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Random Variables
Discrete Probability Distributions
Continuous Random Variables
Lecture 3: Probability Distributions and
Probability Densities - 1
Assist. Prof. Dr. Emel YAVUZ DUMAN
MCB1007 Introduction to Probability and Statistics
İstanbul Kültür University
Random Variables
Discrete Probability Distributions
Outline
1
Random Variables
2
Discrete Probability Distributions
3
Continuous Random Variables
Continuous Random Variables
Random Variables
Discrete Probability Distributions
Outline
1
Random Variables
2
Discrete Probability Distributions
3
Continuous Random Variables
Continuous Random Variables
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Definition 1
If S is a sample space with a probability measure and X is a
real-valued function defined over the elements of S, then X is
called a random variable (or stochastic variable).
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Definition 1
If S is a sample space with a probability measure and X is a
real-valued function defined over the elements of S, then X is
called a random variable (or stochastic variable).
In this course we shall always denote random variables by capital
letters such as X , Y etc., and their values by the corresponding
lowercase letters such as x and y , respectively.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 2
Suppose that a coin is tossed twice so that the sample space is
S = {HH, HT , TH, TT }. Let X represent the number of heads
that can come up.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 2
Suppose that a coin is tossed twice so that the sample space is
S = {HH, HT , TH, TT }. Let X represent the number of heads
that can come up. With each sample point we can associate a
number for X as shown in the table:
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 2
Suppose that a coin is tossed twice so that the sample space is
S = {HH, HT , TH, TT }. Let X represent the number of heads
that can come up. With each sample point we can associate a
number for X as shown in the table:
Sample Point
Probability
x
HH
HT
TH
TT
1
4
1
4
1
4
1
4
2
1
1
0
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 2
Suppose that a coin is tossed twice so that the sample space is
S = {HH, HT , TH, TT }. Let X represent the number of heads
that can come up. With each sample point we can associate a
number for X as shown in the table:
Sample Point
Probability
x
HH
HT
TH
TT
1
4
1
4
1
4
1
4
2
1
1
0
Thus, for example, in the case of HH (i.e., 2 heads), X = 2 while
for TH (1 head), X = 1. It follows that X is a random variable.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 2
Suppose that a coin is tossed twice so that the sample space is
S = {HH, HT , TH, TT }. Let X represent the number of heads
that can come up. With each sample point we can associate a
number for X as shown in the table:
Sample Point
Probability
x
HH
HT
TH
TT
1
4
1
4
1
4
1
4
2
1
1
0
Thus, for example, in the case of HH (i.e., 2 heads), X = 2 while
for TH (1 head), X = 1. It follows that X is a random variable.
Also, we can write P(X = 2) = 14 , P(X = 1) = 14 + 14 = 12 , and
P(X = 0) = 14 .
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 3
A balanced coin is tossed four times. List the elements of the
sample space that are presumed to be equally likely, as this is what
we mean by a coin being balanced, and the corresponding values x
of the random variable X , the total number of heads.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 3
A balanced coin is tossed four times. List the elements of the
sample space that are presumed to be equally likely, as this is what
we mean by a coin being balanced, and the corresponding values x
of the random variable X , the total number of heads.
Solution. If H and T stand for heads and tails, the results are as
shown in the following table:
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 3
A balanced coin is tossed four times. List the elements of the
sample space that are presumed to be equally likely, as this is what
we mean by a coin being balanced, and the corresponding values x
of the random variable X , the total number of heads.
Solution. If H and T stand for heads and tails, the results are as
shown in the following table:
Random Variables
Elements of
sample space
HHHH
HHHT
HHTH
HTHH
THHH
HHTT
HTHT
HTTH
Discrete Probability Distributions
Probability
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
x
4
3
3
3
3
2
2
2
Elements of
sample space
THHT
THTH
TTHH
HTTT
THTT
TTHT
TTTH
TTTT
Continuous Random Variables
Probability
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
x
2
2
2
1
1
1
1
0
Random Variables
Elements of
sample space
HHHH
HHHT
HHTH
HTHH
THHH
HHTT
HTHT
HTTH
Discrete Probability Distributions
Probability
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
x
4
3
3
3
3
2
2
2
Elements of
sample space
THHT
THTH
TTHH
HTTT
THTT
TTHT
TTTH
TTTT
Continuous Random Variables
Probability
1
4
Thus, we can write P(X = 0) = 16
, P(X = 1) = 16
,
6
4
1
.
P(X = 2) = 16 , P(X = 3) = 16 and P(X = 4) = 16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
1
16
x
2
2
2
1
1
1
1
0
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 4
Two socks are selected at random and removed in succession from
a drawer containing five brown socks and three green socks. List
the elements of the sample space, the corresponding probabilities,
and the corresponding values x of the random variable X is the
number of brown socks selected.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 4
Two socks are selected at random and removed in succession from
a drawer containing five brown socks and three green socks. List
the elements of the sample space, the corresponding probabilities,
and the corresponding values x of the random variable X is the
number of brown socks selected.
