduced as a separate, richer object rather than derived automatically from the marginals.

Can I always recover the joint PDF by differentiating the joint CDF?

Wherever the joint CDF FX,Y(x,y)F_{X,Y}(x,y) is twice continuously differentiable, yes: fX,Y(x,y)=∂2FX,Y/∂x ∂yf_{X,Y}(x,y) = \partial^2 F_{X,Y}/\partial x\,\partial y, by the corollary above. This follows from applying the Fundamental Theorem of Calculus twice to the double-integral definition of FX,YF_{X,Y}. However, if FX,YF_{X,Y} is not differentiable everywhere (e.g., mixed discrete-continuous distributions, or densities with jump discontinuities), this formula only holds at points of differentiability, and a genuinely careful treatment requires measure theory beyond the pointwise-density picture.

Is it true that FX,Y(x,y)=FX(x)FY(y)F_{X,Y}(x,y) = F_X(x) F_Y(y) in general?

No, this equality holds if and only if XX and YY are independent — it is not a general identity satisfied by every joint CDF. In general FX,Y(x,y)F_{X,Y}(x,y) can be anywhere between max⁡(0,FX(x)+FY(y)−1)\max(0, F_X(x)+F_Y(y)-1) and min⁡(FX(x),FY(y))\min(F_X(x),F_Y(y)), the Fréchet–Hoeffding bounds, depending on the dependence structure between XX and YY. Assuming the product formula without verifying independence is one of the most common errors when first working with joint distributions.

Does a joint PMF or PDF always exist for a pair of random variables?

Not necessarily as a PMF or PDF in the elementary sense described here. If XX is discrete and YY is continuous, neither a joint PMF nor a joint PDF (in the form given here) applies directly, since the pair is neither purely discrete nor purely continuous. More generally, some joint distributions are "singular," concentrating probability on lower-dimensional sets (like a line) with no density with respect to area. The joint CDF, however, always exists for any pair of random variables, which is why it is the most universally applicable of the three objects.

How do I check whether a proposed table of numbers is a legitimate joint PMF?

Two conditions are necessary and sufficient: every entry must be nonnegative, and the entries must sum to exactly 11 over all possible pairs of values. If either condition fails — a negative entry, or a total different from 11 — the table cannot represent an actual joint PMF of any pair of random variables. This is directly analogous to checking that a single-variable PMF's probabilities are nonnegative and sum to 11, just extended over the two-dimensional grid of value-pairs.

Why is the rectangle probability formula written with four terms rather than two?

Because a rectangle (a1,a2]×(b1,b2](a_1,a_2]\times(b_1,b_2] in the plane is not simply the difference of two "quadrant" events the way an interval (a,b](a,b] on the line is the difference of two half-lines. The joint CDF only gives probabilities of lower-left quadrants {X≤x,Y≤y}\{X\le x, Y\le y\}, and expressing a rectangle in terms of quadrants requires the inclusion–exclusion argument carried out in the proof, which naturally produces four corner terms with alternating signs.

How does this connect to what comes next — marginals, independence, sums?

The joint CDF/PMF/PDF is the raw object from which every other multivariate concept is derived. Marginal Distributions are obtained by "integrating out" or "summing out" one variable from the joint distribution. Independence of Random Variables is the special case where the joint distribution factors as a product of marginals. Sums of Random Variables and Convolution computes the distribution of X+YX+Y by integrating the joint density over the diagonal strips {x+y≤t}\{x+y \le t\}. None of these can even be stated without first fixing what a joint distribution is, which is exactly the role this article plays.

Before this

Where this leads