4. Random Vectors

1 The General Case: Joint cdf
2 The General Case: Marginal cdf
3 The General Case: Independence
4 Discrete Random Variables
5 Continuous Random Variables

1 The General Case: Joint cdf

Multivariate random variable: A $k -$ dimensional random variable (or: random vector) is a function with domain $S$ and codomain $R^{k} :$ $(X_{1}, \dots, X_{k}) : s \in S \to (X_{1} (s), \dots, X_{k} (s)) \in R^{k} .$ The function $(X_{1} (s), \dots, X_{k} (s))$ is usually written for simplicity as $(X_{1}, \dots, X_{k}) .$

Remark: If $k = 2$ we have the bivariate random variable or two dimensional random variable $(X, Y) : s \in S \to (X (s), Y (s)) \in R^{2} .$

Joint cumulative distribution function: Let $(X, Y)$ be a bivariate random variable. The real function of two real variables with domain $R^{2}$ and defined by $F_{X, Y} (x, y) = P (X \leq x, Y \leq y)$ is the joint cumulative distribution function of the two dimensional random variable $(X, Y) .$

Example 1.1 Random Experiment: Roll two different dice (one red and one green) and write down the number of dots on the upper face of each die.

Random vector: $(X_{r e d}, X_{g r e e n})$ , where $X_{i}$ is the of dots the $i$ die, with $i =$ green or red.

Some probabilities: $\begin{aligned} P (X_{r e d} = 2, X_{g r e e n} = 4) = \frac{1}{36} \\ P (X_{r e d} + X_{g r e e n} > 10) = \frac{1}{12} \\ P (\frac{X_{r e d}}{X_{g r e e n}} \leq 2) = P (\frac{X_{r e d}}{X_{g r e e n}} = 1) + P (\frac{X_{r e d}}{X_{g r e e n}} = 2) \\ = \frac{6}{36} + \frac{3}{36} = \frac{1}{4} \end{aligned}$

Example 1.2 Random experiment: Two different and fair coins are tossed once.

Random vector: $(X_{1}, X_{2})$ , where $X_{i}$ represents the number of heads obtained with coin $i$ , with $i = 1, 2$ .

Some probabilities: $\begin{aligned} P (X_{1} = 0, X_{2} = 0) & = P (X_{1} = 0, X_{2} = 1) = P (X_{1} = 1, X_{2} = 0) \\ = P (X_{1} = 1, X_{2} = 1) = \frac{1}{4} \\ P (X_{1} + X_{2} \geq 1) & = \frac{3}{4} \end{aligned}$

Properties of the joint cumulative distribution function:

$0 \leq F_{X, Y} (x, y) \leq 1$
$F_{X, Y} (x, y)$ is non decreasing with respect to $x$ and $y :$
- $Δ_{x} > 0 \Rightarrow F_{X, Y} (x + Δ x, y) \geq F_{X, Y} (x, y)$
- $Δ_{y} > 0 \Rightarrow F_{X, Y} (x, y + Δ y) \geq F_{X, Y} (x, y)$
$lim_{x \to - \infty} F_{X, Y} (x, y) = 0$ , $lim_{y \to - \infty} F_{X, Y} (x, y) = 0$ and

$lim_{x \to + \infty, y \to + \infty} F_{X, Y} (x, y) = 1$
$P (x_{1} < X \leq x_{2}, y_{1} < Y \leq y_{2}) = F_{X, Y} (x_{2}, y_{2}) - F_{X, Y} (x_{1}, y_{2}) - F_{X, Y} (x_{2}, y_{1}) + F_{X, Y} (x_{1}, y_{1}) .$
$F_{X, Y} (x, y)$ is right continuous with respect to $x$ and $y$ : $lim_{x \to a^{+}} F_{X, Y} (x, y) = F_{X, Y} (a, y)$ and $lim_{y \to b^{+}} F_{X, Y} (x, y) = F_{X, Y} (x, b) .$

Example 1.3 Random experiment: Two different and fair coins are tossed once.

Random vector: $(X_{1}, X_{2})$ , where $X_{i}$ represents the number of heads obtained with coin $i$ , with $i = 1, 2$ .

