Joint Distributions #

Probability & StatisticsDifficulty: ★★★☆☆Depth: 6Unlocks: 11

Distributions over multiple random variables. Marginal and conditional.

Interactive Visualization #

⏮◀◀▶▶STEP0.25x1xZOOM

t=0s

Core Concepts #

-Joint distribution: a single function/measure assigning probability or density to each tuple of outcomes for multiple random variables
-Marginal distribution: the distribution of a subset of variables obtained by summing or integrating out the others
-Conditional distribution: the distribution of one variable (or subset) given fixed values of the others

Key Symbols & Notation #

p_{X,Y}(x,y) (joint pmf/pdf)p_{X|Y}(x|y) (conditional pmf/pdf)

Essential Relationships #

-Factorization/marginalization identity: p_{X,Y}(x,y) = p_{X|Y}(x|y) * p_Y(y); marginals obtained by summing/integrating the joint, and Bayes' rule follows by algebraic rearrangement

Prerequisites (2) #

Common Distributions6 atoms Multivariable Calculus6 atoms

Unlocks (2) #

Mutual Informationlvl 3 Covariance and Correlationlvl 3

Advanced Learning Details

Graph Position #

Depth Cost

Fan-Out (ROI)

Bottleneck Score

Chain Length

Cognitive Load #

Atomic Elements

Total Elements

Percentile Level

Atomic Level

All Concepts (16) #

- Joint probability mass function (joint PMF): probability assigned to each pair of discrete outcomes (X=x, Y=y).
- Joint probability density function (joint PDF): density over pairs of continuous values (X=x, Y=y).
- Joint cumulative distribution function (joint CDF): function giving P(X <= x and Y <= y).
- Support of a joint distribution: the set of value pairs (x,y) where the joint PMF/PDF is positive or nonzero.
- Normalization of joint distributions: the sum (discrete) or integral (continuous) of the joint over its support equals 1.
- Marginal distribution obtained by summing or integrating the joint over the other variable(s) to get the distribution of a subset.
- Conditional distribution of one variable given another, defined from the joint and marginal (for events with positive marginal probability/density).
- Marginalization as projection: computing marginals by integrating/summing out unwanted variables (viewed geometrically as projection onto axes).
- Law of total probability expressed via conditional distributions for computing marginals from conditionals.
- Bayes rule in the joint/conditional setting: updating between conditional and marginal distributions.
- Independence of random variables defined in terms of the joint distribution factorizing into the product of marginals.
- Consequences of independence: conditional equals marginal, factorization test, and simplifications for expectations and variances.
- Joint distributions over more than two variables and marginalizing over arbitrary subsets (multivariate joint distributions).
- Expectation of functions of multiple variables: computing E[g(X,Y,...)] via multiple sums/integrals of g times the joint.
- Relationship between joint CDF and joint PDF/PMF: differentiation (for continuous) or differences (for discrete) produce joint density/mass.
- Support constraints and conditional support: conditioning can restrict the domain where conditional distributions are defined.

Teaching Strategy #

Multi-session curriculum - substantial prior knowledge and complex material. Use mastery gates and deliberate practice.

Most real-world uncertainty is not about one variable at a time. Weather affects demand; demand affects price; price affects sales. Joint distributions are the language that lets you assign probabilities to these tuples of outcomes—and then extract the pieces you care about via marginals and conditionals.

TL;DR:

A joint distribution p_{X,Y}(x,y) assigns probability (discrete) or density (continuous) to pairs (x,y). Marginals come from summing/integrating out the other variable, e.g. p_X(x) = ∑_y p_{X,Y}(x,y) or f_X(x) = ∫ f_{X,Y}(x,y) dy. Conditionals come from “renormalizing a slice,” e.g. p_{X|Y}(x|y) = p_{X,Y}(x,y)/p_Y(y) (when p_Y(y) > 0). Independence is the special case p_{X,Y}(x,y) = p_X(x)p_Y(y).

What Is a Joint Distribution? #

Why we need it #

Single-variable distributions (Bernoulli, Poisson, Normal, …) answer questions like “How likely is X = 3?” But many systems involve multiple random variables simultaneously.

Examples:

•X = number of visitors to a website, Y = number of purchases.
•X = a patient’s test result, Y = underlying condition.
•X = temperature, Y = electricity demand.

