Vetores de variáveis aleatórias

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0.0 0.5 1.0 1.5
0.
0
0.
5
1.
0
1.
5
w1
w
2
Figure 7: The square represents Ω in Example 7.2.
7 Vectors of random variables
The previous sections focused on a single random variable. As a result, they didn’t study how one such number
can bring information about another one. This section focuses on vectors of random variables. In this case, the
dependency between each coordinate can be studied.
In order to avoid repetition, this section uses special notation. X = (X1, . . . , Xd) denotes a vector of random
variables, X1, . . . , Xd. Similarly, x = (x1, . . . , xd) denotes a vector of numbers in Rd. Also, X−i and x−i denote,
respectively, (X1, . . . , Xi−1, Xi+1, . . . , Xd) and (x1, . . . , xi−1, xi+1, . . . , xd). finally, Xi1 and xi1 denote, respectively,
(X1, . . . , Xi) and (x1, . . . , xi).
7.1 Cumulative distribution function
Definition 7.1. The cumulative distribution function (cdf) of X, FX(x) is such that, for every x ∈ Rd,
FX(x) := P(X1 ≤ x1, . . . , Xd ≤ xd)
Example 7.2. Let Ω = [0, 1]2. For each i ∈ {1, 2} and 0 < ai < bi < 1, let
P ({w ∈ Ω : w ∈ [a1, b1]× [a2, b2]}) = (b1 − a1) · (b2 − a2)
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0.0 0.2 0.4 0.6 0.8 1.0
0.
0
0.
2
0.
4
0.
6
0.
8
1.
0
w1
w
2
Figure 8: The line represents Ω in Example 7.3.
Define X1(w) = w1 and X2(w) = w2. Observe that
FX(x) = P (X1(w) ≤ x1, X2(w) ≤ x2) Definition 7.1
= P (w1 ≤ x1, w2 ≤ x2) Xi(w) = wi
= P (0 ≤ w1 ≤ min(1, x1), 0 ≤ w2 ≤ min(1, x2)) w ∈ [0, 1]2
=
0 , if x1 < 0 or x2 < 0min(1, x1) ·min(1, x2) , otherwise.
Example 7.3. Let Ω = {w ∈ [0, 1]2 : w1 = 1−w2}. For each 0 < a < b < 1, let P (a ≤ w1 ≤ b) = b− a. Also, let
X1(w) = w1 and X2(w) = w2. Observe that
FX(x) = P (X1(w) ≤ x1, X2(w) ≤ x2) Definition 7.1
= P (w1 ≤ x1, w2 ≤ x2) Xi(w) = wi
= P (w1 ≤ x1, 1− w1 ≤ x2) w1 = 1− w2
= P (1− x2 ≤ w1 ≤ x1)
= P (max(0, 1− x2) ≤ w1 ≤ min(1, x1)) w1 ∈ [0, 1]
=

0 , if x1 < 0 or x2 < 0
0 , if x1 < 1− x2
min(1, x1))−max(0, 1− x2) , otherwise.
Lemma 7.4. Some of the useful properties of a cdf are
1. For every x ∈ Rd, 0 ≤ FX(x) ≤ 1.
2. For every 1 ≤ i ≤ d, limxi→−∞ FX(x) = 0.
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3. For every 1 ≤ i ≤ d, limxi→∞ FX(x) = FX−i(x−i).
Example 7.5. Let (X1, X2) be such as in Example 7.2. Observe that limx2→∞ F(X1,X2)(x1, x2) = max(0,min(1, x1)).
That is, X1 has a uniform distribution over (0, 1). Similarly, limx1→∞ F(X1,X2)(x1, x2) = max(0,min(1, x2)).
Therefore, X2 has the same distribution as X2.
Similarly, let (X1, X2) be such as in Example 7.3. In this case, one also obtains limx2→∞ F(X1,X2)(x1, x2) =
max(0,min(1, x1)) and limx1→∞ F(X1,X2)(x1, x2) = max(0,min(1, x2)). That is, each of X1 and X2 have the same
marginal distributions in Example 7.2 and in Example 7.3.
