Download 16 Linear Functions of a Discrete Random

“Teach A Level Maths”
Statistics 1
Linear Functions of a
Discrete Random
Variable
© Christine Crisp
Linear Functions of a Discrete Random Variable
Statistics 1
Edexcel
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Linear Functions of a Discrete Random Variable
We have already looked at the effect of scaling ( or
coding ) data.
If x gives the values of a variable and y is a linear
function of that variable, we found the following
results:
If y  ax  b, the mean of y changes in the same
way, so
y  ax  b
However, adding a constant has no effect on the
standard deviation, so
s y  as x
If we want variance instead of standard deviation,
we square the result giving:
s y  a 2 s x2
2
Linear Functions of a Discrete Random Variable
We get similar results if we are dealing with a
discrete random variable ( which is used as a model
for the data ).
e.g. The probability table for a random variable X
is shown below:
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5)
Solution:
(a) E(X) is the alternative notation for the mean, m
So,
E ( X )  m   xP( X  x )
 1 0  4  2  0  5  3  0  1
 1 7
Linear Functions of a Discrete Random Variable
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5)
E( X )  1  7
(b) Find E(4X).
We start with a table of values of 4x.
x
4x
1
4
2
8
3
12
The probability that
4x = 4 is the same as the
probability that x = 1,
and so on, so we get
Linear Functions of a Discrete Random Variable
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5)
E( X )  1  7
(b) Find E(4X).
We start with a table of values of 4x.
x
4x
P(X = x)
So,
1
4
0·4
2
8
0·5
3
12
0·1
The probability that
4x = 4 is the same as the
probability that x = 1,
and so on, so we get
E (4 X )  4  0  4  8  0  5  12  0  1
 68
Linear Functions of a Discrete Random Variable
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5)
E( X )  1  7
(b) Find E(4X).
We start with a table of values of 4x.
x
4x
P(X = x)
So,
1
4
0·4
2
8
0·5
3
12
0·1
The probability that
4x = 4 is the same as the
probability that x = 1,
and so on, so we get
E (4 X )  4  0  4  8  0  5  12  0  1
 68
So, E (4 X ) 
Linear Functions of a Discrete Random Variable
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5)
E( X )  1  7
(b) Find E(4X).
We start with a table of values of 4x.
x
4x
P(X = x)
So,
1
4
0·4
2
8
0·5
3
12
0·1
The probability that
4x = 4 is the same as the
probability that x = 1,
and so on, so we get
E (4 X )  4  0  4  8  0  5  12  0  1
 68
So, E (4 X )  4 E ( X )
Linear Functions of a Discrete Random Variable
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5)
E( X )  1  7
E (4 X )  4 E ( X )  6  8
(c) E(5) simply means the “expected value of 5” or
the mean value of 5. So, E(5) = 5
(d) For E(4X + 5) we need
4x + 5
P(X = x)
9
0·4
13
0·5
17
0·1
So, E (4 X  5)  9  0  4  13  0  5  17  0  1

 3  6  6  5  1  7  11  8
E (4 X  5)
Linear Functions of a Discrete Random Variable
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5)
E( X )  1  7
E (4 X )  4 E ( X )  6  8
(c) E(5) simply means the “expected value of 5” or
the mean value of 5. So, E(5) = 5
(d) For E(4X + 5) we need
4x + 5
P(X = x)
9
0·4
13
0·5
17
0·1
So, E (4 X  5)  9  0  4  13  0  5  17  0  1

 3  6  6  5  1  7  11  8
E (4 X  5) =
Linear Functions of a Discrete Random Variable
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5)
E( X )  1  7
E (4 X )  4 E ( X )  6  8
(c) E(5) simply means the “expected value of 5” or
the mean value of 5. So, E(5) = 5
(d) For E(4X + 5) we need
4x + 5
P(X = x)
9
0·4
13
0·5
17
0·1
So, E (4 X  5)  9  0  4  13  0  5  17  0  1

