Chapter 4: Continuous Random Variables
Definition:
Let 𝑌 denote a random variable (r. v.) then the distribution function of 𝑌 is
denoted by 𝑭(𝒚) and is defined by
𝑭(𝒚) = 𝑷 ( 𝒀 ≤ 𝒚 ),
𝐟𝐨𝐫 − ∞ < 𝒚 < ∞.
𝐹( 𝑦 ) is also called the cumulative distribution function of 𝑌 and is
abbreviated as CDF.
Example 4.1
Suppose that 𝑌 has a binomial distribution with 𝑛 = 2 and 𝑝 = 1/2. Find
𝐹(𝑦).
2 1 𝑦 1 2−𝑦
2 1 2
The probability function of Y is ( ) ( ) ( )
= ( ) ( ) for 𝑦 = 0, 1, 2.
2
𝑦 2
𝑦 2
1
1
1
𝑝(0) = ,
𝑝(1) = , 𝑝(2) =
4
2
4
for 𝑦 < 0,
0,
1/4, for 0 ≤ 𝑦 < 1,
𝐹(𝑦) = 𝑃(𝑌 ≤ 𝑦) = {
3/4, for 1 ≤ 𝑦 < 2,
for 𝑦 ≥ 2.
1
1
Properties of a CDF,
If 𝐹( 𝑦 ) is a distribution function, then
1.
𝑭(−∞) = 𝐥𝐢𝐦 𝑭(𝒚) = 𝟎
2.
𝑭(∞) = 𝐥𝐢𝐦 𝑭(𝒚) = 𝟏
3.
𝒚→−∞
𝒚→∞
𝑭( 𝒚 ) is a non-decreasing function.
That is, if 𝑦1 < 𝑦2 , 𝑡ℎ𝑒𝑛 𝐹(𝑦1 ) ≤ 𝐹(𝑦2 )
Definition:
The random variable 𝑌 with CDF 𝐹(𝑦) is said to be continuous if 𝐹(𝑦) is
continuous on (− ∞ , ∞ ).
Let 𝐹(𝑦) be the CDF of a continuous random variable 𝑌. The probability
density function (pdf) of Y is defined as
𝒅𝑭(𝒚)
𝒇(𝒚) = 𝑭′ (𝒚) =
𝒅𝒚
By the fundamental theorem of calculus,
2
𝒚
𝑭(𝒚) = ∫ 𝒇(𝒕)𝒅𝒕
−∞
Properties of a pdf,
If 𝑓(𝑦)is a density function for a continuous random variable, then
1.
𝑓(𝑦) ≥ 0 for all 𝑦, −∞ < 𝑦 < ∞
2.
∫−∞ 𝑓(𝑦)𝑑𝑦 = 1
∞
Let 𝑌 be a random variable with pdf 𝑓(𝑦)then
𝑏
𝑃(𝑎 ≤ 𝑌 ≤ 𝑏) = ∫ 𝑓(𝑦)𝑑𝑦
𝑎
3
Note that 𝑃(𝑎 ≤ 𝑌 ≤ 𝑏) = 𝑃(𝑎 < 𝑌 < 𝑏) = 𝑃(𝑎 ≤ 𝑌 < 𝑏) + 𝑃(𝑎 < 𝑌 ≤ 𝑏)
(WHY?)
Example 4.2
Suppose that
0,
𝐹(𝑦) = {𝑦,
1,
for 𝑦 < 0
for 0 ≤ 𝑦 ≤ 1
for 𝑦 > 1
Find the pdf for 𝑌 and graph it.
Example 4.3
Y’s pdf is
𝑓(𝑦) = {
3𝑦 2 , 0 ≤ 𝑦 ≤ 1
0,
elsewhere
Find CDF of Y. Graph both 𝑓(𝑌)and 𝐹(𝑌).
4
0,
𝐹(𝑦) = {𝑦 3 ,
1
𝑦<0
0≤𝑦≤1
𝑦>1
Exercise 4.11 on page 167:
Suppose random variable 𝑌 has pdf 𝑓(𝑦) given by the following:
a.
𝑐𝑦, if 0 ≤ 𝑦 ≤ 2
𝑓(𝑦) = {
0,
otherwise
Find 𝑐 so that 𝑓(𝑦) is a valid pdf.
b.
Find 𝐹(𝑌)
c.
Graph 𝑓(𝑦)and 𝐹(𝑦)
d.
Use 𝐹(𝑌)to find 𝑃(1 ≤ 𝑌 ≤ 2)
e.
Use 𝑓(𝑦) and geometry to find 𝑃(1 ≤ 𝑌 ≤ 2)
f.
Find 𝑃(𝑌 ≥ 1 | 𝑌 ≤ 3)
Solve:
∞
2
a. By property of pdf, ∫−∞ 𝑓(𝑦)𝑑𝑦 = 1. So, ∫0 𝑐𝑦𝑑𝑦 = 1
2
2
2
𝑦2
22
∫ 𝑐𝑦𝑑𝑦 = 𝑐 ∫ 𝑦𝑑𝑦 = 𝑐 ∙ ] = 𝑐 ∙ ( − 0) = 𝑐 ∙ 2 = 1
2 0
2
0
0
∴𝑐=
1
2
𝑦
0
𝑦
0
𝑦1
b. 𝐹(𝑦) = ∫−∞ 𝑓(𝑡)𝑑𝑡 = ∫−∞ 𝑓(𝑡)𝑑𝑡 + ∫0 𝑓(𝑡)𝑑𝑡 = ∫−∞ 0𝑑𝑡 + ∫0
1 1 2 𝑦 𝑦2
=0+ ∙ 𝑡 ] =
2 2 0
4
0,
𝑦2
𝐹(𝑦) = {
4
1,
5
if − ∞ < 𝑦 < 0
if 0 ≤ 𝑦 < 2
if 𝑦 ≥ 2
2
𝑡𝑑𝑡
c.
d.
𝑃(1 ≤ 𝑌 ≤ 2) = 𝐹 (2) − 𝐹 (1) =
22
4
−
12
4
=
e. The shaded part is a trapezoid.
The area is
1
2
1
3
2
4
[( + 1) ∙ 1] =
f. 𝑃(𝑌 ≥ 1 |𝑌 ≤ 3) =
𝐹 (3)−𝐹(1)
𝐹(3)
=
𝑃(𝑌≥1 and 𝑌≤3)
𝑃(𝑌≤3)
1
4
1−
1
=
3
4
Do Examples 4.4 & 4.5 on page 165yourself.
6
=
𝑃(1≤𝑌≤3)
𝑃(𝑌≤3)
=
3
4