If two random variables \(X\) and \(Y\) are defined on the same sample space, the pair \((X,Y)\) is called a bivariate (two-dimensional) random variable. It can be discrete or continuous.
Let \(X\) and \(Y\) be two random variables defined on the sample space \(S\). The random vector \((X,Y)\) assigns to each outcome a point in the two-dimensional space \(R^{2}\), and for that reason \((X,Y)\) is called a two-dimensional random variable or a bivariate random variable.
Three standard examples:
In each case one observation produces two numbers, and the pair is what is being modelled — not either number on its own.
A bivariate distribution can be defined for two kinds of random variable, discrete and continuous. In each case there are three concepts:
Sections 2 to 5 below set out all three, for the discrete and the continuous case separately. Everything else on this page — independence, the distribution function, the Jacobian method — is built on them.
Two coins are tossed. \(X\) = number of heads on first coin, \(Y\) = total number of heads. \((X,Y)\) takes values (0,0),(0,1),(1,1),(1,2) — a bivariate discrete r.v.
\(X\) = height (cm), \(Y\) = weight (kg) of a randomly chosen person. Both are continuous, so \((X,Y)\) is a bivariate continuous r.v.
For a discrete bivariate r.v. \((X,Y)\), the joint PMF is
\[ p(x,y) \;=\; P(X = x,\; Y = y). \]Let \(X\) and \(Y\) be two discrete random variables taking the values \(x_{1}, x_{2},\ldots, x_{m}\) and \(y_{1}, y_{2},\ldots, y_{n}\) respectively. The function \(p_{ij}\) defined on \(X = x_{i}\) and \(Y = y_{j}\) is the joint probability mass function of \((X,Y)\). It is written \(P(X = x_{i},\, Y = y_{j})\), or \(P(X = x_{i} \cap Y = y_{j})\), or \(P(x_{i}, y_{j})\), and is laid out in the bivariate table below.
The table is the whole of the discrete case in one picture: the interior holds the joint probabilities, the right-hand column and the bottom row hold the two marginals, and the corner cell holds the total, which is 1.
| Y \ X | \(x_{1}\) | \(x_{2}\) | … | \(x_{i}\) | … | \(x_{m}\) | Marginals of \(Y\) |
|---|---|---|---|---|---|---|---|
| \(y_{1}\) | \(p_{11}\) | \(p_{21}\) | … | \(p_{i1}\) | … | \(p_{m1}\) | \(p_{\cdot 1}\) |
| \(y_{2}\) | \(p_{12}\) | \(p_{22}\) | … | \(p_{i2}\) | … | \(p_{m2}\) | \(p_{\cdot 2}\) |
| ⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ||
| \(y_{j}\) | \(p_{1j}\) | \(p_{2j}\) | … | \(p_{ij}\) | … | \(p_{mj}\) | \(p_{\cdot j}\) |
| ⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ||
| \(y_{n}\) | \(p_{1n}\) | \(p_{2n}\) | … | \(p_{in}\) | … | \(p_{mn}\) | \(p_{\cdot n}\) |
| Marginals of \(X\) | \(p_{1\cdot}\) | \(p_{2\cdot}\) | … | \(p_{i\cdot}\) | … | \(p_{m\cdot}\) | \(p_{\cdot\cdot} = 1\) |
Read the dot as “summed over”: \(p_{i\cdot}\) is row \(i\) of \(X\) summed over every \(y\), \(p_{\cdot j}\) is column \(j\) of \(Y\) summed over every \(x\), and \(p_{\cdot\cdot}\) is everything summed over both.
Given joint PMF table:
| X \ Y | 1 | 2 | 3 | Row sum (Marginal of X) |
|---|---|---|---|---|
| 0 | 0.10 | 0.20 | 0.10 | 0.40 |
| 1 | 0.20 | 0.30 | 0.10 | 0.60 |
| Col sum (Marg. Y) | 0.30 | 0.50 | 0.20 | 1.00 |
All entries ≥ 0 and total = 1; valid joint PMF.
\(p(x,y) = k(x+y),\; x = 1,2;\; y = 1,2,3.\)
\(\sum p = k[(1+1)+(1+2)+(1+3)+(2+1)+(2+2)+(2+3)] = k(2+3+4+3+4+5) = 21k = 1 \Rightarrow k = 1/21\).
For continuous \((X,Y)\), the joint PDF \(f(x,y)\) satisfies:
When the region \(A\) is the rectangle \(x \in (a,b)\), \(y \in (c,d)\), the double integral has explicit limits — this is the form used in every worked problem below:
\[ P(a < X < b,\; c < Y < d) \;=\; \int_{a}^{b}\!\!\int_{c}^{d} f(x,y)\,dx\,dy . \]\(f(x,y) = k(x+y),\; 0 \le x,y \le 1\).
\(\int_0^1 \int_0^1 k(x+y) dx dy = k \int_0^1 (1/2 + y) dy = k(1/2 + 1/2) = k = 1\).
Hence \(k = 1\) — the function \(f(x,y) = x+y\) on the unit square.
For \(f(x,y) = 4xy,\; 0 \le x \le 1,\; 0 \le y \le 1\):
\(P(X \le \tfrac12,\; Y \le \tfrac12) = \displaystyle\int_0^{1/2}\!\!\int_0^{1/2} 4xy\,dy\,dx = \int_0^{1/2} 4x\Big[\tfrac{y^2}{2}\Big]_0^{1/2} dx = \int_0^{1/2} \tfrac{x}{2}\,dx = \tfrac{1}{16}.\)
Shortcut: \(f = 4xy = (2x)(2y)\) factorises, so \(X\) and \(Y\) are independent and \(P = P(X\le\tfrac12)\,P(Y\le\tfrac12) = \tfrac14 \cdot \tfrac14 = \tfrac{1}{16}\).