 3  6  6  5  1  7  11  8
E (4 X  5) = 4 E ( X ) 
Linear Functions of a Discrete Random Variable
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5)
E( X )  1  7
E (4 X )  4 E ( X )  6  8
(c) E(5) simply means the “expected value of 5” or
the mean value of 5. So, E(5) = 5
(d) For E(4X + 5) we need
4x + 5
P(X = x)
9
0·4
13
0·5
17
0·1
So, E (4 X  5)  9  0  4  13  0  5  17  0  1

 3  6  6  5  1  7  11  8
E (4 X  5) = 4 E ( X )  5
Linear Functions of a Discrete Random Variable
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) E(X), (b) E(4X), (c) E(5), (d) E(4X + 5)
E( X )  1  7
E (4 X )  4 E ( X )  6  8
(c) E(5) simply means the “expected value of 5” or
the mean value of 5. So, E(5) = 5
(d) For E(4X + 5) we need
4x + 5
P(X = x)
9
0·4
13
0·5
17
0·1
So, E (4 X  5)  9  0  4  13  0  5  17  0  1

 3  6  6  5  1  7  11  8
E (4 X  5) = 4 E ( X )  5
Linear Functions of a Discrete Random Variable
The results we have found can be generalised to
give
E (aX  b)  aE( X )  b
e.g. The probability distribution for the r.v. X is
given by
x
2
4
6
8
10
P( X  x)
1
12
1
4
5
12
1
6
1
12
Find (a) E(X), (b) Hence find E(2X - 3)
Solution:
( X ) of
 the
xP(question
X  x ) means that we
“Hence” in (a)
partE(b)
1
1 rather than1using
35
must use the answer
to
part
(a)
 2   4   . . .  10 

 5 56
12 of 42X - 3.
12 6
the values and probabilities
26
 35 
(b) E ( 2 X - 3)  2 E ( X ) - 3  2  - 3 
 8 23
 6 
3
Linear Functions of a Discrete Random Variable
Before we can find an expression for Var (aX  b) it’s
useful to define a new notation for Var ( X ) .
We know that
Var ( X )   2   x 2 P ( X  x ) - m 2
Using an extension of the notation for  xP( X  x )
( i.e. E(X) ), we write
E( X 2 )   x 2 P( X  x)
So,
Var( X )  E ( X 2 ) - m 2
This, in turn, can be written as
Var( X )  E ( X 2 ) - ( E ( X )) 2
and, because the last term is clumsy, we write
Var( X )  E ( X 2 ) - E 2 ( X )
Linear Functions of a Discrete Random Variable
We must be very careful if we use
Var( X )  E ( X 2 ) - E 2 ( X )
This, . . . is the mean of X 2 and this . . .
is the mean of X which has been squared.
Linear Functions of a Discrete Random Variable
We’ll now find a formula for Var (aX  b)
e.g.
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) Var(X) (b) Var(5) (c) Var(4X) (d) Var(4X+5)
We already have the following results:
E ( X )  1  7 , E (4 X )  6  8, E (4 X  5)  11  8
(a) Var ( X )  E ( X 2 ) - E 2 ( X ) 
2
2
x
P
(
X