However, (X1, X2) has a different distribution in Example 7.2 and in Example 7.3. Indeed, while in Example 7.2,
P (X1 = 1−X2) = 0, in Example 7.2 P (X1 = 1−X2) = 1.
7.2 Probability density function
Although the cdf has all the information about its corresponding vector of random variables, it can be cumbersome
to use it to calculate certain probabilities. It is often easier to use the probability density function, which is defined
in the following way:
Definition 7.6. The probability density function (pdf) of X, fX(x) is such that,
fX(x) =
P (X1 = x1, . . . , Xd = xd) , if X is discrete.dFX(x)
dx , if X is continuous.
Observe that, if X is continuous, then, for every x ∈ Rd, P (X1 = x1, . . . , Xd = xd) = 0. As a result, this
quantity isn’t useful for describing continuous vectors of random variables. In this case, the pdf is defined as
dFX(x)
dx . Intuitively,
dFX(x)
dx describes how much probability is concentrated in a region around x.
The pdf is useful for calculating probabilities regarding X. In order to show how to perform these calculations,
we will overload the meaning of
∫
A fX(x)dx. When X is discrete,
∫
A fX(x)dx will denote
∑
x∈A fX(x). When X
is continuous,
∫
A fX(x)dx will denote the usual Riemann integral of fX. As a result, we can obtain the following
general result:
Lemma 7.7.
P (X ∈ A) =
∫
A
fX(x)dx
Note that Lemma 7.7 can be used to obtain FX(x) from fX(x). Indeed,
FX(x) = P (X1 ≤ x1, . . . , Xd ≤ xd)
=
∫
∏d
i=1(−∞,xi]
fX(x)dx Lemma 7.7
Similarly, Definition 7.6 can be used to obtain fX(x) from FX(x). Indeed, each of FX(x) and fX(x) completely
describes X.
Example 7.8. Let (X,Y ) be a vector of continuous random variables such that
fX,Y (x, y) = cI(0 ≤ x ≤ 1)I(0 ≤ y ≤ 1)
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Lemma 7.7 can be used to compute the value of c. Observe that
1 = P ((X,Y ) ∈ R2) =
∫
R2
f(X,Y )(x, y)dxdy Lemma 7.7
=
∫
R2
cI(0 ≤ x ≤ 1)I(0 ≤ y ≤ 1)dxdy
=
∫ 1
0
∫ 1
0
cdxdy = c
Hence, c = 1. Also,
F(X,Y )(x, y) = P (X ≤ x, Y ≤ y)
= P ((X,Y ) ∈ (−∞, x]× (−∞, y])
=
∫
(−∞,x]×(−∞,y]
I(0 ≤ x ≤ 1)I(0 ≤ y ≤ 1)dxdy Lemma 7.7
=
0 , if x < 0 or y < 0min(1, x) min(1, y) , otherwise
Hence (X,Y ) has the same cdf as in Example 7.3
Example 7.9. Let (X,Y ) be a continuous vector of random variables such that
fX,Y (x, y) = cxyI(x ≥ 0)I(y ≥ 0)I(x+ y ≤ 1)
1 = P ((X,Y ) ∈ R2) =
∫
R2
f(X,Y )(x, y)dxdy Lemma 7.7
=
∫
R2
cxyI(x ≥ 0)I(y ≥ 0)I(x+ y ≤ 1)dxdy
=
∫ 1
0
∫ 1−x
0
cxydydx
= c
∫ 1
0
x
∫ 1−x
0
ydy
= c
∫ 1
0
x
(1− x)2
2
dx
=
c
2
∫ 1
0
(x− 2x2 + x3)dx
=
c
2
(
x2
2
− 2x
3
3
+
x4
4
)
∣∣∣∣1
0
=
c
2
(
1
2
− 2
3
+
1
4
)
=
c
2
· 1
12
=
c
24
Hence, c = 24. Also,
119
HHHHHHX
Y
0 1
0 c 0.4
1 0.3 0.1
Table 6: Table of f(X,Y )(x, y)
P (X < Y ) = P ((X,Y ) ∈ {(x, y) ∈ R2 : x < y})
=
∫
{(x,y)∈R2:x<y}
fX,Y (x, y)d(x, y) Lemma 7.7
=
∫ 1
0
∫ min(1−x,x)
0
24xydydx
=
∫ 0.5
0
∫ x
0
24xydydx+
∫ 1
0.5
∫ 1−x
0
24xydydx
=
∫ 0.5
0
24x
x2
2
dx+
∫ 1
0.5
24x
(1− x)2
2
dx
= 3x4
∣∣∣∣0.5
0
+
∫ 1
0.5
12(x− 2x2 + x3)dx
=
3
16
+ 12(
x2
2
− 2x
3
3
+
x4
4
)
∣∣∣∣1
0.5
=
3
16
+ 1− 3(1
2
− 1
3
+
1
16
) =
1
2
Example 7.10. Let (X,Y ) be a discrete vector of random variables with pdf given by table 6.