Joint cumulative distribution function: $F_{X_{1}, X_{2}} (x_{1}, x_{2}) = P (X_{1} \leq x_{1}, X_{2} \leq x_{2}) = {\begin{cases} 0, & x_{1} < 0 \\ 0, & x_{2} < 0 \\ \frac{1}{4}, & 0 \leq x_{1} < 1, 0 \leq x_{2} < 1 \\ \frac{1}{2}, & 0 \leq x_{1} < 1, x_{2} \geq 1 \\ \frac{1}{2}, & 0 \leq x_{2} < 1, 0 \leq x_{1} < 1 \\ 1, & x_{1} \geq 1, x_{2} \geq 1 \end{cases}$

2 The General Case: Marginal cdf

The (marginal) cumulative distribution functions of $X$ and $Y$ can be obtained form the Joint cumulative distribution functions of $(X, Y) :$

The Marginal cumulative distribution function of $X :$ $F_{X} (x) = P (X \leq x) = P (X \leq x, Y \leq + \infty) = lim_{y \to + \infty} F_{X, Y} (x, y) .$
The Marginal cumulative distribution function of $Y :$ $F_{Y} (y) = (Y \leq y) = P (X \leq + \infty, Y \leq y) = lim_{x \to + \infty} F_{X, Y} (x, y) .$

Remark: The joint distribution uniquely determines the marginal distributions, but the inverse is not true.

Example 2.1 Random experiment: Two different and fair coins are tossed once.

Random vector: $(X_{1}, X_{2})$ , where $X_{i}$ represents the number of heads obtained with coin $i$ , with $i = 1, 2$ .

Joint cululative distribution function: $F_{X_{1}, X_{2}} (x_{1}, x_{2}) = {\begin{cases} 0, & x_{1} < 0 or x_{2} < 0 \\ \frac{1}{4}, & 0 \leq x_{1} < 1, 0 \leq x_{2} < 1 \\ \frac{1}{2}, & 0 \leq x_{1} < 1, x_{2} \geq 1 \\ \frac{1}{2}, & 0 \leq x_{2} < 1, 0 \leq x_{1} < 1 \\ 1, & x_{1} \geq 1, x_{2} \geq 1 \end{cases}$ Marginal cumulative distribution function for $X_{1}$ : $F_{X_{1}} (x_{1}) = lim_{x_{2} \to + \infty} F_{X_{1}, X_{2}} (x_{1}, x_{2}) = {\begin{cases} 0, & x_{1} < 0 \\ \frac{1}{2}, & 0 \leq x_{1} < 1 \\ 1, & x_{1} \geq 1 \end{cases}$

Example 2.2 Let $(X, Y)$ be a jointly distributed random variable with CDF: $F_{X, Y} (x, y) = {\begin{cases} 1 - e^{- x} - e^{- y} + e^{- x - y}, & x \geq 0, y \geq 0 \\ 0, & x < 0, y < 0 \end{cases} .$ Marginal cumulative distribution function of the random variable $X$ is: $F_{X} (x) = {\begin{cases} 1 - e^{- x}, & x \geq 0 \\ 0, & x < 0 \end{cases} .$

Marginal cumulative distribution function of the random variable Y is: $F_{Y} (y) = {\begin{cases} 1 - e^{- y}, & y \geq 0 \\ 0, & y < 0 \end{cases} .$

3 The General Case: Independence

Definition: The jointly distributed random variables $X$ and $Y$ are said to be independent if and only if for any two sets $B_{1} \in R,$ $B_{2} \in R$ we have $P (X \in B_{1}, Y \in B_{2}) = P (X \in B_{1}) P (Y \in B_{2})$

Remark: Independence implies that $F_{X, Y} (x, y) = F_{X} (x) F_{Y} (y),$ for any $(x, y) \in R^{2} .$

Theorem: If $X$ and $Y$ are independent random variables and if $h (X)$ and $g (Y)$ are two functions of $X$ and $Y$ respectively, then the random variables $U = h (X)$ and $V = g (Y)$ are also independent random variables.

Example 3.1 Random experiment: Two different and fair coins are tossed once.

Random vector: $(X_{1}, X_{2})$ , where $X_{i}$ represents the number of heads obtained with coin $i$ , with $i = 1, 2$ .