Solution. If B and G stand for brown and green, then we have
following probabilities
20
5 3
15
5 4
P(BB) = · = , P(BG ) = · = ,
8 7
56
8 7
56
15
3 2
6
3 5
P(GB) = · = , and P(GG ) = · = ,
8 7
56
8 7
56
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 4
Two socks are selected at random and removed in succession from
a drawer containing five brown socks and three green socks. List
the elements of the sample space, the corresponding probabilities,
and the corresponding values x of the random variable X is the
number of brown socks selected.
Solution. If B and G stand for brown and green, then we have
following probabilities
20
5 3
15
5 4
P(BB) = · = , P(BG ) = · = ,
8 7
56
8 7
56
15
3 2
6
3 5
P(GB) = · = , and P(GG ) = · = ,
8 7
56
8 7
56
and the results are shown in the following table:
Elements of Sample Space
Probability
x
BB
20/56
2
BG
15/56
1
GB
15/56
1
GG
6/56
0
Random Variables
Discrete Probability Distributions
Continuous Random Variables
In all of the examples of this section we have limited our discussion
to discrete sample space, and hence to discrete random variable,
namely, random variables whose range is finite or countably
infinite. Continuous random variables defined over continuous
sample spaces will be taken up in the third section.
Random Variables
Discrete Probability Distributions
Outline
1
Random Variables
2
Discrete Probability Distributions
3
Continuous Random Variables
Continuous Random Variables
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Definition 5
If X is a discrete random variable, the function given by
f (x) = P(X = x)
for each x within the range of X is called the probability
distribution (or probability function) of X .
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Definition 5
If X is a discrete random variable, the function given by
f (x) = P(X = x)
for each x within the range of X is called the probability
distribution (or probability function) of X .
Based on the postulates of probability, it immediately follows that
Theorem 6
A function can serve as the probability distribution of a discrete
random variable X if and only if its values, f (x), satisfy the
conditions
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Definition 5
If X is a discrete random variable, the function given by
f (x) = P(X = x)
for each x within the range of X is called the probability
distribution (or probability function) of X .
Based on the postulates of probability, it immediately follows that
Theorem 6
A function can serve as the probability distribution of a discrete
random variable X if and only if its values, f (x), satisfy the
conditions
1
f (x) ≥ 0 for each value within its domains;
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Definition 5
If X is a discrete random variable, the function given by
f (x) = P(X = x)
for each x within the range of X is called the probability
distribution (or probability function) of X .
Based on the postulates of probability, it immediately follows that
Theorem 6
A function can serve as the probability distribution of a discrete
random variable X if and only if its values, f (x), satisfy the
conditions
1
2
f (x) ≥ 0 for each value within its domains;
x f (x) = 1, where the summation extends over all the
values within its domain.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 7
Find a formula for the probability distribution of the total number
of heads obtained in four tosses of a balanced coin.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 7
Find a formula for the probability distribution of the total number
of heads obtained in four tosses of a balanced coin.
1
, P(X = 1) =
Solution. We know that P(X = 0) = 16
6
4
1
.
P(X = 2) = 16 , P(X = 3) = 16 and P(X = 4) = 16
4
16 ,
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 7
Find a formula for the probability distribution of the total number
of heads obtained in four tosses of a balanced coin.
1
4
, P(X = 1) = 16
,
Solution. We know that P(X = 0) = 16
6
4
1
P(X = 2) = 16 , P(X = 3) = 16 and P(X = 4) = 16 . Observing
that the numerators of these
4 fractions,
4 4 1, 4,46, 4, and 1, are
4 five
the binomial coefficients 0 , 1 , 2 , 3 , and 4 , we find that
the formula for the probability distribution can be written as
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 7
Find a formula for the probability distribution of the total number
of heads obtained in four tosses of a balanced coin.
1
4
, P(X = 1) = 16
,
Solution. We know that P(X = 0) = 16
6
4
1
P(X = 2) = 16 , P(X = 3) = 16 and P(X = 4) = 16 . Observing
that the numerators of these
4 fractions,
4 4 1, 4,46, 4, and 1, are
4 five
the binomial coefficients 0 , 1 , 2 , 3 , and 4 , we find that
the formula for the probability distribution can be written as
4
f (x) =
x
16
for x = 0, 1, 2, 3, 4.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 8
Check whether the function given by
f (x) =
x +2
for x = 1, 2, 3, 4, 5
25
can serve as the probability distribution of a discrete random
variable.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 8
Check whether the function given by
f (x) =
x +2
for x = 1, 2, 3, 4, 5
25
can serve as the probability distribution of a discrete random
variable.
Solution. Substituting the different values of x, we get f (1) =
4
5
6
7
, f (3) = 25
, f (4) = 25
, and f (5) = 25
.
f (2) = 25
3
25 ,
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 8
Check whether the function given by
f (x) =
x +2
for x = 1, 2, 3, 4, 5
25
can serve as the probability distribution of a discrete random
variable.
Solution. Substituting the different values of x, we get f (1) =
4
5
6
7
, f (3) = 25
, f (4) = 25
, and f (5) = 25
. Since these
f (2) = 25
values are all nonnegative, the first condition of Theorem 6 is
satisfied,
3
25 ,
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 8
Check whether the function given by
f (x) =
x +2
for x = 1, 2, 3, 4, 5
25
can serve as the probability distribution of a discrete random
variable.