The marginal distribution of one variable is obtained by summing (or integrating) the joint distribution over the other variable.
Marginal probability function of \(X\). Written \(P_X(x)\), or \(P(X = x_i)\), or \(p_{i\cdot}\). Fix \(X = x_i\) and let \(Y\) be anything at all. The event \(X = x_i\) is the union of the disjoint events \((X = x_i,\, Y = y_j)\) over \(j = 1,\ldots,n\), so the probabilities add:
\[ P(X = x_i) = P(X = x_i,\, Y = y_1) + P(X = x_i,\, Y = y_2) + \cdots + P(X = x_i,\, Y = y_n) \] \[ = p_{i1} + p_{i2} + \cdots + p_{in} \;=\; \sum_{j=1}^{n} p_{ij} \;=\; \sum_{j=1}^{n} P(x_i, y_j), \qquad i = 1, 2, \ldots, m. \]That is exactly the row sum in the bivariate table above.
Marginal probability function of \(Y\). Written \(P_Y(y)\), or \(P(Y = y_j)\), or \(p_{\cdot j}\). The same argument along a column gives
\[ P(Y = y_j) = p_{1j} + p_{2j} + \cdots + p_{mj} \;=\; \sum_{i=1}^{m} p_{ij} \;=\; \sum_{i=1}^{m} P(x_i, y_j), \qquad j = 1, 2, \ldots, n. \]Discrete. The marginal of \(X\) must satisfy
\[ (i)\;\; p_{i\cdot} = P(X = x_i) \ge 0 \;\;\forall\, i; \qquad (ii)\;\; \sum_{i=1}^{m} p_{i\cdot} = \sum_{i=1}^{m} P(X = x_i) = 1, \]and the marginal of \(Y\) must satisfy
\[ (i)\;\; p_{\cdot j} = P(Y = y_j) \ge 0 \;\;\forall\, j; \qquad (ii)\;\; \sum_{j=1}^{n} p_{\cdot j} = \sum_{j=1}^{n} P(Y = y_j) = 1. \]Continuous. The same two conditions, with the sum replaced by an integral:
\[ (i)\;\; f(x) \ge 0 \;\;\forall\, x; \qquad (ii)\;\; \int_{-\infty}^{\infty} f(x)\,dx = 1, \] \[ (i)\;\; f(y) \ge 0 \;\;\forall\, y; \qquad (ii)\;\; \int_{-\infty}^{\infty} f(y)\,dy = 1. \]So a marginal is a genuine one-variable distribution in its own right, which is what makes it worth computing.
From Example 1 of Section 2: marginal of \(X\) is \(p_X(0) = 0.40,\; p_X(1) = 0.60\). Marginal of \(Y\) is \(p_Y(1) = 0.30,\; p_Y(2) = 0.50,\; p_Y(3) = 0.20\).
For \(f(x,y) = x+y,\; 0 \le x,y \le 1\):
\(f_X(x) = \int_0^1 (x+y) dy = x + 1/2\) for \(0 \le x \le 1\).
\(f_Y(y) = \int_0^1 (x+y) dx = 1/2 + y\) for \(0 \le y \le 1\).
The conditional distribution of \(Y\) given \(X = x\) describes the probability of \(Y\) when \(X\) takes a fixed value.
In the discrete case the conditional probability function of \(Y\) given \(X\) is written \(P(Y = y_j \mid X = x_i)\) or \(P(y \mid x)\), and is defined as the joint over the marginal:
\[ P(Y = y_j \mid X = x_i) \;=\; \frac{P(X = x_i,\, Y = y_j)}{P(X = x_i)}, \qquad P(X = x_i \mid Y = y_j) \;=\; \frac{P(X = x_i,\, Y = y_j)}{P(Y = y_j)} . \]Each is itself a probability distribution, which is the content of the two properties:
\[ (i)\;\; P(Y = y_j \mid X = x_i) \ge 0 \;\;\forall\, j; \qquad (ii)\;\; \sum_{j=1}^{n} P(Y = y_j \mid X = x_i) = 1, \] \[ (i)\;\; P(X = x_i \mid Y = y_j) \ge 0 \;\;\forall\, i; \qquad (ii)\;\; \sum_{i=1}^{m} P(X = x_i \mid Y = y_j) = 1. \]In the continuous case the same two properties hold with integrals:
\[ (i)\;\; f(y \mid x) \ge 0 \;\;\forall\, y; \qquad (ii)\;\; \int_{-\infty}^{\infty} f(y \mid x)\,dy = 1, \] \[ (i)\;\; f(x \mid y) \ge 0 \;\;\forall\, x; \qquad (ii)\;\; \int_{-\infty}^{\infty} f(x \mid y)\,dx = 1. \]Property (ii) is the one to use as a check: divide a row of the table by its own row total and the row must come back to 1. Several of the worked problems below end with exactly that check.
Written without subscripts, the whole of sections 2 to 5 is four lines:
Joint: \(P(x,y)\). Marginals: \(P(x)\), \(P(y)\). Conditionals: \(P(x \mid y)\), \(P(y \mid x)\).
\[ \text{Marginals:}\qquad P(x) = \sum_{y} P(x,y), \qquad P(y) = \sum_{x} P(x,y) \] \[ \text{Conditionals:}\qquad P(x \mid y) = \frac{P(x,y)}{P(y)},\;\; P(y) > 0, \qquad P(y \mid x) = \frac{P(x,y)}{P(x)},\;\; P(x) > 0 \]For continuous variables, read \(f\) for \(P\) and \(\int\) for \(\sum\); nothing else changes.
From Section 2 Example 1, find conditional distribution of \(Y\) given \(X = 1\).