x
)
m

 12  0  4  2 2  0  5  3 2  0  1 - 1  7 2  0  41
(b) Var (5)  0
N.B. Variance measures
variability or spread. There is
no spread if we just have 5.
Linear Functions of a Discrete Random Variable
We’ll now find a formula for Var (aX  b)
e.g.
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) Var(X) (b) Var(5) (c) Var(4X) (d) Var(4X+5)
E ( X )  1  7 , E (4 X )  6  8, E (4 X  5)  11  8
Var ( X )  0  41
(c) Var (4 X )  E (4 X ) 2 - E 2 (4 X )
Linear Functions of a Discrete Random Variable
We’ll now find a formula for Var (aX  b)
e.g.
4x
P(X = x)
4
0·4
8
0·5
12
0·1
Find (a) Var(X) (b) Var(5) (c) Var(4X) (d) Var(4X+5)
E ( X )  1  7 , E (4 X )  6  8, E (4 X  5)  11  8
Var ( X )  0  41
(c) Var (4 X )  E (4 X ) 2 - E 2 (4 X )
 4 2  0  4  8 2  0  5  122  0  1 - 6  8 2
Linear Functions of a Discrete Random Variable
We’ll now find a formula for Var (aX  b)
e.g.
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) Var(X) (b) Var(5) (c) Var(4X) (d) Var(4X+5)
E ( X )  1  7 , E (4 X )  6  8, E (4 X  5)  11  8
Var ( X )  0  41
(c) Var (4 X )  E (4 X ) 2 - E 2 (4 X )
 4 2  0  4  8 2  0  5  122  0  1 - 6  8 2
 6  56
Linear Functions of a Discrete Random Variable
We’ll now find a formula for Var (aX  b)
e.g.
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) Var(X) (b) Var(5) (c) Var(4X) (d) Var(4X+5)
E ( X )  1  7 , E (4 X )  6  8, E (4 X  5)  11  8
Var ( X )  0  41
(c) Var (4 X )  E (4 X ) 2 - E 2 (4 X )
 4 2  0  4  8 2  0  5  122  0  1 - 6  8 2
 6  56  16Var ( X )
Linear Functions of a Discrete Random Variable
We’ll now find a formula for Var (aX  b)
e.g.
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) Var(X) (b) Var(5) (c) Var(4X) (d) Var(4X+5)
E ( X )  1  7 , E (4 X )  6  8, E (4 X  5)  11  8
Var ( X )  0  41
(c) Var (4 X )  E (4 X ) 2 - E 2 (4 X )
 4 2  0  4  8 2  0  5  122  0  1 - 6  8 2
 6  56  16Var ( X )
(d) Var (4 X  5)  E (4 X  5) 2 - E 2 (4 X  5)
Linear Functions of a Discrete Random Variable
We’ll now find a formula for Var (aX  b)
e.g.
4x + 5
P(X = x)
9
0·4
13
0·5
17
0·1
Find (a) Var(X) (b) Var(5) (c) Var(4X) (d) Var(4X+5)
E ( X )  1  7 , E (4 X )  6  8, E (4 X  5)  11  8
Var ( X )  0  41
(c) Var (4 X )  E (4 X ) 2 - E 2 (4 X )
 4 2  0  4  8 2  0  5  122  0  1 - 6  8 2
 6  56  16Var ( X )
(d) Var (4 X  5)  E (4 X  5) 2 - E 2 (4 X  5)
 9 2  0  4  132  0  5  172  0  1 - 11  8 2
Linear Functions of a Discrete Random Variable
We’ll now find a formula for Var (aX  b)
e.g.
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) Var(X) (b) Var(5) (c) Var(4X) (d) Var(4X+5)
E ( X )  1  7 , E (4 X )  6  8, E (4 X  5)  11  8
Var ( X )  0  41
(c) Var (4 X )  E (4 X ) 2 - E 2 (4 X )
 4 2  0  4  8 2  0  5  122  0  1 - 6  8 2
 6  56  16Var ( X )
(d) Var (4 X  5)  E (4 X  5) 2 - E 2 (4 X  5)
 9 2  0  4  132  0  5  172  0  1 - 11  8 2
 6  56  16Var ( X )
Linear Functions of a Discrete Random Variable
We’ll now find a formula for Var (aX  b)
e.g.
x
P(X = x)
1
0·4
2
0·5
3
0·1
Find (a) Var(X) (b) Var(5) (c) Var(4X) (d) Var(4X+5)
E ( X )  1  7 , E (4 X )  6  8, E (4 X  5)  11  8
Var ( X )  0  41
(c) Var (4 X )  E (4 X ) 2 - E 2 (4 X )
 4 2  0  4  8 2  0  5  122  0  1 - 6  8 2
 6  56  16Var ( X )
(d) Var (4 X  5)  E (4 X  5) 2 - E 2 (4 X  5)
 9 2  0  4  132  0  5  172  0  1 - 11  8 2
 6  56  16Var ( X )
Linear Functions of a Discrete Random Variable
So,
Var (4 X  5)  4 2 Var ( X )
The general result is
Var (aX  b)  a 2 Var ( X )
This is just like the rule for coding data:
y  ax  b  s y  a 2 s x2
2
Multiplying by a constant results in the variance
being multiplied by the square of the constant.
Adding a constant has no effect on the variance
since it does not change the spread of the values.
Linear Functions of a Discrete Random Variable
SUMMARY
The mean of a linear function of a discrete
random variable, X, is given by
•
E (aX  b)  aE( X )  b
•
The variance of X, is given by
Var ( X )   2   x 2 P ( X  x ) - m 2
which can be written as
Var( X )  E ( X 2 ) - E 2 ( X )
•
The variance of a linear function of X is given by
Var (aX  b)  a 2 Var ( X )
Linear Functions of a Discrete Random Variable
e.g. The probability distribution for the r.v. X is
given by
x
2
4
6
8
10
P( X  x)
1
12
1
4
5
12
1
6
1
12
Find (a) Var(X), (b) Hence find Var(2X - 3)
Solution:
35
(a) We already have the result E ( X ) 
6
1
1
1
115
2
2
2
2
E ( X )  2   4   . . .  10  
12
4
12
3
2
155
115
35


2
2
-  
 4 11
So, Var( X )  E ( X ) - E ( X ) 
36
3  6 
36
 155  155
2
(b) Var (2 X - 3)  2 Var ( X )  4 Var ( X )  4 
 17 92

9
 36 
Linear Functions of a Discrete Random Variable
Exercise
1. The probability distribution for the r.v. X is
given by
x
1
2
3
4
5
P( X  x) 0  1 0  2 0  4 0  2 0  1
Find (a) E(X) (b) E(3X + 1) (c) Var(X) (d) Var(3X  1)
Solution: (a) E(X) = 3. ( We can get this directly
from the symmetry of the table. )
(b) E(3X + 1) = 3E(X) + 1 = 10
(c) Var ( X )  E ( X 2 ) - E 2 ( X )
 12  0  1  2 2  0  2  . . . 5 2  0  1 - 3 2  1  2
(d) Var (3 X  1)  9Var ( X )  9(1  2)  10  8
Linear Functions of a Discrete Random Variable
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Linear Functions of a Discrete Random Variable
SUMMARY
The mean of a linear function of a discrete
random variable, X, is given by
•
E (aX  b)  aE( X )  b
•
The variance of X, is given by
Var ( X )   2   x 2 P ( X  x ) - m 2
which can be written as
Var( X )  E ( X 2 ) - E 2 ( X )
•
The variance of a linear function of X is given by
Var (aX  b)  a 2 Var ( X )
Linear Functions of a Discrete Random Variable
e.g. The probability distribution for the r.v. X is
given by
x
2
4
6
8
10
P( X  x)
1
12
1
4
5
12
1
6
1
12
Find (a) E(X) (b) E(2X - 3) (c) Var(X) (d) Var(2X - 3)
Solution: (a) E ( X )   xP( X  x )
1
1
1
 2   4   . . .  10 
12
4
12
35

 5 56
6
(b) E (2 X - 3)  2 E ( X ) - 3
 35 
26
 2  - 3 
 8 23
3
 6 
Linear Functions of a Discrete Random Variable
(c) Var ( X )  E ( X 2 ) - E 2 ( X )
1
1
1
2
2
E ( X )  2   4   . . .  10 
12
4
12
115

3
2
115  35 
So, Var ( X ) 
- 
3  6 
155

 4 11
36
36
(d) Var ( 2 X - 3)  2 2 Var ( X )  4Var ( X )
155 155
 4

 17 92
36
9
2
2