Solution. Substituting the different values of x, we get f (1) =
4
5
6
7
, f (3) = 25
, f (4) = 25
, and f (5) = 25
. Since these
f (2) = 25
values are all nonnegative, the first condition of Theorem 6 is
satisfied, and since
f (1) + f (2) + f (3) + f (4) + f (5) =
3
25 ,
4
5
6
7
3
+
+
+
+
=1
25 25 25 25 25
the second conditions of Theorem 6 is satisfied.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 8
Check whether the function given by
f (x) =
x +2
for x = 1, 2, 3, 4, 5
25
can serve as the probability distribution of a discrete random
variable.
Solution. Substituting the different values of x, we get f (1) =
4
5
6
7
, f (3) = 25
, f (4) = 25
, and f (5) = 25
. Since these
f (2) = 25
values are all nonnegative, the first condition of Theorem 6 is
satisfied, and since
f (1) + f (2) + f (3) + f (4) + f (5) =
3
25 ,
4
5
6
7
3
+
+
+
+
=1
25 25 25 25 25
the second conditions of Theorem 6 is satisfied. Thus, the given
function can serve as the probability distribution of a random
variable having the range {1, 2, 3, 4, 5}.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 9
Suppose that a pair of fair dice are to be tossed, and let the
random variable X denote the sum of the points. Obtain the
probability distribution for X .
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 9
Suppose that a pair of fair dice are to be tossed, and let the
random variable X denote the sum of the points. Obtain the
probability distribution for X .
Solution. The random variable X is the sum of the coordinates for
each point. Thus for (3, 2) we have X = 5. Using the fact that all
36 sample points are equally probable, so that each sample point
has probability 1/36.
First Die
Second
Die
The one is red is the sum
Random Variables
x
f (x)
Discrete Probability Distributions
Continuous Random Variables
2
3
4
5
6
7
8
9
10
11
12
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
Random Variables
Discrete Probability Distributions
Figure 1 : Probability Bar Chart
Continuous Random Variables
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Definition 10
If X is a discrete random variable, the function given by
f (t) for − ∞ < x < ∞
F (x) = P(X ≤ x) =
t≤x
where f (t) is the value of the probability distribution of X at t, is
called the distribution function, or the cumulative distribution, of
X.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Based on the postulates of probability and some of their immediate
consequences, it follows that
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Based on the postulates of probability and some of their immediate
consequences, it follows that
Theorem 11
The values F (x) of the distribution function of a discrete random
variable X satisfy the conditions
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Based on the postulates of probability and some of their immediate
consequences, it follows that
Theorem 11
The values F (x) of the distribution function of a discrete random
variable X satisfy the conditions
1
F (−∞) = 0 and F (∞) = 1;
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Based on the postulates of probability and some of their immediate
consequences, it follows that
Theorem 11
The values F (x) of the distribution function of a discrete random
variable X satisfy the conditions
1
F (−∞) = 0 and F (∞) = 1;
2
if a < b, then F (a) ≤ F (b) for any real numbers a and b.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Based on the postulates of probability and some of their immediate
consequences, it follows that
Theorem 11
The values F (x) of the distribution function of a discrete random
variable X satisfy the conditions
1
F (−∞) = 0 and F (∞) = 1;
2
if a < b, then F (a) ≤ F (b) for any real numbers a and b.
If we are given the probability distribution of a discrete random
variable, the corresponding distribution function is generally easy
to find.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 12
Find the distribution function of the total of heads obtained in four
tosses of a balanced coin.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 12
Find the distribution function of the total of heads obtained in four
tosses of a balanced coin.
1
4
, f (1) = 16
, f (2) =
Solution. Given f (0) = 16
1
f (4) = 16 from Example 3, it follows that
F (0) = f (0) =
6
16 ,
f (3) =
1
,
16
F (1) = f (0) + f (1) =
5
,
16
F (2) = f (0) + f (1) + f (2) =
11
,
16
15
,
16
F (4) = f (0) + f (1) + f (2) + f (3) + f (4) = 1.
F (3) = f (0) + f (1) + f (2) + f (3) =
4
16 ,
and
Random Variables
Discrete Probability Distributions
Hence, the distribution function
⎧
⎪
0
⎪
⎪
⎪
⎪
⎪1
⎪
16
⎪
⎪
⎨5
F (x) = 16
11
⎪
⎪
16
⎪
⎪
⎪
15
⎪
⎪
⎪ 16
⎪
⎩1
is given by
for
for
for
for
for
for
x < 0,
0 ≤ x < 1,
1 ≤ x < 2,
2 ≤ x < 3,
3 ≤ x < 4,
x ≥ 4.
Continuous Random Variables
Random Variables
Discrete Probability Distributions
Hence, the distribution function
⎧
⎪
0
⎪
⎪
⎪
⎪
⎪1
⎪
16
⎪
⎪
⎨5
F (x) = 16
11
⎪
⎪
16
⎪
⎪
⎪
15
⎪
⎪
⎪ 16
⎪
⎩1
Continuous Random Variables
is given by
for
for
for
for
for
for
x < 0,
0 ≤ x < 1,
1 ≤ x < 2,
2 ≤ x < 3,
3 ≤ x < 4,
x ≥ 4.