\(p_X(1) = 0.60\); \(p_{Y|X}(1|1) = 0.20/0.60 = 1/3,\; p_{Y|X}(2|1) = 0.30/0.60 = 1/2,\; p_{Y|X}(3|1) = 0.10/0.60 = 1/6\).
(They sum to 1 ✓.)
For \(f(x,y) = x+y\) on the unit square: \(f_X(x) = x + 1/2\).
\(f_{Y|X}(y|x) = (x+y)/(x+1/2)\) for \(0 \le y \le 1\).
Verify: \(\int_0^1 (x+y)/(x+1/2)\,dy = (x + 1/2)/(x+1/2) = 1\) ✓.
\(X\) and \(Y\) are independent if their joint distribution is the product of marginals:
Discrete: \(p(x,y) = p_X(x) \cdot p_Y(y)\) for all \(x,y\).
Continuous: \(f(x,y) = f_X(x) \cdot f_Y(y)\) for all \(x,y\).
Equivalently, the conditional equals the marginal: \(f_{Y|X}(y|x) = f_Y(y)\).
The definition above can be given the other way round, in terms of conditionals, and the two forms are equivalent. Let \(X\) and \(Y\) have joint density \(f(x,y)\), marginals \(f(x)\) and \(f(y)\), and conditionals \(f(x \mid y)\) and \(f(y \mid x)\). Then \(X\) and \(Y\) are stochastically independent if
\[ P(y) = P(y \mid x) \quad\text{and}\quad P(x) = P(x \mid y) \qquad \text{(discrete)}, \] \[ f(y) = f(y \mid x) \quad\text{and}\quad f(x) = f(x \mid y) \qquad \text{(continuous)}. \]Proof that this gives the product form. Let \(X\) and \(Y\) be continuous and suppose \(f(y \mid x)\) does not depend on \(x\). Start from the marginal of \(Y\) and write the joint as marginal times conditional:
\[ f(y) \;=\; \int_{-\infty}^{\infty} f(x,y)\,dx \;=\; \int_{-\infty}^{\infty} f(x)\, f(y \mid x)\,dx . \]Since \(f(y \mid x)\) does not depend on \(x\), it comes outside the integral:
\[ = f(y \mid x) \int_{-\infty}^{\infty} f(x)\,dx \;=\; f(y \mid x), \]the last step because \(\int_{-\infty}^{\infty} f(x)\,dx = 1\). Therefore
\[ f(y) = f(y \mid x) \qquad\text{whenever } f(y \mid x) \text{ does not depend on } x, \]which is what it means for \(X\) and \(Y\) to be independent. The joint density can then be written
\[ f(x,y) \;=\; f(x)\, f(y \mid x) \;=\; f(x)\, f(y) \qquad \text{(continuous)}, \]and in the same way \(P(x,y) = P(x)\,P(y)\) for the discrete case. The factorisation stated at the top of this section is therefore earned, not assumed — and it is the form actually used to test independence in the problems below.
| X \ Y | 1 | 2 | pX |
|---|---|---|---|
| 0 | 0.12 | 0.18 | 0.30 |
| 1 | 0.28 | 0.42 | 0.70 |
| pY | 0.40 | 0.60 | 1.00 |
Check: \(p(0,1) = 0.12 = 0.30 \times 0.40\); \(p(1,2) = 0.42 = 0.70 \times 0.60\). Yes — independent.
For \(f(x,y) = x + y\) on unit square: \(f_X(x) = x + 1/2\), \(f_Y(y) = y + 1/2\).
Product = \((x+1/2)(y+1/2)\) which is not equal to \(x+y\). Hence \(X,Y\) are not independent.
The joint distribution function of \(X\) and \(Y\), written \(F_{XY}(x,y)\) or \(F(x,y)\), is defined as \(F_{XY}(x,y) = P(X \le x,\, Y \le y)\). Writing that out:
For discrete variables,
\[ F(x,y) \;=\; \sum_{-\infty}^{x}\;\sum_{-\infty}^{y} P(X = x,\, Y = y). \]For continuous variables,
\[ F(x,y) \;=\; \int_{-\infty}^{x}\!\!\int_{-\infty}^{y} f(x,y)\,dx\,dy . \]The distribution function is the one object that covers both cases with the same definition, which is why the properties below need no separate discrete and continuous statements.
The same four, written as the textbook states them.
1. The rectangle formula. If \(a_1, a_2, b_1, b_2\) are real numbers, then
\[ P(a_1 < X \le b_1,\; a_2 < Y \le b_2) \;=\; F_{XY}(b_1, b_2) + F_{XY}(a_1, a_2) - F_{XY}(a_1, b_2) - F_{XY}(b_1, a_2). \]Two corners are added and two subtracted, because the two subtracted rectangles overlap in the corner at \((a_1, a_2)\) and that overlap has to be put back.
2. Limits. The joint distribution function lies between 0 and 1:
\[ 0 \le F(x,y) \le 1 . \]3. The four corner values.
\[ F(-\infty, -\infty) = 0, \qquad F(\infty, \infty) = 1, \] \[ F(-\infty, y) = 0, \qquad F(x, -\infty) = 0 . \]4. Recovering the density. If \(f(x,y)\) is the joint p.d.f. of \((X,Y)\), then
\[ f(x,y) \;=\; \frac{\partial^{2} F(x,y)}{\partial x\, \partial y} . \]Property 4 is the one that makes the distribution function a working tool rather than a definition: it is the exact inverse of the integration in Problem 2 below, which builds \(F(x,y)\) out of \(f(x,y)\) and could be differentiated straight back.