Observe that
1 = P ((X,Y ) ∈ {0, 1}2)
=
∫
R2
f(X,Y )(x, y)dydx Lemma 7.7
=
∑
x∈{0,1}
∑
y∈{0,1}
f(X,Y )(x, y)
= c+ 0.4 + 0.3 + 0.1
Hence, c = 0.2. Also,
P (X = 1) =
∫
{(x,y):x=1}
fX,Y (x, y)dydx Lemma 7.7
=
∑
y∈{0,1}
fX,Y (1, y) = 0.3 + 0.1 = 0.4
Conclude that X ∼ Bernoulli(0.4). Similarly, Y ∼ Bernoulli(0.5).
Finally, one can obtain the probability density function of every sub-vector in X by using Lemma 7.11.
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Lemma 7.11 (Law of total probability for random variables).∫ ∞
−∞
fX(x)dxi = fX−i(x−i)
For example, in order to obtain fX1 , one can integrate out every variable in fX(x) except for x1.
7.2.1 Exercises
Exercise 7.12. Let (X,Y ) be a continuous pair of random variables such that
f(X,Y ) = c · exp(−|x|) exp(−2y)I(y ≥ 0)
(a) Determine the value of c.
(b) Determine F(X,Y )(x, y).
(c) Determine FX(x) and FY (y).
(d) Find P (X < Y ).
Solution:
(a)
1 = P ((X,Y ) ∈ R2) =
∫
R2
f(X,Y )(x, y)dydx Lemma 7.7
=
∫ ∞
−∞
∫ ∞
0
c exp(−|x|) exp(−2y)dydx
= c
∫ ∞
−∞
−2−1 exp(−|x|) exp(−2y)
∣∣∣∣∞
0
dx
= c2−1
∫ ∞
−∞
exp(−|x|)dx
= c2−1
(∫ 0
−∞
exp(x) +
∫ ∞
0
exp(−x)
)
= c2−1
(
exp(x)
∣∣∣∣0
−∞
− exp(−x)
∣∣∣∣∞
0
)
= c
Therefore, c = 1.
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(b) If y < 0, then F(X,Y )(x, y) = 0. For y ≥ 0, let A = {(u, v) : u ≤ x, v ≤ y}. Obtain
F(X,Y )(x, y) = P (X ≤ x, Y ≤ y)
=
∫
A
f(X,Y )(x, y)dydx Lemma 7.7
=
∫ x
−∞
∫ y
0
exp(−|x|) exp(−2y)dydx
=
∫ x
−∞
exp(−|x|)dx ·
∫ y
−∞
exp(−2y)dy
=
(∫ min(0,x)
−∞
exp(x)dx+∫ x
min(0,x)
exp(−x)dx
)
·
∫ y
−∞
exp(−2y)dy
=
(
exp(x)
∣∣∣∣min(0,x)
−∞
− exp(−x)xmin(0,x)
)
· −2−1 exp(−2y)
∣∣∣∣y
0
= (exp(min(0, x)) + exp(−min(0, x))− exp(−x)) (2−1 − 2−1 exp(−2y))
=
exp(x)(2−1 − 2−1 exp(−2y)) , if x < 0(2− exp(−x))(2−1 − 2−1 exp(−2y)) , otherwise
(c)
FX(x) = lim
y→∞F(X,Y )(x, y) Lemma 7.4.3
=
2−1 exp(x) , if x < 01− exp(−x) , otherwise
FY (y) = lim
x→∞F(X,Y )(x, y) Lemma 7.4.3
= 1− exp(−2y)
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(d) Let A = {(x, y) ∈ R2 : x < y}.