Observe that this distribution function is defined not only for the
values taken on by the given random variable, but for all real
5
and F (100) = 1,
numbers. For instance, we can write F (1.7) = 16
although the probabilities of getting at most 1.7 heads or at most
100 heads in four tosses of a balanced coin may not be of any real
significance.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 13
Find the distribution function of the random variable X of Example
4 and plot its graph.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 13
Find the distribution function of the random variable X of Example
4 and plot its graph.
Solution. Based on the probabilities given in the following table
Elements of Sample Space
Probability
x
BB
20/56
2
BG
15/56
1
GB
15/56
1
GG
6/56
0
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 13
Find the distribution function of the random variable X of Example
4 and plot its graph.
Solution. Based on the probabilities given in the following table
Elements of Sample Space
Probability
x
we can write f (0) =
6
56 ,
f (1) =
BB
20/56
2
15
56
+
15
56
BG
15/56
1
=
30
56 ,
GB
15/56
1
GG
6/56
0
and f (2) =
20
56 ,
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 13
Find the distribution function of the random variable X of Example
4 and plot its graph.
Solution. Based on the probabilities given in the following table
Elements of Sample Space
Probability
x
we can write f (0) =
that
6
56 ,
f (1) =
F (0) = f (0) =
BB
20/56
2
15
56
+
15
56
BG
15/56
1
=
30
56 ,
GB
15/56
1
GG
6/56
0
and f (2) =
20
56 ,
6
,
56
36
,
56
F (2) = f (0) + f (1) + f (2) = 1.
F (1) = f (0) + f (1) =
so
Random Variables
Discrete Probability Distributions
Hence, the distribution function
⎧
⎪
0
⎪
⎪
⎪
⎨6
F (x) = 56
36
⎪
⎪
⎪ 56
⎪
⎩1
of X is given by
for
for
for
for
x < 0,
0 ≤ x < 1,
1 ≤ x < 2,
x ≥ 2.
Continuous Random Variables
Random Variables
Discrete Probability Distributions
Hence, the distribution function
⎧
⎪
0
⎪
⎪
⎪
⎨6
F (x) = 56
36
⎪
⎪
⎪ 56
⎪
⎩1
Continuous Random Variables
of X is given by
for
for
for
for
x < 0,
0 ≤ x < 1,
1 ≤ x < 2,
x ≥ 2.
F (x)
1
36/56
6/56
0
x
1
2
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 14
Find the distribution function of the random variable that has the
probability distribution
f (x) =
x
for x = 1, 2, 3, 4, 5.
15
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 14
Find the distribution function of the random variable that has the
probability distribution
f (x) =
Solution. Since f (1) =
5
, then
f (5) = 15
x
for x = 1, 2, 3, 4, 5.
15
1
15 ,
F (x) =
f (2) =
⎧
⎪
0
⎪
⎪
⎪
⎪
1
⎪
⎪
15
⎪
⎪
⎨3
15
6
⎪
⎪
15
⎪
⎪
⎪ 10
⎪
⎪
15
⎪
⎪
⎩1
2
15 ,
for
for
for
for
for
for
f (3) =
3
15 ,
x < 1,
1 ≤ x < 2,
2 ≤ x < 3,
3 ≤ x < 4,
4 ≤ x < 5,
x ≥5
f (4) =
4
15 ,
and
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Theorem 15
If the range of a random variable X consists of the values
x1 < x2 < x3 < · · · < xn , then f (x1 ) = F (x1 ) and
f (xi ) = F (xi ) − F (xi −1 ) for i = 2, 3, · · · , n.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 16
If X has the distribution function F (1) = 0.25, F (2) = 0.61,
F (3) = 0.83, and F (4) = 1 for x = 1, 2, 3, 4, find the probability
distribution of X .
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 16
If X has the distribution function F (1) = 0.25, F (2) = 0.61,
F (3) = 0.83, and F (4) = 1 for x = 1, 2, 3, 4, find the probability
distribution of X .
Solution. We have
f (1) = F (1) = 0.25,
f (2) = F (2) − F (1) = 0.61 − 0.25 = 0.36,
f (3) = F (3) − F (2) = 0.83 − 0.61 = 0.22,
f (4) = F (4) − F (3) = 1 − 0.83 = 0.17.
Random Variables
Discrete Probability Distributions
Example 17
If X has the distribution function
⎧
⎪
0 for
⎪
⎪
⎪
1
⎪
⎪
⎨ 4 for
F (x) = 12 for
⎪
⎪
⎪ 3 for
⎪
⎪
4
⎪
⎩
1 for
x < −1,
− 1 ≤ x < 1,
1 ≤ x < 3,
3 ≤ x < 5,
x ≥ 5.
find
1
P(X ≤ 3), P(X = 3), P(X < 3);
2
P(X ≥ 1);
3
P(−0.4 < X < 4);
4
P(X = 5);
5
the probability distribution of X .
Continuous Random Variables
Random Variables
Discrete Probability Distributions
⎧
⎪
0
⎪
⎪
⎪
1
⎪
⎪
⎨4
F (x) =
1
2
⎪
⎪
3
⎪
⎪
⎪
4
⎪
⎩
1
for
for
for
for
for
x < −1,
− 1 ≤ x < 1,
1 ≤ x < 3,
3 ≤ x < 5,
x ≥ 5.