Are these random variables independent? $F_{X_{1}} (x_{1}) = {\begin{cases} 0, & x_{1} < 0 \\ \frac{1}{2}, & 0 \leq x_{1} < 1 \\ 1, & x_{1} \geq 1 \end{cases}, F_{X_{2}} (x_{2}) = {\begin{cases} 0, & x_{2} < 0 \\ \frac{1}{2}, & 0 \leq x_{2} < 1 \\ 1, & x_{2} \geq 1 \end{cases}$

One can easily verify that $F_{X_{1}, X_{2}} (x_{1}, x_{2}) = F_{X_{1}} (x_{1}) \times F_{X_{1}} (x_{1})$ . $F_{X_{1}, X_{2}} (x_{1}, x_{2}) = {\begin{cases} 0, & x_{1} < 0 or x_{2} < 0 \\ \frac{1}{4}, & 0 \leq x_{1} < 1, 0 \leq x_{2} < 1 \\ \frac{1}{2}, & 0 \leq x_{1} < 1, x_{2} \geq 1 \\ \frac{1}{2}, & 0 \leq x_{2} < 1, 0 \leq x_{1} < 1 \\ 1, & x_{1} \geq 1, x_{2} \geq 1 \end{cases}$

Example 3.2 Let $(X, Y)$ be a jointly distributed random variable with CDF: $F_{X, Y} (x, y) = {\begin{cases} 1 - e^{- x} - e^{- y} + e^{- x - y}, & x \geq 0, y \geq 0 \\ 0, & x < 0, y < 0 \end{cases} .$ Marginal cumulative distribution function of the random variable $X$ and $Y$ are: $\begin{aligned} F_{X} (x) = lim_{y \to + \infty} F_{X, Y} (x, y) = {\begin{cases} 1 - e^{- x}, & x \geq 0 \\ 0, & x < 0 \end{cases} \\ F_{Y} (y) = lim_{x \to + \infty} F_{X, Y} (x, y) = {\begin{cases} 1 - e^{- y}, & y \geq 0 \\ 0, & y < 0 \end{cases} . \end{aligned}$

$X$ and $Y$ are independent random variables because: $F_{X, Y} (x, y) = F_{X} (x) F_{Y} (y)$

since $1 - e^{- x} - e^{- y} + e^{- x - y} = (1 - e^{- y}) (1 - e^{- x}) .$

4 Discrete Random Variables

4.1 Joint pmf

Let $D_{(X, Y)}$ be the set of of discontinuities of the joint cumulative distribution function $F_{(X, Y)} (x, y),$ that is $D_{(X, Y)} = {(x, y) \in R^{2} : P (X = x, Y = y) > 0}$

Definition: $(X, Y)$ is a two dimensional discrete random variable if and only if $\sum_{(x, y) \in D_{(X, Y)}} P (X = x, Y = y) = 1.$

Remark: As in the univariate case, a multivariate discrete random variable can take a finite number of possible values $(x_{i}, y_{j}),$ where $i = 1, 2, . . ., k_{1} and j = 1, 2, . . ., k_{2},$ where $k_{1}$ and $k_{2}$ are finite integers, or a countably infinite $(x_{i}, y_{j}),$ where $i = 1, 2, . . .$ and $j = 1, 2, . . .$ . For the sake of generality we consider the latter case. That is $D_{(X, Y)} = {(x_{i}, y_{j}), i = 1, 2, . . ., j = 1, 2, . . .}$

Joint probability mass function: If $X$ and $Y$ are discrete random variables, then the function given by $f_{X, Y} (x, y) = P (X = x, Y = y)$ for $(x, y) \in D_{(X, Y)}$ is called the joint probability mass function of $(X, Y)$ (joint pmf) or joint probability distribution of the random variables $X$ and $Y .$

Theorem: A bivariate function $f_{X, Y} (x, y)$ can serve as joint probability mass function of the pair of discrete random variables $X$ and $Y$ if and only if its values satisfy the conditions:

$f_{X, Y} (x, y) \geq 0$ for any $(x, y) \in R^{2}$
$\sum_{(x, y) \in D_{(X, Y)}} f_{X, Y} (x, y) = \sum_{i = 1}^{\infty} \sum_{j = 1}^{\infty} f_{X, Y} (x_{i}, y_{j}) = 1$

Remark: We can calculate any probability using this function. For instance $P ((x, y) \in B) = \sum_{(x, y) \in B} f_{X, Y} (x, y)$

