Module II
Lecture 4
Special Probability Distributions
Certain probability distributions occur with such regularity in real-life
situations that they have been given their own names and it is worth studying their
properties.
In this section, we look at three probability distributions that arise in almost
every aspect of business.
The Binomial Distribution
Consider the following situations:
a) You audit a transaction, it is either in compliance with procedures or it is
not;
b) You hire a person, that person is either a female or a male;
c) You visit a customer, it either leads to a sale or it doesn’t;
d) You lower the price of a product, sales either increase or they don’t;
e) You have a model for the stock market, you predict that it will go up at least
30 points, it either goes up 30 or more points or it doesn’t;
f) You have an intermittent problem on your company’s network, on any given
day the problem either appears or doesn’t appear.
All of these situations are ones where the Binomial Distribution may be
applicable.
There is a canonical definition for the binomial distribution. This is, a set of
assumptions, which, if they hold, indicate that the binomial distribution may be
applied to a particular situation.
Let us suppose that in a given situation only one of two possible things can
occur. For example, if we flip a fair coin then we can only get the outcomes heads or
tails. Flipping the coin is called an experiment in statistical jargon, and heads or
tails are called the possible outcomes of the experiment. Each repetition of the
experiment is called a trial.
The binomial distribution applies in the following situation:
a) The outcome of any trial can only take on two possible values, say success
and failure;
b) There is a constant probability p of success on each trial;
c) The experiment is repeated n times (i.e. n trials are conducted);
d) The trials are statistically independent (i.e. the outcome of past trials does
not affect subsequent trials);
then if x equals the number of successes in the n trials, we have:
P( x ) 
for x = 0, 1, 2, …… n.
n!
p x ( 1  p )n x
x! ( n  x )!
For example, if we flipped a fair coin ten times, and let x equal the number of
heads, the above formula would give the following probabilities:
x
P(x)
0
1
2
3
4
5
6
7
8
9
10
0.000977
0.009766
0.043945
0.117188
0.205078
0.246094
0.205078
0.117188
0.043945
0.009766
0.000977
Sum
1
Graphically, the probability distribution looks like:
Binomial Distribution n = 10, p = .50
0.3
0.25
Probability
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
x
6
7
8
9
10
If we used a biased coin so that the probability of getting a head is only .3,
then the probability distribution would look like:
Binomial Distribution n = 10, p = .30
0.3
0.25
Probability
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
x
6
7
8
9
10
If the coin were extremely biased so that the probability of heads was only
.05, then the distribution would look like:
Binomial Distribution n = 10, p = .05
0.7
0.6
Probability
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
x
6
7
8
9
10
EXCEL allows one to compute the binomial probability distribution directly.
The form of the function is:
=binomdist(x, n, p, condition),
where x is the value of interest, n is the number of trials, p is the probability of
success and condition is either “false” or “true”.
If you specify the following command,
=binomdist(3, 10, .50, false),
then EXCEL will compute the probability that x = 3.
If you use the command,
=binomdist(3, 10, .50, true),
then Excel will compute the probability that x  3 . In other words EXCEL will
accumulate the probabilities for x = 0, x = 1, x = 2, and x = 3 and report the total.
The following table shows the use of both conditions in the case where n = 10,
and p = .5:
x
P(x)
P(<=x)
0
1
2
3
4
5
6
7
8
9
10
0.000977
0.009766
0.043945
0.117188
0.205078
0.246094
0.205078
0.117188
0.043945
0.009766
0.000977
0.000977
0.010742
0.054688
0.171875
0.376953
0.623047
0.828125
0.945313
0.989258
0.999023
1.000000
Sum
1
One can show that,
E(x) = np,
and,
SD( x )  np( 1  p )
If instead of x, the number of successes, we are interested in
p̂  x / n
that is the proportion of successes in n trials, then one can show that
E ( p̂ )  p
and,
SD( p̂ ) 
p( 1  p )
n
In our case,
E(x) = 10 * .5 = 5,
and,
SD( x )  (.5 ) * (.5 ) * 10  1.581139
Let us apply the binomial distribution to a more practical problem then
flipping coins.
Suppose that you are going to hire 10 persons from a pool of qualified
candidates which is 30 % women. You find that only 1 woman, and 9 men were
hired. Is this evidence that the firm is discriminating against women in hiring?
The first question is to determine if the binomial distribution is applicable.
Clearly each hire can only be a man or woman so there are only two possible
outcomes.
The hires are probably independent of one another.
