Transcript Chapter 16:The Study of Randomness

Chapter 6
Random Variables
Random Variables and Expected Value
1
Betting on Death!

Many people in America have life insurance policies.
Although you might not want to think of it this way,
when you purchase a life insurance policy, you’re
betting that you will die sooner rather than later…
 Although
this is a bet that people really don’t want to win,
it is a bet that they are willing to take just to be sure that
their families are financially secure in the event of death.

Most families depend on the income of one or more people in the
household. What would happen if that income suddenly
disappeared? Life insurance help us handle such disasters.
 When
you purchase a life insurance policy, it’s in your
best interests that the company makes a profit and does
well; why do you think that is?
2
Betting on Death!

Question:
You purchase a policy that charges only $50 a year.
If it pays $10,000 for death and $5000 for a
permanent disability, is the company likely to make a
profit?
 Actuaries
at for the company have determined the
following probabilities in any given year:
 P (Death) = 1/1000
 P (Permanently disabled) = 2/1000
 P (Healthy) = 997/1000
 We’ll come back to this problem later on…
3
Random Variables

A Random Variable is a variable whose values are
numbers that are determined by an outcome of a
random event.
Note: Random variables are denoted by
capital letters, while the values of
random variables are denoted with
lowercase letters (small letters)
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Discrete Random Variables and Exp. Value
A discrete random variable has a countable number
of outcomes. In other words, it is possible for you
to count and make a list of all of the possible
outcomes. Discrete random variables take on only integer
values. Suppose, for example, that we flip a coin
and count the number of heads. The number of
heads results from a random process - flipping a
coin. And the number of heads is represented by
an integer value.
The mean of the discrete random variable, X, is also called
the expected value of X. Notationally, the expected value of
X is denoted by E(X). It is what we expect to happen.
The formula for expected value is:
 X  E( X )   x  P( x)
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Examples

In the experiment of flipping three coins, consider the
outcomes and define the random variable X as the
number of heads that appear.
 The
outcomes are {no heads, 1 head, 2 heads, or 3 heads}
 X has values in the set: {0, 1, 2, 3

When rolling two dice and finding the sum, determine
the outcomes and the random variable Y.
 The
outcomes are {(1,1), (1,2), (1,3), (1,4), (1,5), etc…}
 Y has values in the set: {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

In our life insurance example, what are the outcomes
and random variables Z if we define them as the
possible payments.
 The outcomes are {die, disabled, healthy}
 Z has values in the set: {$10,000, $5000, $0}
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Back to Betting on Death

So, will the company make a profit for any given year?
How much will they make or lose?


These questions are answered by finding the expected value.
Policyholder
Outcome
Payout
x
Probability
P(X = x)
Die
$10,000
1/1000
Disability
$5000
2/1000
Healthy
$0
997/1000
The expected Value is:
 1 
 2 
 997 
 X  E ( X )   x  P( x)  $10,000
  $5000
  $0

 1000
 1000
 1000
 $10  $10  $0  $20
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Back to Betting on Death

So, what does this mean?
 The expected value for the company is a
payout, on average, of $20 per customer per
year.
 Since each customer pays $50 per year, the
company expects to make a profit of $30 per
customer per year.
 It’s important to note that the insurance
company will never really pay anyone $20; it
only pays $10,000, $5000, or $0. $20 is the
expected average payout given a large
number of policy holders based on the LLN. 8
Labor Costs

A car’s air conditioner recently needed to be
repaired at the auto shop. The mechanic said
that he could for $60 in 75% of the cases by
drawing down and recharging the coolant. If
that fails, it will cost an additional $140 to
replace the unit.
 What
are the outcomes, random variables, and the
probability distribution?
Outcome
Cost
Probability
x
P(X = x)
Quick fix works
$60
¾ =.75
Replace unit
$200
¼ = .25
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Labor Costs

A car’s air conditioner recently needed to be
repaired at the auto shop. The mechanic said
that it could for $60 in 75% of the cases by
drawing down and recharging the coolant. If
that fails, it will cost an additional $140 to
replace the unit.
 What
is the expected value of the cost of this
repair?
 X  E( X )   x  P( x)  $600.75  $2000.25
 $45  $50  $95
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Labor Costs

A car’s air conditioner recently needed to be
repaired at the auto shop. The mechanic said
that it could for $60 in 75% of the cases by
drawing down and recharging the coolant. If
that fails, it will cost an additional $140 to
replace the unit.
 What
does this mean in context of this problem?
 Car
owners with this problem will spend an
average of $95 to get their car fixed at this auto
shop.
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Got to Love Those Aces

It takes $5 to play a game
 From
a standard 52 card deck of cards, if you get an
ace of hearts, you get $100
 If you get any other ace, you get $10.
 If you get any other heart, you get your $5 back.
 If you get any other card, you lose.
 Make a probability distribution for this game.
 Make a histogram of the probability distribution.
 What is the expect value of this game and is it worth it
to play this game?
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Got to Love Those Aces


First, you want to determine your possible winnings
(let’s include the $5 cost) and probabilities:
Outcome
X = Payout
Probability: P(X = x)
Ace of Hearts
Other Aces
$95
$5
1/52 = .0192
3/52 = .0577
Other Hearts
Other Cards
$0
-$5
12/52 = .2308
36/52 = .6923
Now, we can find the expected value, E(X):
 X  E( X )   x  P( x)  $950.0192  $5.0577  $0(.2308)  $5(.6923)
 $1.82  $0.29  $0  $3.46  $1.35

Is this game worth playing?
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Probability Histogram

We can use histograms to display probability distributions as
well as distributions of data.
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Continuous Random Variable

Continuous random variables, in contrast, can take
on any value within a range of values.
A continuous random variable is not countable. In other
words, you cannot list every single possible outcome.
For example, the amount of water can you put into a
5-gallon container – there are an infinite number of
possibilities.
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Example
Which of the following is a discrete random
variable?