Marginal distribution function of \(X\), written \(F_X(x)\). Let \(Y\) run over everything:
\[ F_X(x) \;=\; P(X \le x) \;=\; P(X \le x,\; Y < \infty) \;=\; \sum_{y} P(X \le x,\, Y = y) \;=\; F(x, \infty). \]Its three properties:
\[ (1)\;\; P(a < X \le b) = F(b) - F(a); \qquad (2)\;\; 0 \le F(x) \le 1; \qquad (3)\;\; F(-\infty) = 0,\;\; F(\infty) = 1. \]Marginal distribution function of \(Y\), written \(F_Y(y)\) or \(F(y)\):
\[ F_Y(y) \;=\; P(Y \le y) \;=\; P(X < \infty,\; Y \le y) \;=\; \sum_{x} P(X = x,\, Y \le y) \;=\; F(\infty, y), \] \[ (1)\;\; P(c < Y \le d) = F(d) - F(c); \qquad (2)\;\; 0 \le F(y) \le 1; \qquad (3)\;\; F(-\infty) = 0,\;\; F(\infty) = 1. \]Pushing one argument to \(\infty\) is the distribution-function version of summing a row of the bivariate table: both say “let the other variable be anything”.
Conditional distribution function of \(Y\) given \(X\), written \(F_{Y|X}(y \mid x)\) or \(F(y \mid x)\):
\[ F_{Y|X}(y \mid x) \;=\; P(Y \le y \mid X = x), \] \[ (1)\;\; P(c < Y \le d \mid X) = F(d \mid x) - F(c \mid x); \qquad (2)\;\; 0 \le F(y \mid x) \le 1; \qquad (3)\;\; F(-\infty \mid x) = 0,\;\; F(\infty \mid x) = 1. \]Conditional distribution function of \(X\) given \(Y\), written \(F_{X|Y}(x \mid y)\) or \(F(x \mid y)\):
\[ F_{X|Y}(x \mid y) \;=\; P(X \le x \mid Y = y), \] \[ (1)\;\; P(a < X \le b \mid y) = F(b \mid y) - F(a \mid y); \qquad (2)\;\; 0 \le F(x \mid y) \le 1; \qquad (3)\;\; F(-\infty \mid y) = 0,\;\; F(\infty \mid y) = 1. \]Each conditional distribution function is an ordinary one-variable distribution function with the other variable held fixed, which is why its three properties are the familiar three.
For \(f(x,y) = 1\) on unit square: \(F(x,y) = xy\) for \(0 \le x,y \le 1\).
For \(f(x,y) = e^{-(x+y)},\; x,y \ge 0\):
\(F(x,y) = (1-e^{-x})(1-e^{-y})\). Marginals: each is Exp(1); they are independent (product form).
If \((X, Y)\) has joint pdf \(f_{X,Y}(x,y)\) and we define new variables \(U = g_1(X,Y),\ V = g_2(X,Y)\) through a one-to-one transformation with inverse \(x = h_1(u,v),\ y = h_2(u,v)\), then the joint pdf of \((U,V)\) is
where \(J\) is the Jacobian of the inverse transformation. The marginal of \(U\) (or \(V\)) is then obtained by integrating out the other variable.
Let \(X, Y\) be i.i.d. Exp\((\lambda)\), so \(f(x,y) = \lambda^2 e^{-\lambda(x+y)}\) for \(x,y \ge 0\). Put \(U = X + Y,\ V = X\). Then \(x = v,\ y = u - v\), so \(J = 1\) and
\(f_{U,V}(u,v) = \lambda^2 e^{-\lambda u}\) for \(0 \le v \le u\). Integrating over \(v\) from \(0\) to \(u\):
\[ f_U(u) = \int_0^{u}\lambda^2 e^{-\lambda u}\,dv = \lambda^2 u\, e^{-\lambda u}, \quad u \ge 0, \]which is the Gamma\((2, \lambda)\) density. Hence the sum of two i.i.d. exponentials is Gamma-distributed — a result the Jacobian method delivers cleanly.
Eight problems in the order the textbook sets them. Between them they use every definition in sections 2 to 7 at least once: finding a constant from the total-mass condition, the joint distribution function, both marginals, both conditionals, a probability over a slanted region, a conditional probability, and two independence checks that come out opposite ways.
Source note. These are numbered 1 to 8 here. The textbook numbers them 8 to 15, because its Problems 1 to 7 work the discrete bivariate tables. Every answer below has been recomputed independently; where the printed statement of a support interval is loose, it is corrected here and the original is footnoted.
The joint probability density function of \(X\) and \(Y\) is
\[ f(x,y) = \tfrac18\,(6-x-y), \qquad 0 < x < 2, \quad 2 < y < 4 . \]Find (i) \(P(X < 1 \cap Y < 3)\), (ii) \(P(X+Y < 3)\), (iii) \(P(X < 1 \mid Y < 3)\).