The major problem is whether or not the probability of success, p, is constant
from trial to trial. If there were only a total of 20 applicants, 6 women and 14 men,
then if you hired one of the women on the first hire, that would leave 5 women and
14 men which would mean the probability of hiring a woman for the second hire
could only be
5 / 19 = .2632,
which is very large change from the initial probability of .30. On the other hand if
the hiring pool consisted of 100 applicants, 30 women and 70 men, then if you hired
one of the women on the first hire, that would leave 29 women and 70 men, which
would mean the probability of hiring a woman for the second hire would be
29 / 99 = .2929
which is a very small change.
The actual probability distribution to use is called the hypergeometric
distribution. However it is well know that if the probability p in the binomial
distribution does not change much from trial to trial, then the results from the
hypergeometric distribution and the binomial distribution are almost identical.
Assuming that the probability of hiring a woman does not change much over
the 10 hires, then we can reasonably assume that the probability is approximately
constant over the 10 hires and the assumptions of the binomial distribution are
approximately fulfilled.
The next problem we face is that any value between 0 and 10 can possibly
occur. Indeed the probability distribution for this situation, i.e. binomial with n =
10, and p = .3, is given in the table below:
Female
Hires
Prob
Cumulative
Prob
0
1
2
3
4
5
6
7
8
9
10
0.028248
0.121061
0.233474
0.266828
0.200121
0.102919
0.036757
0.009002
0.001447
0.000138
0.000006
0.028248
0.149308
0.382783
0.649611
0.849732
0.952651
0.989408
0.998410
0.999856
0.999994
1.000000
Sum
1.000000
Notice that although any value is possible the values are not all equally
probable. For example it would not at all seem odd if we hired 3 women or 2 women
or 4 women since these values all have reasonably high probabilities. On the other
hand, it would seem odd if we hired 10 women since the probability of this outcome
is approximately 6 chances in 1, 000, 000. Almost as rare as winning the Texas
Lottery!!
Statistical logic works like this:
a) define what you think is a rare event (most users of statistics define rare
as 1 chance in 20 [.05] or 1 chance in 100 [.01]);
b) if the probability of the observed result or anything more extreme is less
than what you define as rare, then the assumed value of p is suspect.
In our case we observed 1 female hire. More extreme is to hire 0 women.
Therefore we want the probability of observing 1 or fewer women. This can be
obtained directly from the above table or by using the =binomdist(1, 10, .3, true)
command.
The result is a probability of .149308. This is roughly a chance of 1 in 7
which most people would not think is rare. Accordingly this data would not be
suggestive of disproportionate hiring of women. Of course if it happened more than
once, it might be indicative. Suppose a month later the same thing happens. The
probability of hiring 1 woman in 10 hires from a pool that is 30% women twice
would be:
(.149308) * (.149308) = .0223
which for many people would make one suspect of a fair hiring environment.
Graphically, the probability distribution we would expect for x = the number
of women hired when you are hiring for 10 positions (n) from a pool of qualified
applicants which is 30% female ( p = .30) is given below with the observed and more
extreme value highlighted:
Number of Women Hires
0.300000
Probability
0.250000
0.200000
0.150000
0.100000
0.050000
0.000000
0
1
2
3
4
5
x
6
7
8
9
10
Now let us leave the percentage of women in the qualified pool the same as
previously (i.e. p =.3) but now hire for 50 positions (n=50). And assume again we
only hire 10% women (x=5). Then the probability distribution would look like:
Number of Women Hires
Probability
0.15
0.1
0.05
x=5
x
As can be seen, hiring only 5 women in 50 hires from a pool of 30 % women,
is a relatively rare event. Using the binomdist function I can compute:
P(x<= 5) = binomdist(5, 50, .3, true) = .00072.
This amounts to a chance of approximately 7 in 10,000 which is highly
improbable. Accordingly, we would be suspect in this situation that women are
being hired proportionate to their representation in the applicant pool
.
48
44
40
36
32
28
24
20
16
12
8
4
0
0
It is very easy to simulate the binomial distribution. Suppose we wish to
simulate values of x from a binomial distribution with n = 10 and p = .4.
For each trial, we could use the statement:
=if(rand()<=.4, 1, 0)
This would generate a value of 1 approximately 40% of the time. If we repeated the
above statement 10 times and added up the results, this would be equivalent to
taking a sample of n = 10 and observing x where x followed the binomial
distribution with p = .4.
The above procedure, however, is cumbersome. Notice that what we really
did was to divide the range of values between 0 and 1 into two regions. The first
went from 0 to .4. If the random number fell in this range, we said that the outcome
should be 1. If the random number fell in the range .4 to 1 we said the outcome
should be 0.