Example 4.1 Let $X$ and $Y$ be the random variables representing the population of monthly wages of husbands and wives in a particular community. Say, there are only three possible monthly wages in euros: $0$ , $1000$ , $2000$ . The joint probability mass function is

	$X = 0$	$X = 1000$	$X = 2000$
$Y = 0$	$0.05$	$0.15$	$0.10$
$Y = 1000$	$0.10$	$0.10$	$0.30$
$Y = 2000$	$0.05$	$0.05$	$0.10$

The probability that a husband earns $2000$ euros and the wife earns $1000$ euros is given by $\begin{aligned} f_{X, Y} (2000, 1000) & = P (X = 2000, Y = 1000) \\ = 0.30 \end{aligned}$

4.2 Joint cdf

Joint cumulative distribution function: If $X$ and $Y$ are discrete random variables, the function given by $F_{X, Y} (x y) = \sum_{s \leq x} \sum_{t \leq y} f_{X, Y} (s, t)$ for $(x, y) \in R^{2}$ is called the joint distribution function or joint cumulative distribution of $X$ and $Y .$

4.3 Marginal pmf’s

Marginal probability distribution/function: If $Y$ and $X$ are discrete random variables and $f_{X, Y}$ is the value of their joint probability distribution at $(x, y)$ the function given by

$\begin{aligned} P (X = x) & = & {\begin{array}{cc} \sum_{y \in D_{y}} f (x, y) = \sum_{y \in D_{Y}} f_{X, Y} (x, y), & for x \in D_{x} \\ 0, & for x \notin D_{x} \end{array} . \\ P (Y = y) & = & {\begin{array}{cc} \sum_{x \in D_{x}} f (x, y) = \sum_{x \in D_{X}} f_{X, Y} (x, y), & for y \in D_{y} \\ 0, & for y \notin D_{y} \end{array} \end{aligned}$

are respectively is the Marginal probability distribution of the r.v. $X$ and $Y$ , where $D_{x}$ and $D_{y}$ are the range of $X$ and $Y$ respectively.

Example 4.2 $\begin{aligned} P (X = x) = & P (X = x, Y = 0) + P (X = x, Y = 1000) \\ + P (X = x, Y = 2000) \\ P (Y = y) = & P (X = 0, Y = y) + P (X = 1000, Y = y) \\ + P (X = 2000, Y = y) \end{aligned}$

Applying these formulas we have:

	$X = 0$	$X = 1000$	$X = 2000$	$f_{Y} (y)$
$Y = 0$	$0.05$	$0.15$	$0.10$	$0.3$
$Y = 1000$	$0.10$	$0.10$	$0.30$	$0.5$
$Y = 2000$	$0.05$	$0.05$	$0.10$	$0.2$
$f_{X} (x)$	$0.2$	$0.3$	$0.5$	$1$

$\begin{aligned} F_{X, Y} (1000, 1000) = P (X = 0, Y = 0) + P (X = 0, Y = 1000) \\ + P (X = 1000, Y = 0) + P (X = 1000, Y = 1000) \\ F_{X, Y} (0, 1000) = P (X = 0, Y = 0) + P (X = 0, Y = 1000) \end{aligned}$

4.4 Independence

Independence of random variables: Two discrete random variables $X$ and $Y$ are independent if and only if, for all $(x, y) \in D_{X, Y}$ , $P (X = x, Y = y) = P (X = x) P (Y = y) .$

	$X = 0$	$X = 1000$	$X = 2000$	$f_{Y} (y)$
$Y = 0$	$0.05$	$0.15$	$0.10$	$0.3$
$Y = 1000$	$0.10$	$0.10$	$0.30$	$0.5$
$Y = 2000$	$0.05$	$0.05$	$0.10$	$0.2$
$f_{X} (x)$	$0.2$	$0.3$	$0.5$	$1$

Are these two random variables independent?

$\begin{aligned} P (X = 2000, Y = 2000) = P (X = 2000) \times P (Y = 2000) = 0.1 \\ Is this sufficient to say that X and Y are independent? NO! \\ P (X = 0, Y = 0) = 0.05 but P (X = 0) P (Y = 0) = 0.06 \end{aligned}$ thus $X$ and $Y$ are not independent.