To model relationships, we need a distribution over tuples of outcomes.

Definition (discrete) #

Let X and Y be discrete random variables. The joint pmf is

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

It must satisfy:

•Nonnegativity: p_{X,Y}(x,y) ≥ 0
•Normalization:

∑_x ∑_y p_{X,Y}(x,y) = 1

Think of p_{X,Y} as a table: rows are x values, columns are y values, and each cell is the probability of that pair.

Definition (continuous) #

If X and Y are continuous, we use a joint pdf f_{X,Y}(x,y) such that probability is obtained by integrating over regions:

P((X,Y) ∈ A) = ∬_A f_{X,Y}(x,y) dx dy.

Requirements:

•f_{X,Y}(x,y) ≥ 0
•∬_{ℝ²} f_{X,Y}(x,y) dx dy = 1

Crucial subtlety: For continuous variables, f_{X,Y}(x,y) is a density, not a probability. So P(X = x, Y = y) = 0 for exact points, even though f_{X,Y}(x,y) can be positive.

Support (where the mass/density lives) #

A joint distribution often only assigns nonzero probability/density on a particular set:

•Discrete: a set of pairs (x,y)
•Continuous: a region in the plane

For example, if X is uniform on [0,1] and Y = 1 − X, then (X,Y) lies only on the line y = 1 − x. This is a joint distribution, but it’s not a regular 2D pdf (it’s “singular” on a line). In this lesson we focus on standard pmfs and pdfs where the formulas below apply cleanly.

Joint CDF as an alternative #

Another way to specify the joint distribution is the joint CDF:

F_{X,Y}(x,y) = P(X ≤ x, Y ≤ y).

For continuous distributions, you can (when differentiable) recover the joint pdf via partial derivatives:

f_{X,Y}(x,y) = ∂²/∂x∂y F_{X,Y}(x,y).

This connects joint distributions to multivariable calculus: joint CDFs are like “accumulated volume,” joint pdfs are like “density per unit area.”

A mental model #

•Joint pmf/pdf: “How likely is each pair (x,y)?”
•Marginal: “What do I believe about X alone?”
•Conditional: “Given I learned Y = y, what do I believe about X now?”

Those last two are the real workhorses—and they both come directly from the joint.

Core Mechanic 1: Marginal Distributions (Sum/Integrate Out Variables) #

Why marginals matter #

Even if you build a full model of (X,Y), you frequently need statements about just one variable:

•“What is the distribution of demand X regardless of price Y?”
•“What’s the probability Y exceeds a threshold, regardless of X?”

Marginals let you ignore variables by aggregating over all their possible values.

Discrete marginals #

Starting from p_{X,Y}(x,y), the marginal pmfs are

p_X(x) = ∑_y p_{X,Y}(x,y)

p_Y(y) = ∑_x p_{X,Y}(x,y).

Interpretation: Fix x and add up the probabilities of all pairs (x,y) across all y.

This is literally the “row sum” (or “column sum”) of the joint table.

Continuous marginals #

From a joint pdf f_{X,Y}(x,y), the marginal pdfs are

f_X(x) = ∫_{−∞}^{∞} f_{X,Y}(x,y) dy

f_Y(y) = ∫_{−∞}^{∞} f_{X,Y}(x,y) dx.

Interpretation: Fix x and integrate the density along the vertical line at that x.

Regions and supports #

In practice, the integration limits often come from the support.

Example: Suppose f_{X,Y}(x,y) = c on the triangle 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x (and 0 otherwise). Then

•For a given x ∈ [0,1], y ranges from 0 to 1 − x.

f_X(x) = ∫_0^{1−x} c dy = c(1 − x), 0 ≤ x ≤ 1.

The lesson: marginalization is conceptually simple, but you must be careful about the geometry of where the joint density is nonzero.

A table viewpoint (discrete) #

If you have a joint pmf table, marginals are easy bookkeeping.

x \ y	y₁	y₂	y₃	p_X(x)
x₁	p(x₁,y₁)	p(x₁,y₂)	p(x₁,y₃)	row sum
x₂	p(x₂,y₁)	p(x₂,y₂)	p(x₂,y₃)	row sum
p_Y(y)	col sum	col sum	col sum	1

The bottom-right cell becomes 1 because the whole table sums to 1.