(i) Both limits are constants, so the region is a plain rectangle:
\[ P(X < 1 \cap Y < 3) = \int_{0}^{1}\!\!\int_{2}^{3} f(x,y)\,dy\,dx = \frac18 \int_{0}^{1}\!\!\int_{2}^{3} (6-x-y)\,dy\,dx \] \[ = \frac18 \int_{0}^{1}\left[6y - xy - \frac{y^{2}}{2}\right]_{2}^{3} dx = \frac18 \int_{0}^{1}\left(6 - x - \frac52\right) dx = \frac18 \int_{0}^{1}\left(\frac72 - x\right) dx \] \[ = \frac18\left[\frac72 x - \frac{x^{2}}{2}\right]_{0}^{1} = \frac18\left[\frac72 - \frac12\right] = \frac{3}{8} . \](ii) Now the upper limit of \(y\) is \(3-x\), and since \(y\) must also exceed 2 the constraint \(3-x > 2\) forces \(x < 1\):
\[ P(X+Y < 3) = \frac18 \int_{0}^{1}\!\!\int_{2}^{3-x} (6-x-y)\,dy\,dx = \frac18 \int_{0}^{1}\left[6y - xy - \frac{y^{2}}{2}\right]_{2}^{3-x} dx \] \[ = \frac18 \int_{0}^{1}\left[6(1-x) - x(1-x) - \left(\frac{9 + x^{2} - 6x}{2} - 2\right)\right] dx = \frac18 \int_{0}^{1}\left(\frac{x^{2}}{2} - 4x + \frac72\right) dx \] \[ = \frac18\left[\frac{x^{3}}{6} - 4\cdot\frac{x^{2}}{2} + \frac72 x\right]_{0}^{1} = \frac18\left(\frac16 - 2 + \frac72\right) = \frac18 \cdot \frac{10}{6} = \frac{5}{24} . \](iii) By the definition of a conditional probability,
\[ P(X < 1 \mid Y < 3) = \frac{P(X < 1 \cap Y < 3)}{P(Y < 3)} . \tag{1} \]The numerator is part (i). For the denominator, let \(x\) run over its whole range:
\[ P(Y < 3) = \frac18 \int_{0}^{2}\!\!\int_{2}^{3} (6-x-y)\,dy\,dx = \frac18 \int_{0}^{2}\left(\frac72 - x\right) dx = \frac18\left[\frac72 x - \frac{x^{2}}{2}\right]_{0}^{2} = \frac18(7-2) = \frac{5}{8} . \]Substituting into (1),
\[ P(X < 1 \mid Y < 3) = \frac{3/8}{5/8} = \frac{3}{5} . \]The joint p.d.f. of \(X\) and \(Y\) is \(f(x,y) = e^{-(x+y)},\ x \ge 0,\ y \ge 0\). (i) Find the joint distribution function. (ii) Find the marginal density functions. (iii) Find (a) \(P(X > 1)\), (b) \(P(X < Y \mid X < 2Y)\), (c) \(P(1 < X+Y < 2)\). (iv) Check the independence of the random variables.
(i) By the definition of the joint distribution function,
\[ F(x,y) = P(X \le x,\, Y \le y) = \int_{0}^{x}\!\!\int_{0}^{y} e^{-(x+y)}\,dy\,dx = \int_{0}^{x} e^{-x}\left\{\int_{0}^{y} e^{-y}\,dy\right\} dx \] \[ = \int_{0}^{x} e^{-x}\left[\frac{e^{-y}}{-1}\right]_{0}^{y} dx = \int_{0}^{x} e^{-x}\big(1 - e^{-y}\big)\,dx = (1-e^{-y})\left[\frac{e^{-x}}{-1}\right]_{0}^{x} \] \[ \therefore\quad F_{XY}(x,y) = (1 - e^{-x})(1 - e^{-y}) . \](ii) Integrate out the other variable:
\[ f_X(x) = \int_{0}^{\infty} e^{-(x+y)}\,dy = e^{-x}\left[\frac{e^{-y}}{-1}\right]_{0}^{\infty} = e^{-x}\big[-(0-1)\big] = e^{-x}, \quad x \ge 0, \] \[ f_Y(y) = \int_{0}^{\infty} e^{-(x+y)}\,dx = e^{-y}, \quad y \ge 0 . \]Each marginal is the standard exponential.
(iii)(a) Using the marginal,
\[ P(X > 1) = \int_{1}^{\infty} e^{-x}\,dx = \left[\frac{e^{-x}}{-1}\right]_{1}^{\infty} = -(0 - e^{-1}) = \frac{1}{e} . \]The joint p.d.f. gives the same answer, integrating \(y\) from \(0\) to \(\infty\) first.
(iii)(b) On the positive axis \(x < y\) implies \(x < 2y\), so the intersection of the two events is the smaller one:
\[ P(X < Y \mid X < 2Y) = \frac{P(X < Y \cap X < 2Y)}{P(X < 2Y)} = \frac{P(X < Y)}{P(X < 2Y)} . \tag{1} \] \[ P(X < Y) = \int_{0}^{\infty}\!\!\int_{0}^{y} e^{-(x+y)}\,dx\,dy = \int_{0}^{\infty} e^{-y}\big(1 - e^{-y}\big)\,dy = \int_{0}^{\infty}\big(e^{-y} - e^{-2y}\big)\,dy \] \[ = \left[\frac{e^{-y}}{-1} - \frac{e^{-2y}}{-2}\right]_{0}^{\infty} = -(0-1) + \left(0 - \frac12\right) = 1 - \frac12 = \frac12 . \] \[ P(X < 2Y) = \int_{0}^{\infty}\!\!\int_{0}^{2y} e^{-(x+y)}\,dx\,dy = \int_{0}^{\infty} e^{-y}\big(1 - e^{-2y}\big)\,dy = \int_{0}^{\infty}\big(e^{-y} - e^{-3y}\big)\,dy \] \[ = \left[\frac{e^{-y}}{-1} - \frac{e^{-3y}}{-3}\right]_{0}^{\infty} = -(0-1) + \left(0 - \frac13\right) = 1 - \frac13 = \frac23 . \]From (1),
\[ P(X < Y \mid X < 2Y) = \frac{1/2}{2/3} = \frac{3}{4} . \](iii)(c) The strip \(1 < x+y < 2\) meets the first quadrant in two pieces, so the integral splits at \(x = 1\):
\[ P(1 < X+Y < 2) = \int_{0}^{1}\!\!\int_{1-x}^{2-x} e^{-(x+y)}\,dy\,dx \;+\; \int_{1}^{2}\!\!\int_{0}^{2-x} e^{-(x+y)}\,dy\,dx \] \[ = \int_{0}^{1} e^{-x}\big[e^{x-1} - e^{x-2}\big]\,dx + \int_{1}^{2} e^{-x}\big[1 - e^{x-2}\big]\,dx \] \[ = \int_{0}^{1}\big(e^{-1} - e^{-2}\big)\,dx + \int_{1}^{2}\big(e^{-x} - e^{-2}\big)\,dx \] \[ = \big(e^{-1} - e^{-2}\big)\big[x\big]_{0}^{1} + \left[\frac{e^{-x}}{-1}\right]_{1}^{2} - e^{-2}\big[x\big]_{1}^{2} \] \[ = e^{-1} - e^{-2} + \big(e^{-1} - e^{-2}\big) - e^{-2} = 2e^{-1} - 3e^{-2} = \frac{2}{e} - \frac{3}{e^{2}} . \](iv) \(f(x)\,f(y) = e^{-x}\cdot e^{-y} = e^{-(x+y)} = f(x,y)\), so the joint factorises into the marginals and \(X\) and \(Y\) are stochastically independent. Note the joint distribution function in (i) factorises too, which is the same statement one level up.