4.5 Conditional pmf’s

Conditional probability mass function of $Y$ given $X$ :A conditional probability function of a discrete random variable $Y$ given another discrete variable $X$ taking a specific value is defined as $\begin{aligned} f_{Y | X = x} (y) = & P (Y = y | X = x) = \frac{P (Y = y, X = x)}{P (X = x)} \\ = & \frac{f_{X, Y} (x, y)}{f_{X} (x)}, {} f_{X} (x) > 0, for a fixed x . \end{aligned}$ The conditional probability function of $X$ given $Y$ is defined by

$f_{X | Y = y} (x) = \frac{f_{X, Y} (x, y)}{f_{Y} (y)}, {} f_{Y} (y) > 0.$

Remarks:

The conditional probability functions satisfy all the properties of probability functions, and therefore $\sum_{i = 1}^{\infty} f_{Y | X} (y_{i}) = 1.$
If $X$ and $Y$ are independent $f_{Y | X = x} (y) = f_{y} (y)$ and $f_{X | Y = y} (x) = f_{X} (x)$

example Consider the joint probability function

	$X = 0$	$X = 1000$	$X = 2000$	$f_{Y} (y)$
$Y = 0$	$0.05$	$0.15$	$0.10$	$0.3$
$Y = 1000$	$0.10$	$0.10$	$0.30$	$0.5$
$Y = 2000$	$0.05$	$0.05$	$0.10$	$0.2$
$f_{X} (x)$	$0.2$	$0.3$	$0.5$	$1$

Compute $P (Y = y | X = 0)$ .

$\begin{aligned} P (Y = 0 | X = 0) & = & \frac{P (Y = 0, X = 0)}{P (X = 0)} = \frac{0.05}{0.2} = 0.25 \\ P (Y = 1000 | X = 0) & = & \frac{P (Y = 1000, X = 0)}{P (X = 0)} = \frac{0.1}{0.2} = 0.5 . \\ P (Y = 2000 | X = 0) & = & \frac{P (Y = 2000, X = 0)}{P (X = 0)} = \frac{0.05}{0.2} = 0.25 . \end{aligned}$

4.6 Conditional cdf’s

Definition: The conditional CDF of $Y$ given $X$ is defined by

$F_{Y | X = x} (y) = P (Y \leq y | X = x) = \sum_{y^{'} \in D_{Y} \land y^{'} \leq y} \frac{P (Y = y^{'}, X = x)}{P (X = x)}$ for a fixed $x$ , with $P (X = x) > 0$ .

Remark: It can be checked that $F_{Y | X = x}$ is indeed a CDF.

Exercice: Verify that $F_{Y | X = x}$ is non-decreasing and and $lim_{y \to + \infty} F_{Y | X = x} (y) = 1$ .

$\begin{aligned} 1) F_{Y | X = x} & (y + δ) - F_{Y | X = x} (y) \\ = \frac{P (Y \leq y + δ, X = x)}{P (X = x)} - \frac{P (Y \leq y, X = x)}{P (X = x)} \geq 0. \\ 2) lim_{y \to + \infty} & F_{Y | X = x} (y) = lim_{y \to + \infty} P (Y \leq y | X = x) \\ = lim_{y \to + \infty} & \frac{P (Y \leq y, X = x)}{P (X = x)} = \frac{P (Y \leq \infty, X = x)}{P (X = x)} = \frac{P (X = x)}{P (X = x)} = 1. \end{aligned}$

Example 4.3 Consider the conditional probability of $Y$ given that $X = 0$ previously deduced: $\begin{aligned} P (Y = 0 | X = 0) & = \frac{P (Y = 0, X = 0)}{P (X = 0)} = \frac{0.05}{0.2} = 0.25 \\ P (Y = 1000 | X = 0) & = \frac{P (Y = 1000, X = 0)}{P (X = 0)} = \frac{0.1}{0.2} = 0.5 . \\ P (Y = 2000 | X = 0) & = \frac{P (Y = 2000, X = 0)}{P (X = 0)} = \frac{0.05}{0.2} = 0.25 . \end{aligned}$

Then the conditional CDF of $Y$ given that $X = 0$ is

$F_{Y | X = 0} (y) = {\begin{cases} 0, & y < 0 \\ 0.25, & 0 \leq y < 1000 \\ 0.75, & 1000 \leq y < 2000 \\ 1, & y \geq 0 \end{cases} .$