P (X < Y ) = P ((X,Y ) ∈ A)
=
∫
A
f(X,Y )(x, y)dydx Lemma 7.7
=
∫ ∞
−∞
∫ ∞
max(0,x)
exp(−|x|) exp(−2y)dydx
=
∫ ∞
−∞
−2−1 exp(−|x|) exp(−2y)
∣∣∣∣∞
max(0,x)
dx
=
∫ ∞
−∞
2−1 exp(−|x|) exp(−2 max(0, x))dx
= 2−1
(∫ 0
−∞
exp(x)dx+
∫ ∞
0
exp(−x) exp(−2x)dx
)
= 2−1
(
exp(x)
∣∣∣∣0
−∞
− 3−1 exp(−3x)
∣∣∣∣∞
0
)
= 2−1(1 + 3−1) =
2
3
Exercise 7.13. Let (X,Y ) be a continuous pair of random variables such that
f(X,Y ) = c sin(x+ y)I(0 ≤ x ≤ 2−1pi)I(0 ≤ y ≤ 2−1pi)
(a) Determine the value of c.
(b) Determine F(X,Y )(x, y).
(c) Determine FX(x) and FY (y).
(d) Find P (X < Y ).
Exercise 7.14 (Ross (2015; p.244)). Let (X,Y ) be continuous pair of random variables with pdf f(X,Y )(x, y) =
cI(x2 + y2 < 1).
(a) Find the value of c.
(b) Compute P (X2 + Y 2 ≤ k), for each k ∈ R.
Exercise 7.15 (Ross (2015; p.290)). Let (X,Y ) be a continuous pair of random variables such that f(X,Y )(x, y) =
c(y2 − x2) exp(−y)I(x < |y|)I(y > 0).
(a) Determine the value of c.
(b) Determine F(X,Y )(x, y).
(c) Determine FX(x) and FY (y).
(d) Determine fX(x) and fY (y).
Exercise 7.16 (Ross (2015; p.255)). Let (X,Y, Z) be a continuous triple of random variables with pdf f(X,Y,Z) =
I(0 ≤ x, y, z ≤ 1). For each k ∈ R, find P (X ≤ kY Z).
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Exercise 7.17. Let (X,Y ) be a discrete pair of random variables such that f(X,Y )(x, y) = cxyI((x, y) ∈ {1, . . . , 10}2).
(a) Determine the value of c.
(b) Determine F(X,Y )(x, y).
(c) Determine FX(x) and FY (y).
(d) Determine fX(x) and fY (y).
(e) Compute P (X ≤ Y ).
Exercise 7.18 (Ross (2015; p.290)). Let Ω be the sample space for the roll of two die, Ω = {1, 2, 3, 4, 5, 6}2 and
P the discrete uniform probability over Ω, P ({w}) = 136 , for every w ∈ Ω. Find f(X,Y )(x, y) when
(a) X(w) = w1 and Y (w) = w2.
(b) X(w) = max(w1, w2) and Y (w) = min(w1, w2).
(c) X(w) = w1 + w2 and Y (w) = w1 − w2.
Exercise 7.19 (Ross (2015; p.240)). Prove that
P (x1 < X ≤ x2, y1 < Y ≤ y2) = F(X,Y )(x2, y2) + F(X,Y )(x1, y1)− F(X,Y )(x1, y2)− F(X,Y )(x2, y1)
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