Solution.
1
3
4
1
3 1
P(X = 3) = − =
4 2
4
1
P(X < 3) =
2
P(X ≤ 3) =
Continuous Random Variables
Random Variables
Discrete Probability Distributions
⎧
⎪
0
⎪
⎪
⎪
1
⎪
⎪
⎨4
F (x) =
2
1
2
⎪
⎪
3
⎪
⎪
⎪
4
⎪
⎩
1
for
for
for
for
for
Continuous Random Variables
x < −1,
− 1 ≤ x < 1,
1 ≤ x < 3,
3 ≤ x < 5,
x ≥ 5.
P(X ≥ 1) = 1 − P(X < 1) = 1 −
1
4
= 34 .
Random Variables
Discrete Probability Distributions
⎧
⎪
0
⎪
⎪
⎪
1
⎪
⎪
⎨4
F (x) =
1
2
⎪
⎪
3
⎪
⎪
⎪
4
⎪
⎩
1
for
for
for
for
for
Continuous Random Variables
x < −1,
− 1 ≤ x < 1,
1 ≤ x < 3,
3 ≤ x < 5,
x ≥ 5.
2
P(X ≥ 1) = 1 − P(X < 1) = 1 −
3
P(−0.4 < X < 4) =
3
4
−
1
4
= 12 .
1
4
= 34 .
Random Variables
Discrete Probability Distributions
⎧
⎪
0
⎪
⎪
⎪
1
⎪
⎪
⎨4
F (x) =
1
2
⎪
⎪
3
⎪
⎪
⎪
4
⎪
⎩
1
for
for
for
for
for
Continuous Random Variables
x < −1,
− 1 ≤ x < 1,
1 ≤ x < 3,
3 ≤ x < 5,
x ≥ 5.
2
P(X ≥ 1) = 1 − P(X < 1) = 1 −
3
P(−0.4 < X < 4) =
4
P(X = 5) = 1 −
3
1
4 − 4
3
1
4 = 4.
= 12 .
1
4
= 34 .
Random Variables
Discrete Probability Distributions
⎧
⎪
0
⎪
⎪
⎪
1
⎪
⎪
⎨4
F (x) =
1
2
⎪
⎪
3
⎪
⎪
⎪
4
⎪
⎩
1
for
for
for
for
for
x < −1,
− 1 ≤ x < 1,
1 ≤ x < 3,
3 ≤ x < 5,
x ≥ 5.
2
P(X ≥ 1) = 1 − P(X < 1) = 1 −
3
P(−0.4 < X < 4) =
4
5
Continuous Random Variables
1
4
3
1
1
4 − 4 = 2.
P(X = 5) = 1 − 34 = 14 .
f (−1) = 14 , f (1) = 12 − 14 = 14 , f (3)
f (5) = 1 − 34 = 14 , and 0 elsewhere.
= 34 .
=
3
4
−
1
2
= 14 ,
Random Variables
Discrete Probability Distributions
Outline
1
Random Variables
2
Discrete Probability Distributions
3
Continuous Random Variables
Continuous Random Variables
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Continuous Random Variables
So far we have considered discrete random variables that can take
on a finite or countably infinite number of values.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Continuous Random Variables
So far we have considered discrete random variables that can take
on a finite or countably infinite number of values. In applications,
we are often interested in random variables that can take on an
uncountable continuum of values; we call these continuous random
variables.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Continuous Random Variables
So far we have considered discrete random variables that can take
on a finite or countably infinite number of values. In applications,
we are often interested in random variables that can take on an
uncountable continuum of values; we call these continuous random
variables.
Consider modeling the distribution of the age that a person dies at.
Age of death, measured perfectly with all the decimals and no
rounding, is a continuous random variable (e.g., age of death could
be 87.3248583585642 years).
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Continuous Random Variables
So far we have considered discrete random variables that can take
on a finite or countably infinite number of values. In applications,
we are often interested in random variables that can take on an
uncountable continuum of values; we call these continuous random
variables.
Consider modeling the distribution of the age that a person dies at.
Age of death, measured perfectly with all the decimals and no
rounding, is a continuous random variable (e.g., age of death could
be 87.3248583585642 years). Other examples of continuous
random variables include: time until the occurrence of the next
earthquake in İstanbul;
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Continuous Random Variables
So far we have considered discrete random variables that can take
on a finite or countably infinite number of values. In applications,
we are often interested in random variables that can take on an
uncountable continuum of values; we call these continuous random
variables.
Consider modeling the distribution of the age that a person dies at.
Age of death, measured perfectly with all the decimals and no
rounding, is a continuous random variable (e.g., age of death could
be 87.3248583585642 years). Other examples of continuous
random variables include: time until the occurrence of the next
earthquake in İstanbul; the lifetime of a battery;
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Continuous Random Variables
So far we have considered discrete random variables that can take
on a finite or countably infinite number of values. In applications,
we are often interested in random variables that can take on an
uncountable continuum of values; we call these continuous random
variables.