5 Continuous Random Variables

5.1 Joint pdf and joint cdf

Definition: $(X, Y)$ is a two-dimensional continuous random variable with a joint cumulative distribution function $F_{X, Y} (x, y),$ if and only if $X$ and $Y$ are continuous random variables and there is a non-negative real function $f_{X, Y} (x, y),$ such that $F_{X, Y} (x, y) = P (X \leq x, Y \leq y) = \int_{- \infty}^{y} \int_{- \infty}^{x} f_{X, Y} (t, s) d t d s .$

The function $f_{X, Y} (x, y)$ is the joint (probability) density of $X$ and $Y$ .

Remark: Let $A$ be a set in the $R^{2}$ . Then, $P ((X, Y) \in A) = \int \int_{A} f_{X, Y} (t, s) d t d s .$

::: {.example} Joint probability density function of the two dimensional random variable $(P_{1}, S)$ where $P_{1}$ represents the price and $S$ the total sales (in 10000 units).

Joint probability density function: $f_{P_{1}, S} (p, s) = {\begin{cases} 5 p e^{- p s}, & 0.2 < p < 0.4, s > 0 \\ 0, & otherwise \end{cases}$

Joint cumulative distribution function: $\begin{aligned} F_{P_{1}, S} (p, s) & = P (P_{1} \leq p, S \leq s) \\ = {\begin{cases} 0, & p < 0.2 or s < 0 \\ - 1 + 5 p - 5 \frac{e^{- 0.2 s} - e^{- p s}}{s}, & 0.2 < p < 0.4, s \geq 0 \\ 1 - 5 \frac{e^{- 0.2 s} - e^{- 0.4 s}}{s}, & p \geq 0.4, s \geq 0 \end{cases} \end{aligned}$ To get the CDF we need to make the following computations:

If $p < 0.2 or s < 0$ , then $f_{P_{1}, S} (p, s) = 0$ and $\begin{aligned} P (P_{1} \leq p, S \leq s) = \int_{- \infty}^{p} \int_{- \infty}^{s} f_{P_{1}, S} (t, u) d t d u = 0 \end{aligned}$
If $0.2 < p < 0.4$ and $s \geq 0$ , then $\begin{aligned} P (P_{1} \leq p, S \leq s) = \int_{- \infty}^{p} \int_{- \infty}^{s} f_{P_{1}, S} (t, u) d t d u \\ = \int_{0.2}^{p} \int_{0}^{s} f_{P_{1}, S} (t, u) d t d u = - 1 + 5 p - 5 \frac{e^{- 0.2 s} - e^{- p s}}{s} \end{aligned}$
If $p \geq 0.4$ and $s \geq 0$ , then $\begin{aligned} P (P_{1} \leq p, S \leq s) = \int_{- \infty}^{p} \int_{- \infty}^{s} f_{P_{1}, S} (t, u) d t d u \\ = \int_{0.2}^{0.4} \int_{0}^{s} f_{P_{1}, S} (t, u) d t d u = 1 - 5 \frac{e^{- 0.2 s} - e^{- 0.4 s}}{s} \end{aligned}$

Theorem: A bivariate function can serve as a joint probability density function of a pair of continuous random variables $X$ and $Y$ if its values, $f_{X, Y} (x, y)$ , satisfy the conditions:

$f_{X, Y} (x, y) \geq 0,$ for all $(x, y) \in R^{2}$
$\int_{- \infty}^{+ \infty} \int_{- \infty}^{+ \infty} f_{X, Y} (x, y) d x d y = 1.$

Property: Let $(X, Y)$ be a bivariate random variable and $B \in R^{2}$ , then $P ((X, Y) \in B) = \int \int_{B} f_{X, Y} (x, y) d x d y .$

Example 5.1 Let $(X, Y)$ be a continuous bi-dimensional random variable with density function $f_{X, Y}$ given by $f_{X, Y} (x, y) = {\begin{cases} k x + y, & 0 < x < 1, 0 < y < 1 \\ 0, & otherwise \end{cases} .$ Find $k$ .