The joint p.d.f. of \(X\) and \(Y\) is \(f(x,y) = 4xy\,e^{-(x^{2}+y^{2})},\ x \ge 0,\ y \ge 0\). Test whether \(X\) and \(Y\) are independent.
Marginal density of \(X\).
\[ f_X(x) = \int_{0}^{\infty} 4xy\,e^{-(x^{2}+y^{2})}\,dy = 4x\,e^{-x^{2}} \int_{0}^{\infty} y\,e^{-y^{2}}\,dy . \]Put \(y^{2} = t\), so \(2y\,dy = dt\), \(y\,dy = dt/2\), and the limits \(0 \le t < \infty\) are unchanged:
\[ = 4x\,e^{-x^{2}} \int_{0}^{\infty} e^{-t}\,\frac{dt}{2} = 2x\,e^{-x^{2}}\left[\frac{e^{-t}}{-1}\right]_{0}^{\infty} = 2x\,e^{-x^{2}}\big[-(0-1)\big] \] \[ \therefore\quad f_X(x) = 2x\,e^{-x^{2}}, \quad x \ge 0 . \]Marginal density of \(Y\). The same substitution with \(x^{2} = t\) gives
\[ f_Y(y) = \int_{0}^{\infty} 4xy\,e^{-(x^{2}+y^{2})}\,dx = 2y\,e^{-y^{2}}, \quad y \ge 0 . \]The test.
\[ f(x)\,f(y) = 2x\,e^{-x^{2}} \cdot 2y\,e^{-y^{2}} = 4xy\,e^{-(x^{2}+y^{2})} = f(x,y) . \]Since \(f(x,y) = f(x)\,f(y)\), the random variables \(X\) and \(Y\) are independent.
Worth noticing why this one works: the region is the whole quadrant, a product of two intervals, and the density is a product of a function of \(x\) alone and a function of \(y\) alone. When either of those two things fails — as in Problems 4, 5 and 7, where the region is a triangle — independence fails with it.
The joint p.d.f. of \(X\) and \(Y\) is \(f(x,y) = Kxy,\ 1 \le x \le y \le 2\). Find (i) \(K\), (ii) the marginals, (iii) the conditionals.
(i) By the definition of a joint p.d.f. the total mass is 1. Taking \(x\) from 1 to \(y\) and then \(y\) from 1 to 2,
\[ \int\!\!\int f(x,y)\,dx\,dy = 1 \quad\Longrightarrow\quad K\int_{1}^{2} y \left\{\int_{1}^{y} x\,dx\right\} dy = 1 \] \[ K\int_{1}^{2} y\left[\frac{x^{2}}{2}\right]_{1}^{y} dy = 1 \quad\Longrightarrow\quad \frac{K}{2}\int_{1}^{2} y\,(y^{2}-1)\,dy = 1 \quad\Longrightarrow\quad \frac{K}{2}\int_{1}^{2} (y^{3}-y)\,dy = 1 \] \[ \frac{K}{2}\left[\frac{y^{4}}{4} - \frac{y^{2}}{2}\right]_{1}^{2} = 1 \quad\Longrightarrow\quad \frac{K}{2}\left[\frac{15}{4} - \frac{3}{2}\right] = 1 \quad\Longrightarrow\quad \frac{K}{2}\cdot\frac94 = 1 \quad\therefore\quad K = \frac{8}{9} . \](ii) Marginal of \(X\). With \(x\) fixed, \(y\) runs from \(x\) to 2 (read the chain \(1 \le x < y \le 2\) from the middle outwards):
\[ f_X(x) = \int_{x}^{2} \frac89\,xy\,dy = \frac89\,x\left[\frac{y^{2}}{2}\right]_{x}^{2} = \frac89\,x\left(\frac{4-x^{2}}{2}\right) \;\Longrightarrow\; f_X(x) = \frac49\,x\,(4-x^{2}), \quad 1 \le x \le 2 . \]Marginal of \(Y\). With \(y\) fixed, \(x\) runs from 1 to \(y\):
\[ f_Y(y) = \int_{1}^{y} \frac89\,xy\,dx = \frac89\,y\left[\frac{x^{2}}{2}\right]_{1}^{y} = \frac89\,y\left(\frac{y^{2}-1}{2}\right) \;\Longrightarrow\; f_Y(y) = \frac49\,y\,(y^{2}-1), \quad 1 \le y \le 2 . \](iii) Joint over marginal, in each direction:
\[ f(y \mid x) = \frac{f(x,y)}{f(x)} = \frac{\frac89 xy}{\frac49 x (4-x^{2})} = \frac{2y}{4-x^{2}}, \qquad x < y < 2 \;\text{ and }\; 1 \le x \le 2, \] \[ f(x \mid y) = \frac{f(x,y)}{f(y)} = \frac{\frac89 xy}{\frac49 y (y^{2}-1)} = \frac{2x}{y^{2}-1}, \qquad 1 \le x < y \;\text{ and }\; 1 \le y \le 2 . \]The joint p.d.f. of \(X\) and \(Y\) is
\[ f(x,y) = \begin{cases} 8xy, & 0 < x < y < 1 \\ 0, & \text{otherwise.} \end{cases} \](i) Find the marginal p.d.f.s of \(X\) and \(Y\). (ii) Find the conditional p.d.f.s of \(X\) given \(Y\) and of \(Y\) given \(X\). (iii) Examine whether \(X\) and \(Y\) are independent. (iv) Find \(P\!\left(X < \tfrac12 \cap Y < \tfrac14\right)\) and \(P\!\left(X \le \tfrac14 \;\middle|\; \tfrac12 \le Y \le 1\right)\).
