Transcript Expected Value- Random variables

Expected ValueRandom variables
Def. A random variable, X, is a
numerical measure of the outcomes
of an experiment
Example:
Experiment- Two cards randomly selected
Let X be the number of diamonds selected

CC
 DC

S 
 HC
 SC
CD
DD
HD
CH
DH
HH
SD
SH
CS 
DS 

HS 
SS 
Events can be described in terms of
random variables
Example:
 X 1
is the event that exactly one
diamond is selected
 X 1
is the event that at most one
diamond is selected

Probabilities of events can be stated
as probabilities of the corresponding
values of X
P( X  1)  P( F )
6

16
3

8

Example:
 CC

P( X  1)  P  DH
  HS

15

16
CD CH CS DC  

DS HC HD HH  
SC SD SH SS  
In general,
is the probability that X
takes on the value x
P( X  x)
is the probability that X
takes on a value that is less than or
equal to x
P( X  x)
Suppose that X can only assume the
values x1, x2, ... xn. Then
n
 P( X  x )  1
i 1
i

Def. The mean (or expected
value) of X gives the value that we
would expect to observe on average
in a large number of repetitions of
the experiment
n
 X  E ( X )   xi * P( X  xi )
i 1
Important

Concept of Expected value describe
the expected monetary return of
experiment
n
 X  E ( X )   xi * P( X  xi )
i 1
Sum of the values,
weighted by their
respected probabilities
Example (Exercise 13):
An investment in Project A will result in a
loss of $26,000 with probability 0.30,
break even with probability 0.50, or result
in a profit of $68,000 with probability
0.20. An investment in Project B will
result in a loss of $71,000 with
probability 0.20, break even with
probability 0.65, or result in a profit of
$143,000 with probability 0.15. Which
investment is better?

Tools to calculate E(X)-Project A

Random Variable (X)- The amount of
money received from the investment in
Project A
X can assume only x1 ,
X= x1 is the event that we
X= x2 is the event that we
X= x3 is the event that we

x1=$-26,000

x2=$0

x3=$68,000

P(X= x1)=0.3

P(X= x2)= 0.5

P(X= x3)= 0.2

x2 , x3
have Loss
are breaking even
have a Profit
Tools to calculate E(X)-Project B

Random Variable (X)- The amount of
money received from the investment in
Project B
X can assume only x1 ,
X= x1 is the event that we
X= x2 is the event that we
X= x3 is the event that we

x1=$-71,000

x2=$0

x3=$143,000

P(X= x1)=0.2

P(X= x2)= 0.65

P(X= x3)= 0.15

x2 , x3
have Loss
are breaking even
have a Profit
Project A :
E ( X )  0.30  ($26,000)  0.50  $0  0.20  $68,000
 $5800
Project B :
E ( X )  0.20  ($71,000)  0.65  $0  0.15  $143,000
 $7250
Focus on the Project
How can Expected value help us
with the decision on whether or not
to attempt a loan workout?
 Recall:
Events
S- An attempted workout is a Success
F- An attempted workout is a Failure

Tools to calculate E(X)
Random Variable (X)- The amount
of money Acadia receives from a
future loan workout attempt
 X can assume only
Full Value
Default Value
x1=$ 4,000,000
x2=$ 250,000


Using Expected value formula
The sheet Expected Value in the Excel file Loan Focus.xls
performs the following computation for the expected value of X.
E ( X )  $4,000,000  P ( X  $4,000,000)  $250,000  P ( X  $250,000)
 $4,000,000  P ( S )  $250,000  P ( F )
 $4,000,000  (0.464)  $250,000  (0.536)
 $1,991,000
Decision?

Recall
Bank Forecloses a loan if
Benefits of Foreclosure > Benefits of Workout

Bank enters a Loan Workout if
Expected Value Workout > Expected Value Foreclose
Since the expected value of a work out is
$1,991,000 and the “expected value” of
foreclosing is a guaranteed $2,100,000, it
might seem that Acadia Bank should
foreclose on John Sanders’ loan.