Download Continuous random variable.

Eighth lecture
Random
Variables
Consider the experiment of tossing a coin twice.
. If we are interested in the number of heads that
show on the top face, describe the sample space.
Solution:
S={ HH , HT , TH , TT }
2
1
1
0
w
S
X(w)
R
Definition (1):
A random variable is a function that associates a
real number with each element in the sample space.
Remark:
We shall use a capital letter, say X, to denote a
random variable and its corresponding small letter,
x in this case, for one of its values.
Example (1):
Two balls are drawn in succession without replacement from
an urn containing 4 red balls and 3 black balls. The possible
outcomes and the values y of the random variable Y, where Y is
the number of red balls, are
Sample Space
RR
RB
BR
BB
y
2
1
1
0
Example(2)
Two dice are rolled and we define the familiar
sample space
Ω= {(1, 1), (1, 2),….(6, 6)}
containing 36 elements. Let X denote the random
variable whose value for any element of Ω
is the sum of the numbers on the two dice.
Then the range of X is the set containing the 11
values of X:
2,3,4,5,6,7,8,9,10,11,12.
Definition (2):
If a sample space contains a finite number of
possibilities or an unending sequence with as many
elements as there are whole numbers , it is called a
discrete sample space.
Definition (3):
If a sample space contains an infinite number of
possibilities equal to the number of points on a line
segment, it is called a continuous sample space.
Types of random variables:
1. Discrete random variable.
A random variable is called a discrete random variable if its set
of possible outcomes is countable.
2. Continuous random variable.
A random variable is called a continuous random variable when
can take on values on a continuous scale .
Example (3):
Classify the following random variables as discrete or continuous:
X: the number of automobile accidents per year in Virginia.
Y: the length of time to play 18 holes of golf.
M: the amount of milk produced yearly by a particular cow.
N: the number of eggs laid each month by a hen.
P: the number of building permits issued each month in a certain
city.
Q: the weight of grain produced per acre.
Definition (4):
The set of ordered pairs (x, f(x)) is a probability function, probability mass
function, or probability distribution of the discrete random variable X if, for
each possible outcome x,
1  f ( x )  0,
2
f
( x )  1,
x
3  P (X
Example(4):
 x )  f ( x ).
consider random variable X with probabilities
X
0
1
2
3
4
5
P(X=x)
0.05
0.10
0.20
0.40
0.15
0.10
You can observe that the probabilities sum to 1.
The notation P(x) is often used for P(X = x). The notation f(x) is also used. In this
example, P(4) = 0.15. The symbol P (or f) denotes the probability function, also called the
probability mass function
.
The cumulative probabilities are given as F(x) = 𝒊≤𝒙 𝑷(𝒊).
The interpretation is that F(x)
is the probability that X will take a value less than or equal to x.
The function F is called the cumulative distribution function
(CDF). This is the only notation that is commonly
used. For our example 4,
F(3) = P(X ≤ 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
= 0.05 + 0.10 + 0.20
+ 0.40
= 0.75
One can of course list all the values of the CDF easily by taking
cumulative sums:
X
0
1
2 3
4
5
P(X = x) 0.05 0.10 0.20 0.40 0.15 0.10
F(x)
0.05 0.15 0.35 0.75 0.90 1.00
The values of F increase.
Definition(5):
Let X a random variable with probability distribution f(x). The
mean or expected value of X is
  E (x )   x f (x )
x
If X is discrete,
Definition(6):
Let X be a random variable with probability distribution f(x).
The expected value of the random variable g(x) is
Definition(7):
Let X a random variable with probability distribution f(x) and
mean m. The variance of X is
The positive square root of the variance, s, is called standard
deviation of X.
Theorem(1):
The variance of a random variable X is
From Example 4
A-The expected value of X is
E( X ) = 𝑥 𝑥 P(x) = 𝑥 𝑥 P(X= x)
The calculation for this example is
E(X) = 0 × 0.05 + 1 × 0.10 + 2 × 0.20 + 3 × 0.40 + 4 × 0.15 + 5
× 0.10
= 0.00 + 0.10 + 0.40 + 1.20 + 0.60 + 0.50 = 2.80
B- The variance of X, is
This is the expected square of the difference between X and its expected
value, μ. We can calculate this for our example 4:
X
0
1
2
3
4
5
x - 2.8
-2.8
-1.8
-0.8
0.2
1.2
2.2
(x − 2.8)2
7.84
3.24
0.64
0.04
1.44
4.84
P(X=x)
0.05
0.10
0.20
0.40
0.15
0.10
P(X=x) (𝑥−2.8)2
0.392
0.324
0.128
0.016 0.216
0.484
The variance is the sum of the final row. This value is 1.560.
-There’s a simplified method, based on the result:
This is easier because we’ve already found μ
for our example, E(𝑋 2 )= 02 × 0.05 + 12 × 0.10 + 22 × 0.20 + 32 × 0.40 + 42 ×
0.15 + 52 × 0.10 = 9.40.
Then
𝛔𝟐 = 9.40 - (𝟐. 𝟖)𝟐 = 9.40 - 7.84 = 1.56. This is the same number as before
The standard deviation of X is determined from the variance. Specifically,
σ = 𝟏. 𝟓𝟔 ≈ 1.2490,
Example (5):
Suppose that the number of cars X that pass through a car wash
between 4 p.m. and 5 p.m. on any sunny Friday has the
following probability distribution:
x
4
1
12
P(X=x)
5
1
12
6
1
4
7
1
4
8
1
6
9
1
6
Let g(X)=2X-1 represent the amount of money in dollars, paid to
the attendant by the manager. Find the attendant’s expected
earning for this particular time period.
Solution: using definition 6
E(g(X))=E(2X-1)= 9𝑥=4 2𝑥 − 1 𝑓(𝑥)
=
1
1
1
1
1
1
(7)( )+(9)( )+(11)( )+(13)( )+(15)( )+(17)( )
12
12
4
4
6
6
= $12.67
The probability that X assumes a
value between a and b is equal to the
shaded area under the density
function between the ordinates at x=a
and x=b and from integral calculus is
given
f(x)
b
P (a  X  b )   f (x )dx
a
a
b
x
Definition (8):
The function f(x) is a probability density function for the
continuous random variable X, defined over the set of real
number R, if
1  f ( x )  0, for all x  R
2