(i) Read the limits off the chain \(0 < x < y < 1\): for \(x\) fixed, \(x < y < 1\); for \(y\) fixed, \(0 < x < y\).
\[ f_X(x) = \int_{x}^{1} 8xy\,dy = 8x\left[\frac{y^{2}}{2}\right]_{x}^{1} = 4x\,(1-x^{2}), \quad 0 \le x \le 1, \] \[ f_Y(y) = \int_{0}^{y} 8xy\,dx = 8y\left[\frac{x^{2}}{2}\right]_{0}^{y} = 4y\cdot y^{2} = 4y^{3}, \quad 0 \le y \le 1 . \](ii)
\[ f(y \mid x) = \frac{8xy}{4x(1-x^{2})} = \frac{2y}{1-x^{2}}, \qquad x < y < 1 \;\text{ and }\; 0 \le x \le 1, \] \[ f(x \mid y) = \frac{8xy}{4y^{3}} = \frac{2x}{y^{2}}, \qquad 0 < x < y \;\text{ and }\; 0 \le y \le 1 . \](iii)
\[ f(x)\,f(y) = 4x(1-x^{2}) \cdot 4y^{3} = 16\,x\,(1-x^{2})\,y^{3} \neq 8xy = f(x,y) . \]So \(f(x,y) \neq f(x)\,f(y)\) and \(X\) and \(Y\) are not independent. The triangular region alone is enough to settle it: knowing \(Y = y\) restricts \(X\) to \((0,y)\).
(iv) Since \(x < y\) always, the condition \(y < \tfrac14\) already forces \(x < \tfrac14\), so the constraint \(X < \tfrac12\) adds nothing:
\[ P\!\left(X < \tfrac12 \cap Y < \tfrac14\right) = \int_{0}^{1/4}\!\!\int_{0}^{y} 8xy\,dx\,dy = 8\int_{0}^{1/4} y\left[\frac{x^{2}}{2}\right]_{0}^{y} dy = 4\int_{0}^{1/4} y^{3}\,dy = 4\left[\frac{y^{4}}{4}\right]_{0}^{1/4} = \frac{1}{256} . \]For the conditional probability,
\[ P\!\left(X \le \tfrac14 \;\middle|\; \tfrac12 \le Y \le 1\right) = \frac{P\!\left(X \le \tfrac14,\ \tfrac12 \le Y \le 1\right)} {P\!\left(\tfrac12 \le Y \le 1\right)} . \tag{1} \]Here \(x \le \tfrac14 < \tfrac12 \le y\), so \(x < y\) holds automatically and the two ranges are independent of one another as limits:
\[ P\!\left(X \le \tfrac14,\ \tfrac12 \le Y \le 1\right) = \int_{0}^{1/4}\!\!\int_{1/2}^{1} 8xy\,dy\,dx = 8\int_{0}^{1/4} x\left[\frac{y^{2}}{2}\right]_{1/2}^{1} dx \] \[ = 4\int_{0}^{1/4} x\left(1 - \frac14\right) dx = 3\left[\frac{x^{2}}{2}\right]_{0}^{1/4} = \frac32 \cdot \frac{1}{16} = \frac{3}{32} , \] \[ P\!\left(\tfrac12 \le Y \le 1\right) = \int_{1/2}^{1} 4y^{3}\,dy = 4\left[\frac{y^{4}}{4}\right]_{1/2}^{1} = 1 - \frac{1}{16} = \frac{15}{16} . \]From (1),
\[ P\!\left(X \le \tfrac14 \;\middle|\; \tfrac12 \le Y \le 1\right) = \frac{3/32}{15/16} = \frac{1}{10} . \]If the joint p.d.f. of \(X\) and \(Y\) is \(f(x,y) = 6(1-x-y),\ x > 0,\ y > 0\) and \(x+y < 1\), find the marginals.
Marginal of \(X\). The limits of \(y\): \(y > 0\) and \(x+y < 1\) give \(y < 1-x\), so \(0 < y < 1-x\).
\[ f_X(x) = \int_{0}^{1-x} 6(1-x-y)\,dy = 6\left[y - xy - \frac{y^{2}}{2}\right]_{0}^{1-x} \] \[ = 6\left[1-x-x(1-x)-\frac{(1-x)^{2}}{2}\right] = 6\left[x^{2}-2x+1-\frac{(1-x)^{2}}{2}\right] = 6\left[(x-1)^{2}-\frac{(x-1)^{2}}{2}\right] \] \[ \therefore\quad f_X(x) = 3(x-1)^{2}, \qquad 0 < x < 1 . \]Marginal of \(Y\). Symmetrically, \(x > 0\) and \(x+y < 1\) give \(0 < x < 1-y\):
\[ f_Y(y) = \int_{0}^{1-y} 6(1-x-y)\,dx = 6\left[x - \frac{x^{2}}{2} - yx\right]_{0}^{1-y} = 6\left[(1-y)^{2} - \frac{(1-y)^{2}}{2}\right] \] \[ \therefore\quad f_Y(y) = 3(1-y)^{2}, \qquad 0 < y < 1 . \]Check: \(\int_{0}^{1} 3(1-x)^{2}\,dx = \big[-(1-x)^{3}\big]_{0}^{1} = 1\), as a marginal must.