Consider modeling the distribution of the age that a person dies at.
Age of death, measured perfectly with all the decimals and no
rounding, is a continuous random variable (e.g., age of death could
be 87.3248583585642 years). Other examples of continuous
random variables include: time until the occurrence of the next
earthquake in İstanbul; the lifetime of a battery; the annual rainfall
in Ankara.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Because it can take on so many different values, each value of a
continuous random variable winds up having probability zero.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Because it can take on so many different values, each value of a
continuous random variable winds up having probability zero. If I
ask you to guess someone’s age of death perfectly, not
approximately to the nearest millionth year, but rather exactly to
all the decimals, there is no way to guess correctly - each value
with all decimals has probability zero. But for an interval, say the
nearest half year, there is a nonzero chance you can guess
correctly.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Because it can take on so many different values, each value of a
continuous random variable winds up having probability zero. If I
ask you to guess someone’s age of death perfectly, not
approximately to the nearest millionth year, but rather exactly to
all the decimals, there is no way to guess correctly - each value
with all decimals has probability zero. But for an interval, say the
nearest half year, there is a nonzero chance you can guess
correctly. So, we have the following definition:
Definition 18
A random variable X is called a continuous random variable if
its distribution function F is a continuous function on R, or
equivalently, if
P(X = x) = 0, for every x ∈ R.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
The definition of probability in the continuous case presumes for
each random variable the existence of a function, called a
probability density function (pdf), such that areas under the curve
give the probabilities associated with the corresponding intervals
along the horizontal axis.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
The definition of probability in the continuous case presumes for
each random variable the existence of a function, called a
probability density function (pdf), such that areas under the curve
give the probabilities associated with the corresponding intervals
along the horizontal axis. In other words, a probability density
function, integrated form a to b (with a ≤ b), gives the probability
that the corresponding random variable will take on a value on the
interval from a to b.
Definition 19
A function with values f (x), defined over the set of all real
numbers, is called a probability density function (pdf) of the
continuous random variable X if and only if
b
P(a ≤ X ≤ b) =
f (x)dx
a
for any real constants a and b with a ≤ b.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
f (x)
a
b
x
Figure 2 : P(a ≤ X ≤ b) = area under the density curve between a and
b.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
f (x)
a
b
x
Figure 2 : P(a ≤ X ≤ b) = area under the density curve between a and
b.
Theorem 20
If X is a continuous random variable and a and b are real
constants with a ≤ b, then
P(a ≤ X ≤ b) = P(a ≤ X < b) = P(a < X ≤ b) = P(a < X < b).
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Analogous to Theorem 6, let us now state the following properties
of probability densities, which again follow directly from the
postulates of probability.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Analogous to Theorem 6, let us now state the following properties
of probability densities, which again follow directly from the
postulates of probability.
Theorem 21
A function can serve as a probability density of a continuous
random variable X if its values, f (x), satisfy the conditions
1
2
f (x) ≥ 0 for −∞ < x < ∞,
∞
−∞ f (x)dx = 1.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 22
The probability density of the continuous random variable X is
given by
1
for 2 < x < 7,
f (x) = 5
0 elsewhere.
1
2
Draw its graph and verify that the total area under the curve
(above the x-axis) is equal to 1.
Find P(3 < X < 5).
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 22
The probability density of the continuous random variable X is
given by
1
for 2 < x < 7,
f (x) = 5
0 elsewhere.
1
2
Draw its graph and verify that the total area under the curve
(above the x-axis) is equal to 1.
Find P(3 < X < 5).
Solution.
f (x)
1
5
2
7
x
Random Variables
Discrete Probability Distributions
f (x) =
1
1
5
0
Continuous Random Variables
for 2 < x < 7,
elsewhere.
It is clear that f (x) ≥ 0 for all x and
2
7
∞
∞
f (x)dx =
f (x) dx +
f (x) dx +
f (x) dx
−∞
−∞ 2 7 0
=
2
7
1
5
1 7 1
1
dx = x = (7 − 2) = 1.
5
5 2 5
0
Random Variables
Discrete Probability Distributions
f (x) =
1
1
5
0
Continuous Random Variables
for 2 < x < 7,
elsewhere.
It is clear that f (x) ≥ 0 for all x and
2
7
∞
∞
f (x)dx =
f (x) dx +
f (x) dx +
f (x) dx
−∞
−∞ 2 7 0
7
=
2
1
5
1 7 1
1
dx = x = (7 − 2) = 1.
5
5 2 5
0
2
5
P(3 < X < 5) =
f (x)dx =
3
2
= .
5
3
5
1 5 1
1
dx = x = (5 − 3)
5
5 3 5
Random Variables
Discrete Probability Distributions
Example 23
The pdf of the random variable X is given by
√c
for 0 < x < 4,
x
f (x) =
0
elsewhere.
Find
1
the value of c,
2
P(X < 0.25) and P(X > 1).
Continuous Random Variables
Random Variables
Discrete Probability Distributions
f (x) =
√c
x
for 0 < x < 4,
0
elsewhere.
Continuous Random Variables
Solution.