From marginals to probabilities #

Once you have marginals, you can compute 1D probabilities:

•Discrete: P(X ∈ S) = ∑_{x∈S} p_X(x)
•Continuous: P(a ≤ X ≤ b) = ∫_a^b f_X(x) dx

Marginalization generalizes #

For more variables, the idea is identical.

If you have p_{X,Y,Z}(x,y,z), then

p_{X,Y}(x,y) = ∑_z p_{X,Y,Z}(x,y,z)

and similarly for continuous with integrals.

This “sum/integrate out what you don’t need” pattern will show up everywhere later (Bayes nets, latent-variable models, expectation-maximization, variational inference).

Core Mechanic 2: Conditional Distributions (Normalize a Slice) #

Why conditionals matter #

Information changes beliefs. If you observe Y = y, your uncertainty about X should typically shrink or shift.

Conditional distributions formalize this update.

Discrete conditional pmf #

If p_Y(y) > 0, define

p_{X|Y}(x|y) = P(X = x | Y = y) = p_{X,Y}(x,y) / p_Y(y).

This is the key identity:

•The joint splits into “marginal × conditional”:

p_{X,Y}(x,y) = p_{X|Y}(x|y) p_Y(y).

Continuous conditional density #

For continuous variables, we use conditional densities (informally, “density of X given Y = y”):

f_{X|Y}(x|y) = f_{X,Y}(x,y) / f_Y(y), when f_Y(y) > 0.

And similarly

f_{X,Y}(x,y) = f_{X|Y}(x|y) f_Y(y).

“Normalize a slice” intuition #

Fix y.

•The function x ↦ f_{X,Y}(x,y) is a slice of the joint density.
•Its total mass in x is f_Y(y):

f_Y(y) = ∫ f_{X,Y}(x,y) dx.

So dividing by f_Y(y) forces the slice to integrate to 1:

∫ f_{X|Y}(x|y) dx

= ∫ f_{X,Y}(x,y)/f_Y(y) dx

= (1/f_Y(y)) ∫ f_{X,Y}(x,y) dx

= (1/f_Y(y)) f_Y(y)

= 1.

That’s what a conditional density is: the renormalized slice.

Bayes’ rule emerges immediately #

Once you accept

p_{X,Y}(x,y) = p_{X|Y}(x|y)p_Y(y) = p_{Y|X}(y|x)p_X(x),

you can equate them to get Bayes’ rule:

p_{X|Y}(x|y) = p_{Y|X}(y|x) p_X(x) / p_Y(y).

And the denominator is just a marginalization:

p_Y(y) = ∑_x p_{Y|X}(y|x)p_X(x) (discrete)

f_Y(y) = ∫ f_{Y|X}(y|x) f_X(x) dx (continuous).

So “Bayes” is not a separate topic from joint distributions—it’s a rearrangement of the joint.

Independence as a special structure #

X and Y are independent if learning Y tells you nothing about X:

p_{X|Y}(x|y) = p_X(x) for all x,y with p_Y(y) > 0.

Equivalent factorizations:

p_{X,Y}(x,y) = p_X(x)p_Y(y)

f_{X,Y}(x,y) = f_X(x)f_Y(y).

Independence is strong. Many variables are not independent, but may still be uncorrelated or weakly dependent. Later nodes (covariance/correlation, mutual information) quantify dependence in different ways.

Chain rule for more variables #

For three variables (X,Y,Z), the joint can be factored as

p_{X,Y,Z}(x,y,z)

= p_{X|Y,Z}(x|y,z) p_{Y|Z}(y|z) p_Z(z).

More generally, for X₁,…,X_n:

p(x₁,…,x_n) = ∏_{i=1}^n p(x_i | x₁,…,x_{i−1}).

This is not “assumptions”—it’s always true when the conditionals exist. Assumptions enter when you simplify the conditionals (e.g., Markov properties).

Application/Connection: Using Joints to Compute Probabilities, Expectations, and Dependence #

Computing probabilities of events #

With a joint distribution, probabilities of events about (X,Y) become sums/integrals over regions.

Discrete:

P((X,Y) ∈ A) = ∑_{(x,y)∈A} p_{X,Y}(x,y).