f ( x )dx  1
b
3  P (a  X  b )   f (x )dx .
a
Example(6):
Suppose X is a continuous random variable having the probability
density function
5𝑒 −5𝑥 ,
𝑥≥0
𝑓 𝑥 =
0,
𝑥<0
Find P(3≤ x ≤ 4)
4
4
P(3≤ x ≤ 4)= 3 5𝑒 −5𝑥 dx= −𝑒 −5𝑥 = −𝑒 −20 + 𝑒 −15
3
Example 7
Find the value of k for the following probability density
function:
(a) f(x)=1/k , a < x < b
(b) 1- f(x)=k𝑒 −3𝑥
X>0
2- P(0.5 < x < 1)
Definition (9):
The cumulative distribution function F(x) of a continuous
random variable X with density function f(x) is
x
F (x )  P ( X  x )   f (t )dt , for    x  

As an immediate consequence of Definition (9) one can
write the two results,
dF (x )
P (a  X  b )  F (b )  F (a ), and f (x ) 
dx
If the derivative exists.
Example(8):
For the density function of Example (b7) find F(x) and use it to
evaluate P(0.5≤ X ≤ 1)
 To find F(x) for f(x)=3𝑒 −3𝑥 , x>0
𝑥
𝑥
F(x)= −∞ 𝑓 𝑡 𝑑𝑡 = 0 3𝑒 −3𝑥 dt
𝑥
= −𝑒 −3𝑡 = −𝑒 −3𝑥 + 𝑒 0 = −𝑒 −3𝑥 +1
0
−3𝑥
= 1- 𝑒
0 ,
𝐹 𝑥 =
1 − 𝑒 −3𝑥 ,
𝑥≤0
𝑥>0
 P(0.5 ≤ X ≤ 1) = F(1) – F(0.5)
=1- 𝑒 −3 -(1-𝑒 −1.5 )
= - 𝑒 −3 + 𝑒 −1.5 = 0.173
Definition(10):
Let X a random variable with probability distribution f(x).
The mean or expected value of X is
If X is continuous.

  E (x )   x f (x )dx

Example(9):
Let X be the random variable that denotes the life in hours
of a certain electronic device. The probability density
function is
 20,000
, x  100

3
f (x )   x
0,elsewhere
Find the expected life of this type of device.
Solution:
Using definition 10, we have
∞
∞ 𝟐𝟎,𝟎𝟎𝟎
𝟐𝟎,𝟎𝟎𝟎
𝝁 = 𝑬 𝑿 = 𝟏𝟎𝟎 𝒙 𝟑 dx= 𝟏𝟎𝟎 𝟐 dx= 200.
𝒙
𝒙
Definition(11):
Let X be a random variable with probability distribution f(x).
The expected value of the random variable g(x) is

 g ( X )  E  g ( X )   g (x ) f (x )dx

If X is continuous.
Example(10):
Let X be a random variable with density function
Find the expected value of g(X)=4X+3
Solution:
E(4X+3)=
2 4𝑋+3 𝑥 2
−1
3
dx =
1 2
3
(4𝑥
3 −1
+ 3𝑥 2 ) dx=8.
Definition(12):
Let X a random variable with probability distribution f(x)
and mean . The variance of X is
if X is continuous.

s  E ( X   )    (x   )2 f (x )
2
2

The positive square root of the variance, s, is called
standard deviation of X.
Theorem(1):
The variance of a random variable X is
s 2  E (X 2 )   2
Example(11):
The weekly demand for Pepsi, in thousand of liters, from a
local chain of efficiency stores, is a continuous random
variable X having the probability density
2(x  1),1  x  2
f (x )  
0,eleswhere
Find the mean and the variance of X.
Theorem(2):
If a and b are conestants, then
E(aX+b)=aE(X)+b
Corollary(1):
Setting a=0, we see that E(b)=b
Corollary(2):
Setting b=0, we see that E(aX)=aE(X)
Theorem(3):
If a and b are conestants, then
s2aX+b=a2s2X
Corollary(1):
Setting a=1, we see that s2X+b=s2X
Corollary(2):
Setting b=0, we see that s2aX=a2s2X