The average height of a randomly selected group of
boys.
II.
The annual number of sweepstakes winners from
New York City.
III. The number of presidential elections in the 20th
century.
I.
(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III
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Solution

The correct answer is B. The annual number of
sweepstakes winners is an integer value and it
results from a random process; so it is a discrete
random variable. The average height of a group
of boys could be a non-integer, so it is not a
discrete variable. And the number of
presidential elections in the 20th century is an
integer, but it does not vary and it does not
result from a random process; so it is not a
random variable.
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Means and Variances of Random Variables




Recall that the mean, x, of a set of observations is our
ordinary average.
The mean of a random variable X is a weighted
average – it takes into account the fact that not all
outcomes need be equally likely.
The mean of a random variable X is also called the
expected value of X.
The expected value takes probability into account
The Variance of a Random Variable
µ is the mean of X and the VARIANCE of X is
2x = (x1- µx)2p1 + (x2 - µx)2p2 + ….+ (xk - µx)2pk
=
k
(x  )
i 1
i
2
 pi
The standard deviation x of X is the square root of the
variance.
Let’s go back and try to determine the variance and standard
deviation of our insurance policy problem.
The Variance of a Random Variable



Let’s go back and try to determine the variance and
standard deviation of our insurance policy problem.
Recall: μX = E(X) = $20
Policyholder
Outcome
Payout
x
Probability
P(X = x)
Deviation
(x – μ)
Death
$10,000
1/1000
(10,000 – 20) = 9,980
Disability
Neither
$5000
$0
2/1000
997/1000
(5,000 – 20) = 4,980
(0 – 20) = -20
The variance is the expected value of those squared
deviations:
2 
 1 
2
2  997 
Var( X )  99802 

4980

(

20
)




  149,600
 1000
 1000
 1000
The Variance of a Random Variable

To find the standard deviation of our problem, we
find the square root of our variance:
SD( X )  Var( X )  149,600  $386.78
So the insurance company can expect an average
payout of $20 per policy with a standard
deviation of about $386.78.

What does this mean?

The company charges $50 for each policy and expects to
pay $20 per policy, so there is a $30 profit for each policy
(on average). However, a spread of $386.78 is very large
for just $30. Remember, about 68% of the time, the values
fall within one SD of the mean in a normal distribution.
Another Example: Playing a game
Let’s say that you want
to play a spinner game
that costs $5 to play:
The following are the payouts:
You spin a spinner using the
following probabilities
Spin
A
B
C
p
1/3
1/6
1/2
If it lands on A, you get $5
If it lands on B, you get $12
If it lands on C, you get $0
You end up with the
following random variables
Spin
A
B
C
What is the Expected Value of
the game?
X
0
7
-5
 X =E(X)= Expected Value
p
1/3
1/6
1/2
Example: Playing a game
Using the distribution
and random variable in
the table, we get the
following:
Spin
X
p
A
0
1/3
B
7
1/6
C
-5
1/2
The Expected Value of the game:
 X = E(X)= Expected Value = 0(1/3) + 7(1/6) + (-5)(1/2)
= 0 + 7/6 - 5/2 = -8/6  1.33
What does this mean?
This means that in the very long-run, we can expect to lose about
$1.33 per game on average. It is important to note that we will never
actually lose $1.33 (because the payouts are $0, $7, and -$5), but a loss
of $1.33 is our average long-term payout per game.
Example: Playing a game
Using the distribution
and random variable in
the table, we get the
following
Spin
X
p
A
0
1/3
B
7
1/6
C
-5
1/2
Now let’s calculate the variance and standard deviation for the
payouts of this game:
Xi
X
A
0
-1.33
B
7
C
-5
(Xi -  X )2
(Xi - X )2pi
1.33
1.7689
.5896
-1.33
8.33
69.3889
11.5648
-1.33
-3.67
13.4689
6.7344
(Xi -
X)
Example: Playing a game
Now let’s calculate the variance and standard deviation for the
payouts of this game:
Xi
X
(Xi -  X)
(Xi -  X)2
(Xi - X )2pi
A
0
-1.33
1.33
1.7689
.5896
B
7
-1.33
8.33
69.3889
11.5648
C
-5
-1.33
-3.67
13.4689
6.7344
In order to get the variance we add up all the numbers in the last
column:
varianceof X  var(X)  ( X i   X )2 pi 18.8889
standarddeviationof X  SD(X)  var(X)  18.8889  $4.35
The Law of Large Numbers

The Law of Large Numbers states that the longrun relative frequency of repeated independent
events gets closer and closer to the true relative
frequency as the number of trials increases.
 The
LLN ensures us that, in the long-run, we can find
an average value that we expect to happen, namely,
the Expected Value.