1
To satisfy the second condition of Theorem 21, we must have
∞
f (x)dx =
−∞
0
f (x) dx +
−∞ 0
0
4
f (x) dx +
√c
x
∞
4
f (x) dx
0
4
4
− 12 +1 1
x
c
√ dx =
cx − 2 dx = c 1
=
x
−2 + 1
0
0
0
√
= 2c x|40 = 4c = 1
4
and it follows that c = 1/4.
Random Variables
Discrete Probability Distributions
f (x) =
1
√
4 x
for 0 < x < 4,
0
elsewhere.
Continuous Random Variables
2
P(X < 0.25) =
0.25
f (x)dx =
−∞
0.25
=
0
1
= .
4
0
f (x) dx +
−∞ 0
0
0.25
f (x) dx
1
√
4 x
1 √ 0.25 1 √
1
√ dx = 2 x = ( 0.25 − 0)
4
2
4 x
0
Random Variables
Discrete Probability Distributions
f (x) =
1
√
4 x
for 0 < x < 4,
0
elsewhere.
Continuous Random Variables
2
P(X < 1) =
∞
f (x)dx =
1
=
1
1
= .
2
1
4
4
f (x) dx +
1
√
4 x
4
∞
f (x) dx
1 √ 4 1
1
√ dx = 2 x = (2 − 1)
4
2
4 x
1
0
Random Variables
Discrete Probability Distributions
Example 24
If X has the probability density
ke −3x
f (x) =
0
find k and P(0.5 ≤ X ≤ 1).
for x > 0,
elsewhere,
Continuous Random Variables
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 24
If X has the probability density
ke −3x
f (x) =
0
for x > 0,
elsewhere,
find k and P(0.5 ≤ X ≤ 1).
Solution. To satisfy the second condition of Theorem 21, we must
have
t
e −3x f (x)dx =
ke
dx = k lim
t→∞ −3 −∞
0
0
k
k
k
= − lim (e −3t − e 0 ) = − (0 − 1) = = 1
t→∞
3
3
3
∞
∞
and it follows that k = 3.
−3x
Random Variables
3e −3x
f (x) =
0
Discrete Probability Distributions
Continuous Random Variables
for x > 0,
elsewhere,
For the probability we get
P(0.5 ≤ X ≤ 1) =
1
3e
0.5
−3x
1
e −3x dx = 3
= −e −3 + e −1.5
−3 0.5
≈ 0.1733430918.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Definition 25
If X is a continuous random variable and the value of its
probability density at t is f (t), then the function given by
x
f (t)dt for − ∞ < x < ∞
F (x) = P(X ≤ x) =
−∞
is called the distribution function, or the cumulative
distribution, of X .
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Definition 25
If X is a continuous random variable and the value of its
probability density at t is f (t), then the function given by
x
f (t)dt for − ∞ < x < ∞
F (x) = P(X ≤ x) =
−∞
is called the distribution function, or the cumulative
distribution, of X .
The values F (x) of the distribution function of a continuous
random variable X satisfies the conditions: F (−∞) = 0,
F (∞) = 1, and F (a) ≤ F (b) when a < b.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Theorem 26
If f (x) and F (x) are the values of the probability density and the
distribution function of X at x, then
P(a ≤ X ≤ b) = F (b) − F (a)
for any real constants a and b with a ≤ b, and
f (x) =
where the derivative exists.
dF (x)
dx
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 27
Find the distribution function of the random variable X of Example
24, and use it to reevaluate P(0.5 ≤ x ≤ 1).
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 27
Find the distribution function of the random variable X of Example
24, and use it to reevaluate P(0.5 ≤ x ≤ 1).
Solution. For x > 0,
x
f (t)dt =
F (x) =
−∞
0
x
x
3e −3t dt = −e −3t 0 = 1 − e −3x
and since F (x) = 0 for x ≤ 0, we can write
0
for x ≤ 0,
F (x) =
−3x
for x > 0.
1−e
Random Variables
Discrete Probability Distributions
0
F (x) =
1 − e −3x
Continuous Random Variables
for x ≤ 0,
for x > 0.
To determine the probability P(0.5 ≤ X ≤ 1), we use the first part
of Theorem 26, getting
P(0.5 ≤ X ≤ 1) = F (1) − F (0.5)
= (1 − e −3 ) − (1 − e −1.5 )
= −e −3 + e −1.5
≈ 0.1733430918.
This agrees with the result obtained by using the probability
density directly in Example 24.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 28
Find the distribution function of the random variable X whose
probability density is given by
⎧
⎪
for 0 < x < 1,
⎨x
f (x) = 2 − x for 1 ≤ x < 2,
⎪
⎩
0
elsewhere.
Random Variables
Discrete Probability Distributions
⎧
⎪
⎨x
f (x) = 2 − x
⎪
⎩
0
for 0 < x < 1,
for 1 ≤ x < 2,
elsewhere.