Printing note. The source gives the two supports as \(x \ge 0\) and \(y \ge 0\). They must be \(0 < x < 1\) and \(0 < y < 1\): over \(x \ge 0\) the integral of \(3(x-1)^{2}\) diverges, so it could not be a density. The formulae themselves are right, and \(3(x-1)^{2}\) and \(3(1-x)^{2}\) are the same function.
The joint p.d.f. of \(X\) and \(Y\) is \(f(x,y) = A\,e^{-x-y},\ 0 \le x \le y,\ 0 \le y < \infty\). (i) Determine \(A\). (ii) Find the marginals. (iii) Find the conditional p.d.f. of \(Y\) given \(X = 2\). (iv) Check the independence.
(i) The total mass is 1, with \(x\) running from 0 to \(y\):
\[ A\int_{0}^{\infty} e^{-y}\left\{\int_{0}^{y} e^{-x}\,dx\right\} dy = 1 \quad\Longrightarrow\quad A\int_{0}^{\infty} e^{-y}\big(1-e^{-y}\big)\,dy = 1 \] \[ A\int_{0}^{\infty}\big(e^{-y} - e^{-2y}\big)\,dy = 1 \quad\Longrightarrow\quad A\left[\frac{e^{-y}}{-1} - \frac{e^{-2y}}{-2}\right]_{0}^{\infty} = 1 \] \[ A\left[-(0-1) + \frac12(0-1)\right] = 1 \quad\Longrightarrow\quad A\cdot\frac12 = 1 \quad\therefore\quad A = 2 . \](ii) Marginal of \(X\). With \(x\) fixed, the chain \(0 \le x < y < \infty\) puts \(y\) above \(x\):
\[ f_X(x) = \int_{x}^{\infty} 2e^{-x-y}\,dy = 2e^{-x}\left[\frac{e^{-y}}{-1}\right]_{x}^{\infty} = 2e^{-x}\big[-(0-e^{-x})\big] = 2e^{-x}\cdot e^{-x} \] \[ \therefore\quad f_X(x) = 2e^{-2x}, \qquad 0 \le x < \infty . \]Marginal of \(Y\). With \(y\) fixed, \(0 \le x \le y\):
\[ f_Y(y) = \int_{0}^{y} 2e^{-x-y}\,dx = 2e^{-y}\left[\frac{e^{-x}}{-1}\right]_{0}^{y} = 2e^{-y}\big(1 - e^{-y}\big) \] \[ \therefore\quad f_Y(y) = 2\big[e^{-y} - e^{-2y}\big], \qquad 0 \le y < \infty . \](iii)
\[ f(y \mid x) = \frac{f(x,y)}{f(x)} = \frac{2e^{-x-y}}{2e^{-2x}} = e^{x-y}, \qquad x < y < \infty \;\text{ and }\; 0 \le x < \infty , \]so at \(X = 2\),
\[ f(y \mid x = 2) = e^{2-y}, \qquad 2 \le y < \infty . \](iv)
\[ f(x)\,f(y) = 2e^{-2x} \cdot 2\big(e^{-y} - e^{-2y}\big) = 4e^{-2x}\big[e^{-y} - e^{-2y}\big] \neq f(x,y) . \]Therefore \(f(x,y) \neq f(x)\,f(y)\) and \(X\) and \(Y\) are not independent. The conditional in (iii) says the same thing: it still contains \(x\), so it is not the marginal of \(Y\).
The joint p.d.f. of \(X\) and \(Y\) is \(f(x,y) = 2,\ 0 < x < 1,\ 0 < y < x\). (i) Find the marginal density functions of \(X\) and \(Y\). (ii) Find the conditional density functions of \(Y\) given \(X\) and of \(X\) given \(Y\). (iii) Find the marginal distribution functions of \(X\) and \(Y\).
(i) Limits of \(y\): \(0 < y < x\).
\[ f_X(x) = \int_{0}^{x} 2\,dy = 2\big[y\big]_{0}^{x} \;\Longrightarrow\; f_X(x) = 2x, \qquad 0 < x < 1 . \]Limits of \(x\): read \(0 < y < x < 1\) from the middle outwards, giving \(y < x < 1\).
\[ f_Y(y) = \int_{y}^{1} 2\,dx = 2\big[x\big]_{y}^{1} \;\Longrightarrow\; f_Y(y) = 2(1-y), \qquad 0 < y < 1 . \](ii)
\[ f(y \mid x) = \frac{f(x,y)}{f(x)} = \frac{2}{2x} = \frac{1}{x}, \qquad 0 < y < x \;\text{ and }\; 0 < x < 1, \] \[ f(x \mid y) = \frac{f(x,y)}{f(y)} = \frac{2}{2(1-y)} = \frac{1}{1-y}, \qquad y < x < 1 \;\text{ and }\; 0 < y < 1 . \]Both conditionals are flat — uniform on \((0,x)\) and on \((y,1)\) respectively — which is what a constant joint density on a region has to give.
(iii) Integrate each marginal density from the bottom of its range:
\[ F_X(x) = \int_{0}^{x} 2u\,du = \big[u^{2}\big]_{0}^{x} = x^{2}, \qquad 0 < x < 1, \] \[ F_Y(y) = \int_{0}^{y} 2(1-u)\,du = \big[2u - u^{2}\big]_{0}^{y} = 2y - y^{2}, \qquad 0 < y < 1, \]each taking the value 0 below its range and 1 above it. Differentiating them returns \(2x\) and \(2(1-y)\), which is property 4 of section 7 in one dimension.
Source note. Part (iii) above is worked here from the marginals found in part (i).