Solution: From the first condition, we know that $f_{X, Y} (x, y) \geq 0$ . Therefore $k \geq 0$ . Additionally, $\int_{- \infty}^{+ \infty} \int_{- \infty}^{+ \infty} f_{X, Y} (x, y) d x d y = 1.$ This is equivalent to $\int_{0}^{1} \int_{0}^{1} k x + y d x d y = 1 \Leftrightarrow \frac{1 + k}{2} = 1 \Leftrightarrow k = 1.$

::: {.example} Let $(X, Y)$ be a continuous bi-dimensional random variable with density function $f_{X, Y}$ given by $f_{X, Y} (x, y) = {\begin{cases} x + y, & 0 < x < 1, 0 < y < 1 \\ 0, & otherwise \end{cases} .$ Compute $P (X > Y)$ .

Solution: Firstly, we notice that

$\begin{aligned} P (X > Y) & = \int_{0}^{1} \int_{0}^{x} (x + y) d y d x \\ = \int_{0}^{1} \frac{3}{2} x^{2} d x = \frac{1}{2} \end{aligned}$

Properties: Let $(X, Y)$ be a continuous bivariate random variable. If $f_{X, Y}$ represents the density function of $(X, Y)$ and $F_{X, Y}$ represents respectively joint CDF of $(X, Y)$ . Then,

$f_{X, Y} (x, y) = \frac{\partial^{2} F_{X, Y} (x, y)}{\partial x \partial y} = \frac{\partial^{2} F_{X, Y} (x, y)}{\partial y \partial x}, almost everywhere.$

5.2 Marginal pdf’s

Marginal density functions of the random variable $X$ $f_{X} (x) = \int_{- \infty}^{+ \infty} f_{X, Y} (x, v) d v,$
Marginal density functions of the random variable $Y$ $f_{Y} (y) = \int_{- \infty}^{+ \infty} f_{X, Y} (u, y) d u .$

5.3 Marginal cdf’s

Marginal CDF of the random variable $X$ $F_{X} (x) = lim_{y \to + \infty} F_{X, Y} (x, y) = \int_{- \infty}^{x} \underset{= f_{X} (u)}{\underset{⏟}{\int_{- \infty}^{+ \infty} f_{X, Y} (u, y) d y}} d u,$
Marginal CDF of the random variable $Y$ $F_{Y} (y) = lim_{x \to + \infty} F_{X, Y} (x, y) = \int_{- \infty}^{y} \underset{= f_{Y} (v)}{\underset{⏟}{\int_{- \infty}^{+ \infty} f_{X, Y} (x, v) d v}} d x .$

Example 5.2 Joint density function: $f_{P, S} (p, s) = {\begin{cases} 5 p e^{- p s}, & 0.2 < p < 0.4, s > 0 \\ 0, & otherwise \end{cases}$

Marginal density function of $P$ : $\begin{aligned} f_{P} (p) & = \int_{- \infty}^{+ \infty} f_{P, S} (p, s) d s = {\begin{cases} 5 \underset{= 1}{\underset{⏟}{\int_{0}^{+ \infty} p e^{- p s} d s}}, & 0.2 < p < 0.4 \\ 0, & otherwise \end{cases} \\ = {\begin{cases} 5, & 0.2 < p < 0.4 \\ 0, & otherwise \end{cases} \end{aligned}$

Marginal cumulative distribution function: $\begin{aligned} F_{S} (s) & = lim_{p \to + \infty} F_{P, S} (p, s) = {\begin{cases} 0, & s < 0 \\ 1 - 5 \frac{e^{- 0.2 s} - e^{- 0.4 s}}{s^{2}}, & s \geq 0 \end{cases} \end{aligned}$

5.4 Conditional pdf’s

Definition: If $f_{X, Y} (x, y)$ is the joint probability density function of the continuous random variables $X$ and $Y$ and $f_{Y} (y)$ is the marginal density function of $Y$ , the function given by $f_{X | Y = y} (x) = \frac{f_{X, Y} (x, y)}{f_{Y} (y)}, x \in R (for fixed y), f_{Y} (y) > 0$ is the conditional probability density function of $X$ given ${Y = y} .$ Similarly if $f_{X} (x)$ is the marginal density function of $X$ $f_{Y | X = x} (y) = \frac{f_{X, Y} (x, y)}{f_{X} (x)}, y \in R (for fixed x), f_{X} (x) > 0$ is the conditional probability function of $Y$ given ${X = x}$ .

