LECTURE 24: CONTINUOUS RANDOM VARIABLES
MINGFENG ZHAO
March 09, 2015
The cumulative distribution function
Definition 1. A random variables is a function from the set of some outcomes to (−∞, ∞).
Definition 2. The cumulative distribution function, or simply CDF, for a random variable X, denoted by F (x), is the
probability that X assumes a value less than or equal to x:
F (x) = Pr(X ≤ x).
Theorem 1. The cumulative distribution function F (x) satisfies the following four properties:
I. 0 ≤ F (x) ≤ 1 for all x.
II.
lim
x→−∞
F (x) = 0.
III. lim F (x) = 1.
x→∞
IV. F (x) is a non-decreasing function of x.
Remark 1. If a function F (x) satisfies the above four properties in Theorem 1, we can define a random variable X
such that F (x) is the CDF for this random variable X.
Definition 3. A continuous random variable is one that has a continuous cumulative distribution function.
Remark 2. For a continuous random variable X, notice that
lim Pr(x ≤ X ≤ x + ∆x)
∆x→0
=
lim [Pr(X ≤ x + ∆x) − Pr(X ≤ x)]
∆x→0
=
Pr(X ≤ x) − Pr(X ≤ x)
=
0.
That is,
Pr(X = x) = 0,
1
for all x.
2
MINGFENG ZHAO

 0, if x < 0,
NonExample 1. Let F (x) =
, then F (x) can not represent a CDF for a continuous random variable,
 1, if x ≥ 0,
because F (x) is not continuous.
Example 1. Let F (x) =
1
for some constant k > 0, it’s easy to check that F (x) satisfies the above four
1 + e−kx
properties in Theorem 1.
The probability density function
Definition 4. Let F (x) be the CDF for a continuous random variable X, the probability density function, or simply
PDF, of X is given by:
f (x) =
dF (x)
,
dx
wherever the derivative exists.
Theorem 2. Let f (x) be a PDF for a continuous random variable X, then
I. The CDF for X is given by:
Z
x
F (x) = Pr(X ≤ x) =
f (t) dt.
−∞
II. f (x) ≥ 0 for all x.
Z ∞
III.
f (t) dt = 1.
−∞
Z
IV. For any a ≤ b, we have Pr(a ≤ x ≤ b) =
b
f (x) dx.
a

 k(3x2 + 1), if 0 ≤ x ≤ 2,
, then
Example 2. Let f (x) =
 0,
otherwise.
a. Find the value of k that makes the given function a PDF.
b. Let X be a continuous random variable whose PDF is f (x). compute the probability that X is between 1 and
2.
c. Find the CDF of X.
d. Find the probability that X is exactly equal to 1.
Z
∞
Solution: a. Since f (x) is a PDE for some continuous random variable, by Theorem 2, we have
f (x) dx = 1. So
−∞
we get
Z
1
∞
=
f (x) dx
−∞
LECTURE 24: CONTINUOUS RANDOM VARIABLES
3
2
Z
k(3x2 + 1) dx
=
0
Z
=
2
k
(3x2 + 1) dx
0
=
2
k x3 + x 0
=
k(23 + 2)
=
10k.
Since f (x) = 0 when x is outside of [0, 2]
So we get
k=
1
.
10
b. By Theorem 2, then
Z
Pr(1 ≤ X ≤ 2)
2
1
(3x2 + 1) dx
10
1
Z 2
1
(3x2 + 1) dx
10 1
2
1 3
· x + x 1
10
1 3
· 2 + 2 − 13 − 1
10
1
·8
10
4
.
5
=
=
=
=
=
=
c. By Theorem 2, then
Z
x
F (x) =
f (t) dt,
for all x.
−∞

Z x
Z x
 k(3x2 + 1), if 0 ≤ x ≤ 2,
Since f (x) =
, then F (x) ==
f (t) dt =
0 dt = 0 for all x ≤ 0. Since f (t) = 0
 0,
−∞
−∞
otherwise.
Z x
Z 2
when t is out side of [0, 2], by the computation in part a, thenF (x) =
f (t) dt =
f (t) dt = 1 for all x ≥ 2. Now
−∞
for any 0 ≤ x ≤ 2, we have
Z
F (x)
x
=
f (t) dt
−∞
Z x
=
=
1
(3t2 + 1) dt
10
0
Z x
1
(3t2 + 1) dt
10 0
0
4
MINGFENG ZHAO
=
=
x
1
· [t3 + t]0
10
1 3
(x + x).
10
So we get



0,
if x ≤ 0,



1 3
F (x) =
(x + x), if 0 < x < 2,

10



 1,
if x ≥ 2.
d. Since X is a continuous random variable, then
Z
1
f (t) dt = 0.
Pr(X = 1) =
−1
Remark 3. For a continuous random variable X, then the CDF F (x) is continuous, but the PDF f (x) maybe not be
continuous.
Remark 4. Not all continuous random variables have PDFs.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca