Download Handout 11

Expected Values
The Mean of a Random Variable
You should remember that a random variable is a model of a population, so it has
parameters which describe it. As such it has a mean or average, which serves as a
measure of central tendency within the population. Consider our random variable
X
0
1
2
3
p(x)
0.125
0.375
0.375
0.125
This represents a population that looks like this:
2
1
1
0
1
1 2
3 0
2
2 2
3
0 0
1
1 1 2 1 0
0 1
1
3 30
2
2
0
3 2 1 2 2
31
1
0
3
3
2 0 2
22 0 1 1 1 2 3
1 1
1 1
0 3
12 2
How large is this population?
If you select a number from this population at random, what fraction of the time will you
get a 2? A 3?
In the population of “the number of heads,” 3/8ths of the population is made up of 2’s.
3’s make up 1/8th of the population. The mean, μ, of this population is just the weighted
sum:
1
3
3
1
8
8
8
8
 (0.125)(0)  (0.375)(1)  (0.375)( 2)  (0.125)(3)  1.5
  (0)  (1)  (2)  (3)
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Why didn’t we use
 1 N
  xi ?
N
  i 1
 
In general
  E[ X ]   x p( x).
all x
The Variance of a Random Variable
Again, when we are interested in the variance of a random variable, we are just
talking about the variance, σ2, of a population. It provides a measure of dispersion within
the population through the average squared distance around μ, and is therefore computed
by
 2  E[ X   ]2   ( x   ) 2 p( x).
all x
Example: Recall for our random variable, μ = 1.5.
X
0
1
2
3
p(x)
0.125
0.375
0.375
0.125
(X - )
-1.5
-0.5
0.5
1.5
(X - )2
2.25
0.25
0.25
2.25
 2  E[ X   ]2  (0.125)( 2.25)  (0.375)(0.25)  (0.375)(0.25)  (0.125)( 2.25)
 0.75
The standard deviation, σ, is just the square root of the variance:
   2  0.75  0.866
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Some Facts about Expected Values
Given any constant c, and
E[ X ]   x p( x).
all x
(1) E[ f ( X )] 
 f ( x) p( x).
all x
(2) E[c] = c
(3) E[cX] = c E[X]
E[cX ]   cx p( x)
by (1)
all x
 c  x p( x)
all x
 cE[ X ]
(4) E[X + Y] = E[X] +E[Y]
for any random variables X and Y.
(5) E[XY] = E[X]E[Y] if and only if X and Y are independent, that is, for all values x
and y,
P{ X  x  Y  y}  P{ X  x} P{Y  y}.
The other important expectation we talked about was a variance. Let σ2 = Var(X).
Then
(1) Var(c) = 0.
(2) Var(cX) = c2 Var(X).
(3) Var(X + c) = Var(X).
(4) Var(X + Y) = Var(X) + Var(Y)
if and only if X and Y are independent.
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An Example:
X
0
1
2
p(x)
0.25
0.50
0.25
Let W = 2X + 3. Compute E[W] and Var(W) by
(a) developing the distribution of W and using the definitions of expectations and
variances, and
(b) computing E[X] and Var(X) and using the rules cited above.
In (a), to develop the distribution of W, begin by answering the question, if X = 0, what
will W equal?
How often will X = 0?
How often will W = 3?
W = 2X + 3
3
5
7
p(w)
0.25
0.50
0.25
Therefore
E[W ]  W  3(0.25)  5(0.50)  7(0.25)
 5.
Var (W )  (3  5) 2 (0.25)  (5  5) 2 (0.50)  (7  5) 2 (0.25)
2
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To get the same results by (b), note
E[ X ]  0(0.25)  1(0.50)  2(0.25)
1
Var ( X )  (0  1) 2 (0.25)  (1  1) 2 (0.50)  (2  1) 2 (0.25)
 0.50
Then
E[W ]  E[2 X  3]  E[2 X ]  E[3]
 E[2 X ]  3
 2 E[ X ]  3
 2(1)  3
5
by rule (4)
by rule (2)
by rule (3)
and
Var (W )  Var (2 X  3)
 Var (2 X )
 4Var ( X )
 4(0.50)
2
by rule (3)
by rule (2)
What information was required to do the problem by (a)? By (b)?
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Homework for Expectations
(1) For a random variable X with probability distribution:
X
-1
0
1
2
p(x)
0.4
0.3
0.2
0.1
Compute: (a) E[X] (b) Var(X) (c) E[X2] (d) E[X4] (e) Var(X2).
(2) For the random variable X described above, compute the following in two ways:
(a) E[3X]
(b) Var(3X)
(c) E[X + 10]
(d) Var(X + 10)
(3) Arnold Fatuous, a rising matinee idol and stereotyped arch villain, earns $1000 for
every public appearance he makes. For the year to date, Arnold has attended four
supermarket openings and one Tai-Die opening. Arnold’s agent assesses the number of
public appearances that he can attend during the rest of the year is a random variable with
the following probability distribution:
Number of Appearances
0
1
2
3
Probability
0.2
0.3
0.4
0.1
Compute the mean and variance of the annual earnings from public appearances of
Arnold Fatuous.
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