Solution. For 0 < x < 1,
x
f (t)dt =
F (x) =
−∞
x
0
for 1 ≤ x < 2,
x
f (t)dt =
F (x) =
−∞
2 1
Continuous Random Variables
x
t 2 x2
tdt = = ,
2 0
2
1
f (t)dt +
0
x
f (t)dt =
1
1
x
tdt +
0
1
(2 − t)dt
x
x2
1
x2
t2
1
t − 2 + = − + 2x − 1,
= + 2x −
= + 2t −
2 0
2 1
2
2
2
2
and for F (x) = 1 where x ≥ 2.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
So, we have
⎧
⎪
0
⎪
⎪
⎪
⎨ x2
F (x) =
2
⎪
− x2 + 2x − 1
⎪
⎪
⎪
⎩1
2
for
for
for
for
x ≤ 0,
0 < x < 1,
1 ≤ x < 2,
x ≥ 2.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 29
Find a probability density function for the random variable whose
distribution function is given by
⎧
⎪
⎨0 for x ≤ 0,
F (x) = x for 0 < x < 1,
⎪
⎩
1 for x ≥ 1.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 29
Find a probability density function for the random variable whose
distribution function is given by
⎧
⎪
⎨0 for x ≤ 0,
F (x) = x for 0 < x < 1,
⎪
⎩
1 for x ≥ 1.
Solution. Since the given density function is differentiable
everywhere except at x = 0 and x = 1, we differentiate for x < 0,
0 < x < 1, and x > 1, getting 0, 1, and 0.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 29
Find a probability density function for the random variable whose
distribution function is given by
⎧
⎪
⎨0 for x ≤ 0,
F (x) = x for 0 < x < 1,
⎪
⎩
1 for x ≥ 1.
Solution. Since the given density function is differentiable
everywhere except at x = 0 and x = 1, we differentiate for x < 0,
0 < x < 1, and x > 1, getting 0, 1, and 0. Thus, we can write
⎧
⎪
⎨0 for x ≤ 0,
f (x) = 1 for 0 < x < 1,
⎪
⎩
0 for x ≥ 1.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
To fill the gaps at x = 0 and x = 1, we let f (0) and f (1) both
equal to zero. Actually, it does not matter how the probability
density is defined at these points, but there are certain advantages
for choosing the values in such a way that the probability density is
nonzero over an open interval.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
To fill the gaps at x = 0 and x = 1, we let f (0) and f (1) both
equal to zero. Actually, it does not matter how the probability
density is defined at these points, but there are certain advantages
for choosing the values in such a way that the probability density is
nonzero over an open interval. Thus, we can write the probability
density of the original random variable as
1 for 0 < x < 1,
f (x) =
0 elsewhere.
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 30
The distribution function of the random variable Y is given by
1 − y92 for y > 3,
F (y ) =
0
elsewhere.
Find P(Y ≤ 5), P(Y > 8) and a probability density function of Y .
Solution.
P(Y ≤ 5) = F (5) = 1 −
9
52
=
16
25 ,
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 30
The distribution function of the random variable Y is given by
1 − y92 for y > 3,
F (y ) =
0
elsewhere.
Find P(Y ≤ 5), P(Y > 8) and a probability density function of Y .
Solution.
P(Y ≤ 5) = F (5) = 1 −
9
52
=
16
25 ,
P(Y > 8) = 1 − P(Y ≤ 8) = 1 − F (8) = 1 − 1 +
9
64
=
9
64 ,
Random Variables
Discrete Probability Distributions
Continuous Random Variables
Example 30
The distribution function of the random variable Y is given by
1 − y92 for y > 3,
F (y ) =
0
elsewhere.
Find P(Y ≤ 5), P(Y > 8) and a probability density function of Y .
Solution.
P(Y ≤ 5) = F (5) = 1 −
9
52
=
16
25 ,
P(Y > 8) = 1 − P(Y ≤ 8) = 1 − F (8) = 1 − 1 +
d
9
for y > 3,
1
−
= 18
y2
y3
f (y ) = dy
0
elsewhere.
9
64
=
9
64 ,
Random Variables
Discrete Probability Distributions
Example 31
The distribution function of the
⎧
⎪
⎨0
F (x) = x+1
2
⎪
⎩
1
Continuous Random Variables
random variable X is given by
for x < −1,
for − 1 < x < 1,
for x ≥ 1.
Find P − 12 < X < 12 and P(2 < X < 3).
Random Variables
Discrete Probability Distributions
Example 31
The distribution function of the
⎧
⎪
⎨0
F (x) = x+1
2
⎪
⎩
1
Continuous Random Variables
random variable X is given by
for x < −1,
for − 1 < x < 1,
for x ≥ 1.
Find P − 12 < X < 12 and P(2 < X < 3).
Solution.
P − 12 < X < 12 = F 12 − F − 12 =
3
1
1
4 − 4 = 2.
1
+1
2
2
−
− 12 +1
2
=
Random Variables
Discrete Probability Distributions
Example 31
The distribution function of the
⎧
⎪
⎨0
F (x) = x+1
2
⎪
⎩
1
Continuous Random Variables
random variable X is given by
for x < −1,
for − 1 < x < 1,
for x ≥ 1.
Find P − 12 < X < 12 and P(2 < X < 3).
Solution.
P − 12 < X < 12 = F 12 − F − 12 =
3
1
1
4 − 4 = 2.
1
+1
2
2
P(2 < X < 3) = F (3) − F (2) = 1 − 1 = 0.
−
− 12 +1
2
=
Random Variables
Discrete Probability Distributions
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Continuous Random Variables