Continuous:

P((X,Y) ∈ A) = ∬_A f_{X,Y}(x,y) dx dy.

This is often the cleanest way to solve problems like P(X + Y ≤ 1) or P(X ≤ Y).

Expectation with a joint #

Many quantities of interest are functions g(X,Y). The expected value is

E[g(X,Y)] = ∑_x ∑_y g(x,y) p_{X,Y}(x,y) (discrete)

E[g(X,Y)] = ∬ g(x,y) f_{X,Y}(x,y) dx dy (continuous).

Special cases:

•E[X] = ∬ x f_{X,Y}(x,y) dx dy
•E[XY] = ∬ xy f_{X,Y}(x,y) dx dy

A key identity (law of total expectation) links conditionals back to marginals:

E[X] = E[E[X|Y]].

You can see it directly in the discrete case:

E[E[X|Y]]

= ∑_y E[X|Y=y] p_Y(y)

= ∑_y (∑_x x p_{X|Y}(x|y)) p_Y(y)

= ∑_y ∑_x x (p_{X,Y}(x,y)/p_Y(y)) p_Y(y)

= ∑_x ∑_y x p_{X,Y}(x,y)

= E[X].

This is one of the most useful “bridge” theorems in probability.

Dependence: beyond independence #

Independence is all-or-nothing. In practice you often ask: “How dependent are X and Y?”

Two major upcoming tools depend on joints:

Tool	What it measures	Needs from joints
Covariance/correlation	Linear relationship via E[XY] − E[X]E[Y]	Joint moments like E[XY]
Mutual information	Any statistical dependence via KL divergence / entropies	p_{X,Y}, p_X, p_Y

Joint distributions are the raw material those measures are made from.

A geometric view for continuous joints #

For a continuous f_{X,Y}:

•High density regions in the (x,y)-plane correspond to likely pairs.
•Contours (level sets) show the “shape” of dependence.
•If contours are circular (as in an isotropic Gaussian), dependence is weak.
•If contours are elongated along a diagonal, X and Y move together.

Even before computing correlation, you can often “see” dependence from the joint density.

Practical modeling patterns #

Start with a generative story (conditionals):

•Choose Y ~ p_Y
•Then choose X | Y=y ~ p_{X|Y}(

· | y)

This defines a valid joint via p_{X,Y}(x,y) = p_{X|Y}(x|y)p_Y(y).

Or start with a joint and derive everything else:

•Compute p_Y via marginalization
•Compute p_{X|Y} via division

Both are common. In machine learning, you often specify p(y) and p(x|y) (mixture models, Naive Bayes), and marginalize y to get p(x).

A note on vectors and multivariate distributions #

When you have many variables, you often package them into a random vector X ∈ ℝᵈ.

•The “joint distribution of d variables” is exactly the distribution of X.
•A marginal corresponds to selecting some coordinates of X.

You’ll later see multivariate normals, covariance matrices, and transformations of random vectors. Joint distributions are the foundation.

Worked Examples (3) #

Discrete joint pmf → marginals → conditional pmf #

Let X ∈ {0,1} and Y ∈ {0,1,2}. Suppose the joint pmf is:

p_{X,Y}(0,0)=0.10, p(0,1)=0.20, p(0,2)=0.10,

p_{X,Y}(1,0)=0.05, p(1,1)=0.25, p(1,2)=0.30.

(1) Compute p_X and p_Y.

(2) Compute p_{X|Y}(x|1).

(3) Are X and Y independent?

(1) Compute p_X(x) by summing over y.
p_X(0) = p(0,0)+p(0,1)+p(0,2)
= 0.10 + 0.20 + 0.10
= 0.40
p_X(1) = p(1,0)+p(1,1)+p(1,2)
= 0.05 + 0.25 + 0.30
= 0.60
(1) Compute p_Y(y) by summing over x.
p_Y(0) = p(0,0)+p(1,0) = 0.10 + 0.05 = 0.15
p_Y(1) = p(0,1)+p(1,1) = 0.20 + 0.25 = 0.45
p_Y(2) = p(0,2)+p(1,2) = 0.10 + 0.30 = 0.40
Check: 0.15+0.45+0.40 = 1.00 ✓
(2) Compute p_{X|Y}(x|1) = p_{X,Y}(x,1)/p_Y(1).
p_{X|Y}(0|1) = p(0,1)/p_Y(1) = 0.20 / 0.45 = 4/9 ≈ 0.444...
p_{X|Y}(1|1) = p(1,1)/p_Y(1) = 0.25 / 0.45 = 5/9 ≈ 0.555...
Check: 4/9 + 5/9 = 1 ✓
(3) Test independence using p_{X,Y}(x,y) ?= p_X(x)p_Y(y).
For (x,y)=(0,0):
Right side = p_X(0)p_Y(0) = 0.40·0.15 = 0.06
Left side = p(0,0) = 0.10
Not equal ⇒ not independent.

Insight: Marginals are just sums of the joint table. Conditionals are a “renormalized column” (fix y) or row (fix x). Independence fails as soon as any joint cell disagrees with the product of marginals.

Continuous joint pdf on a triangle → find constant and marginals #

Let (X,Y) have joint pdf f_{X,Y}(x,y) = c on the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x, and 0 otherwise.

(1) Find c.

(2) Find the marginal pdf f_X(x).

(3) Find P(X ≤ 1/2).

(1) Normalize the pdf: ∬ f_{X,Y}(x,y) dx dy = 1.
The region is: x from 0 to 1, and for each x, y from 0 to 1−x.
So:
1 = ∫_{x=0}^{1} ∫_{y=0}^{1−x} c dy dx
= ∫_0^1 c(1−x) dx
= c ∫_0^1 (1−x) dx
= c [x − x²/2]_0^1
= c (1 − 1/2)
= c/2
Therefore c = 2.
(2) Compute f_X(x) = ∫ f_{X,Y}(x,y) dy.
For x outside [0,1], f_X(x)=0.
For x ∈ [0,1], y ranges 0 to 1−x:
f_X(x) = ∫_0^{1−x} 2 dy
= 2(1−x), 0 ≤ x ≤ 1.
(3) Compute P(X ≤ 1/2) from f_X.
P(X ≤ 1/2) = ∫_0^{1/2} 2(1−x) dx
= 2 [x − x²/2]_0^{1/2}
= 2 (1/2 − (1/4)/2)
= 2 (1/2 − 1/8)
= 2 (3/8)
= 3/4.

Insight: In continuous problems, most difficulty is identifying the support correctly. Once the limits match the geometry, marginalization is straightforward calculus.

Derive a conditional density from a joint and interpret it #

Assume a joint pdf f_{X,Y}(x,y) = 2y for 0 < x < 1 and 0 < y < 1 (and 0 otherwise).

(1) Find f_Y(y).

(2) Find f_{X|Y}(x|y).

(3) Are X and Y independent?

(1) Marginalize out x:
f_Y(y) = ∫_{x=0}^{1} 2y dx
= 2y ∫_0^1 dx
= 2y, 0 < y < 1.
Check normalization:
∫_0^1 2y dy = [y²]_0^1 = 1 ✓
(2) Condition on Y=y:
f_{X|Y}(x|y) = f_{X,Y}(x,y)/f_Y(y)
= (2y)/(2y)
= 1, for 0 < x < 1.
So X|Y=y ~ Uniform(0,1).
(3) Test factorization:
f_X(x) = ∫_{y=0}^{1} 2y dy = 1, for 0 < x < 1.
So f_X(x)=1 on (0,1).
Now f_X(x)f_Y(y) = 1 · (2y) = 2y.
This equals f_{X,Y}(x,y) for 0<x<1, 0<y<1.
Therefore X and Y are independent.

Insight: A joint can look like it “depends on y,” yet still represent independence if it factors cleanly. The conditional f_{X|Y} revealing a constant 1 is a strong clue: X does not change when you learn Y.

Key Takeaways #

✓
A joint distribution p_{X,Y}(x,y) or f_{X,Y}(x,y) assigns probability mass/density to outcome pairs (x,y).
✓
Marginals come from summing/integrating out the other variable: p_X(x)=∑_y p_{X,Y}(x,y) and f_X(x)=∫ f_{X,Y}(x,y) dy.
✓
Conditionals are normalized slices: p_{X|Y}(x|y)=p_{X,Y}(x,y)/p_Y(y) (when p_Y(y)>0), and similarly for densities.
✓
The joint always factors as joint = conditional × marginal: p_{X,Y}=p_{X|Y}p_Y (and also = p_{Y|X}p_X).
✓
Bayes’ rule is a direct consequence of two ways to factor the same joint.
✓
Independence is equivalent to factorization: p_{X,Y}=p_X p_Y; equivalently p_{X|Y}=p_X.
✓
Probabilities of events about (X,Y) are sums/integrals over regions in the (x,y)-plane.
✓
Future tools like covariance/correlation and mutual information are computed from the joint (plus its marginals/conditionals).