Remark: Note that $P (X \in B | Y = y) = \int_{B} f_{X | Y = y} (x) d x$ for any $B \subset R$ .

Example 5.3 $(X, Y)$ is a random vector with the following joint density function: $f_{X, Y} (x, y) = {\begin{array}{cc} (y + x) & for (x, y) \in (0, 1) \times (0, 1) \\ 0 & otherwise \end{array} .$

Conditional density function of $Y$ given that $X = x$ (with $x \in (0, 1)$ ):

$\begin{aligned} f_{X} (x) & = \int_{0}^{1} (y + x) d y \\ = x + \frac{1}{2} \end{aligned}$

$\begin{aligned} f_{Y | X = x} (y) & = \frac{f_{X, Y} (x, y)}{f_{X} (x)} \\ = \frac{x + y}{x + \frac{1}{2}}, y \in (0, 1) \end{aligned}$

Probability of $Y \geq 0.7 | X = 0.5$ $\begin{aligned} P (Y \geq 0.7 | X = 0.5) & = \int_{0.7}^{1} f_{Y | X = 0.5} (y) d y \\ = \int_{0.7}^{1} (y + 0.5) d y = 0.405. \end{aligned}$

Remark:

The conditional density functions of $X$ and $Y$ verify all the properties of a density function of a univariate random variable.
Note that we can always decompose a joint density function in the following way $f_{X, Y} (x, y) = f_{X} (x) f_{Y | X = x} (y) = f_{Y} (y) f_{X | Y = y} (x) .$

5.5 Independence

If $X$ and $Y$ are independent $f_{Y | X = x} (y) = f_{Y} (y)$ and $f_{X | Y = y} (x) = f_{X} (x) .$

Example 5.4 Consider the conditional density function of $Y$ given that $X = x$ (with $x \in (0, 1)$ ): $\begin{aligned} f_{Y | X = x} (y) & = \frac{f_{X, Y} (x, y)}{f_{X} (x)} = \frac{x + y}{x + \frac{1}{2}}, y \in (0, 1) . \end{aligned}$

$f_{Y | X = x}$ is indeed a density function: $f_{Y | X = x} (y) \geq 0 and \int_{0}^{1} \frac{x + y}{x + \frac{1}{2}} d y = 1.$

Example 5.5 Consider the conditional density function of $Y$ given that $X = x$ (with $x \in (0, 1)$ ) and the marginal density function of $Y$ . $\begin{aligned} f_{Y | X = x} (y) & = \frac{f_{X, Y} (x, y)}{f_{X} (x)} = \frac{x + y}{x + \frac{1}{2}}, y \in (0, 1) \\ f_{Y} (y) & = y + \frac{1}{2}, y \in (0, 1) . \end{aligned}$

The random variables are not independent because $f_{Y | X = x} (y) \neq f_{Y} (y), for some y \in (0, 1) .$

5.6 Conditional cdf’s

Definition: The conditional CDF of $Y$ given $X$ by defined by $F_{Y | X = x} (y) = \int_{- \infty}^{y} f_{Y | X = x} (s) d s = \int_{- \infty}^{y} \frac{f_{Y, X} (s, x)}{f_{X} (x)} d s$ for a fixed $x$ , with $f_{X} (x) > 0$ .

Remark: It can be checked that $F_{Y | X = x}$ is indeed a CDF.

Example 5.6 Consider the conditional density function of $Y$ given that $X = x$ (with $x \in (0, 1)$ ): $\begin{aligned} f_{Y | X = x} (y) & = \frac{f_{X, Y} (x, y)}{f_{X} (x)} = \frac{x + y}{x + \frac{1}{2}}, y \in (0, 1) . \end{aligned}$

For $x \in (0, 1)$ , the conditional cumulative density function is given by: $\begin{aligned} F_{Y | X = x} (y) = {\begin{cases} 0, & y < 0 \\ \frac{y (2 x + y)}{1 + 2 x}, & 0 \leq y < 1 \\ 1, & y \geq 1 \end{cases} \end{aligned}$

where, $\frac{y (2 x + y)}{1 + 2 x} = \int_{0}^{y} \frac{x + s}{x + \frac{1}{2}} d s$ .

Last updated on Jan 1, 0001

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