If we had a random variable with three possible outcomes, then we could
divide the interval between 0 and 1 into three regions with the size of each region
proportional to its probability.
In our case, with n = 10, we have eleven possible values which can occur,
namely the values 0, 1, 2, . . . , 10. We wish to divide these proportionally to their
probability given as:
Binomial Distribution
n = 10
p = .4
x
P(x)
0
1
2
3
4
5
6
7
8
9
10
0.0060466
0.0403108
0.1209324
0.2149908
0.2508227
0.2006581
0.1114767
0.0424673
0.0106168
0.0015729
0.0001049
Sum
1
This can easily be accomplished by computing the cumulative binomial
probability distribution which is shown below:
Cumulative
Binomial Distribution
n = 10
p = .4
x
P(<=x)
0
1
2
3
4
5
6
7
8
9
10
0.0060466
0.0463574
0.1672898
0.3822806
0.6331033
0.8337614
0.9452381
0.9877054
0.9983223
0.9998951
1
We now set up the following rule:
If Random Number is
between
0
0.006047
0.046357
0.16729
0.382281
0.633103
0.833761
0.945238
0.987705
0.998322
0.999895
and
and
and
and
and
and
and
and
and
and
and
0.006047
0.046357
0.16729
0.382281
0.633103
0.833761
0.945238
0.987705
0.998322
0.999895
1
Then x
=
0
1
2
3
4
5
6
7
8
9
10
For example if we generated the random number .188754, this value falls
between .16729 and .382281 so we would say that x = 3 for those 10 trials. If we
generated a second random number, say .99461, then this value falls between
.987705 and .998322 so that we would say that x = 8 for those 10 trials.
The entire process can be automated in further in EXCEL using the function
“LOOKUP”.
This function has three arguments:
The first is the value we wish to look up, in this case this is the random number.
The second argument is the table in which you want to look up the probability,
in our case this is the cumulative probability distribution column (but we must add the
value “0” in the row proceeding the first probability).
Finally, the third argument is the table containing the results, in our case the
values of x.
The entire process is shown in the following screen shot:
The first argument of the function “LOOKUP” is the random number in
column F row 175 (shown in blue). The second argument is the table of the
cumulative binomial distribution (with zero added) shown in green. The third
argument is the actual value of the lookup process, the value of x shown in lavender.
Notice that the ranges of the second and third arguments have the symbol
“$” prefixing both the column and row entry. This is necessary so that if the entries
are copied (as in using a table of random numbers), the relevant look up table
entries remain constant since in EXCEL relative addressing is always used.
The following steps a), b), and c) show the entire process simulating 25 times
the number of success when 10 trials are run.
Simulate Binomial with n = 10 and p = .4
a) Generate Distribution
binomdist(x,10,.4,true)
x
0
0
0.006047
1
0.046357
2
0.16729
3
0.382281
4
0.633103
5
0.833761
6
0.945238
7
0.987705
8
0.998322
9
0.999895
10
1
b) Generate Random Numbers
0.188754
0.105325
0.555951
0.480493
0.131033
0.99461
0.608283
0.207534
0.129214
0.085797
0.280296
0.161509
0.404118
0.220414
0.821619
0.997045
0.394184
0.052352
0.892615
0.335211
0.562853
0.490515
0.363331
0.583843
0.316501
c) Use LOOKUP(x,ProbCOL,ResultCOL) after anchoring results
3
2
4
4
2
8
4
3
2
2
3
2
4
3
5
8
4
2
6
3
4
4
3
4
3
The procedure described above will work for any discrete probability
distribution. In 2001, I discovered and esoteric function in EXCEL which in fact
will do the above procedure in one step, but only for the Binomial Distribution. The
name of this function is “Critbinom” and has the following syntax:
=critbinom(n, p, random number)
The table below shows the application of the above function to the same
random numbers used in the previous example. You will note that the results are
identical to those previously obtained using the lookup table.
a) Generate Random Numbers
0.188754
0.105325
0.555951
0.480493
0.131033
0.99461
0.608283
0.207534
0.129214
0.085797
0.280296
0.161509
0.404118
0.220414
0.821619
0.997045
0.394184
0.052352
0.892615
0.335211
0.562853
0.490515
0.363331
0.583843
0.316501
b) Use CRITBINOM(10, .4, random number)
3
2
4
4
2
8
4
3
2
2
3
2
4
3
5
8
4
2
6
3
4
4
3
4
3
This function replaces a rather complicated three stage procedure with a
simple two step procedure.