Common Mistakes #

✗
Treating a pdf value f_{X,Y}(x,y) as a probability; probabilities require integrating over an area, not reading off a point value.
✗
Forgetting to respect the support when integrating (wrong bounds), especially for triangular or otherwise constrained regions.
✗
Dividing by p_Y(y) or f_Y(y) without checking it’s positive (conditioning on an event of probability 0 is subtle).
✗
Assuming uncorrelated implies independent; independence is stronger and must be checked via factorization or equivalent criteria.

Practice #

easy

Discrete: Suppose p_{X,Y}(x,y) is given by

p(0,0)=0.3, p(0,1)=0.1,

p(1,0)=0.2, p(1,1)=0.4.

Compute (a) p_X, (b) p_{Y|X}(1|0), (c) P(X=Y).

Hint: Marginals are row/column sums. Conditionals divide a joint cell by the corresponding marginal. P(X=Y) adds the diagonal cells.

Show solution

(a) p_X(0)=0.3+0.1=0.4, p_X(1)=0.2+0.4=0.6.

Also p_Y(0)=0.3+0.2=0.5, p_Y(1)=0.1+0.4=0.5.

(b) p_{Y|X}(1|0)=p(0,1)/p_X(0)=0.1/0.4=0.25.

medium

Continuous: Let f_{X,Y}(x,y) = k(x + y) on 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and 0 otherwise.

(a) Find k.

(b) Find f_X(x).

Hint: Use ∬ f = 1 to find k. Then f_X(x)=∫_0^1 k(x+y) dy. For E[X], use E[X]=∫ x f_X(x) dx.

Show solution

(a) Normalize:

1 = ∫_0^1 ∫_0^1 k(x+y) dy dx

= k ∫_0^1 ∫_0^1 (x+y) dy dx.

Compute inner integral:

∫_0^1 (x+y) dy = [xy + y²/2]_0^1 = x + 1/2.

Then

1 = k ∫_0^1 (x+1/2) dx

= k [x²/2 + x/2]_0^1

= k (1/2 + 1/2)

= k.

So k = 1.

(b) f_X(x)=∫_0^1 (x+y) dy = x + 1/2, 0≤x≤1.

= ∫_0^1 (x² + x/2) dx

= [x³/3 + x²/4]_0^1

= 1/3 + 1/4

= 7/12.

hard

Independence check: Let X be Uniform(0,1). Given X=x, let Y|X=x be Uniform(0,x).

(a) Write f_{Y|X}(y|x).

(b) Find the joint density f_{X,Y}(x,y).

Hint: Use f_{X,Y}(x,y)=f_{Y|X}(y|x)f_X(x). Watch the support: 0<y<x<1. Independence would require factoring into f_X(x)f_Y(y).

Show solution

(a) For x∈(0,1), Y|X=x is Uniform(0,x), so

f_{Y|X}(y|x)=1/x for 0<y<x, else 0.

(b) Since f_X(x)=1 on (0,1):

f_{X,Y}(x,y)=f_{Y|X}(y|x)f_X(x)= (1/x)·1 = 1/x on the region 0<y<x<1, else 0.

(c) Not independent. Intuitively, if you learn X is small, Y must be even smaller because 0<Y<X. Formally, the support is triangular (0<y<x<1), which already prevents factorization into a product of a function of x and a function of y over the full unit square.

Connections #

Next steps that build directly on this node:

•Covariance and Correlation — uses E[XY] and the joint to quantify linear dependence.
•Mutual Information — uses p_{X,Y} versus p_X p_Y to quantify any dependence.

Helpful refreshers:

•Common Distributions — joint models often use familiar marginals/conditionals.
•Multivariable Calculus — needed for double integrals and interpreting supports/regions.

Quality: A (4.5/5)

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