The Poisson Distribution
The Poisson Distribution is another distribution which arises in a great
number of business situations. It usually is applicable in situations where random
phenomena occur at a certain rate over a period of time. For example, it describes
the number of people in line at a checkout counter as well as the number of
telephone calls received at a switching point. It, like the Binomial Distribution, has
a canonical definition.
Assume you have an “exposure” variable such as time (it does not have to be
time, but it has to be continuous). Assume that this time period can be divided into
small enough increments, say of width dt, so that in any one of these intervals
something happens or doesn’t happen. For example, consider phone calls during an
hour period. Obviously we can divide the hour into small intervals, maybe of width
1 second, so that we can either receive one call or no call.
Assume the probability of an event occurring in an interval of width dt is
 dt. Assume further that the occurrence or non-occurrence of an event in one
interval is independent of the occurrence or non-occurrence of the same event in
another interval.
Then if one defines,
 = t
where t is the length of the interval, then the probability of x occurrences in the
interval of length t is given by,
P( x ) 
for x = 0, 1, 2, . . . .
e   x
x!
One can show that for the Poisson Distribution with parameter ,
E( x )  
and,
SD( x )  
One can usually easily recognize situations where the Poisson Distribution is
applicable since they usually involve a rate and an exposure.
For example, suppose an office receives, on average, 15 calls per hour. In a
two hour period the office received 45 calls. Is this a rare event? Here the rate is 15
calls per hour and the exposure is a two hour period. This implies that
 = 15 * 2 = 30.
As another example, suppose a manufacturing plant has, on average, 24
accidents per year. In a one month period it has 5 accidents. Is this a rare event?
Here the rate is 24 accidents per year and the exposure is one month ( 1/12 of a
year). This implies that
 = 24 * (1 /12) = 2.
The exposure rate does not have to be time. For example consider the
situation where, on average, a driver has 1 accident per 50,000 miles driven.
Suppose the person drives a vehicle for 100,000 miles and has no accidents is this a
rare event? Here the rate is 1 accident per 50,000 miles driven and the exposure is
100,000 miles so that
 = 100,000 * ( 1 / 50,000) = 2.
The probability distribution for this last case is given below:
x
P(x)
P(<=x)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
0.135335
0.270671
0.270671
0.180447
0.090224
0.036089
0.012030
0.003437
0.000859
0.000191
0.000038
0.000007
0.000001
0.000000
0.135335
0.406006
0.676676
0.857123
0.947347
0.983436
0.995466
0.998903
0.999763
0.999954
0.999992
0.999999
1.000000
1.000000
Note that the probabilities continue on past 12, it is just that they are so small
that they appear in the table as zero.
Graphically, the distribution is as shown below:
Poisson Distribution mu =2
0.30
0.25
Probability
0.20
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
x
8
9
10
11
12
13
Now consider the situation where we are inspecting parts for defects.
Assume we have a defective rate of 1/1000. If we inspect 1000 parts and observe 3
defects, should we worry? In this case  = 1000 * ( 1 /1000) = 1.
As in the binomial case, EXCEL can be used to compute the probability of
this or any more extreme event. The function to use has the form,
=Poisson(x, mu, condition),
where, just as in the case of the binomial distribution, a condition of ‘false’ gives us
the probability of x , and a condition of ‘true’ gives us the probability of being less
than or equal to x.
In this case we want the probability of 3 or anything more extreme, that is
the probability of three or more. We can find,
P ( x  2 )  POISSON ( 2 ,1 , true )  .919699
so that,
P ( x  3 )  1  P ( x  2 )  1  .919699  .080311
which is not rare by the usual standards of .05 or .01.
A picture of the probabilities to be added is shown below:
Defective Rate .001, n = 1000
Probability
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
8
Defects
9
10
11
12
13
14
15
Now assume we scale up the situation and inspect 5,000 parts. Then the
parameter would change to
 = 5,000 * (1 / 1000) = 5.
If we scale up the defective rate to mirror that in the first problem, this
would correspond to observing 5 * 3 = 15 defectives.
Graphically the Poisson Distribution with  = 5 would look like:
24
20
22
16
18
12
14
8
10
6
4
x=15
2
0
0.2
0.18
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
By using the Poisson function in EXCEL, one obtains:
P ( x  15 )  1  P ( x  14 )  1  POISSON ( 14 ,5 , true )  1  .999774  .000226
which is clearly a relatively rare event which might encourage us to improve quality
control.
Actually, I have been misleading you a bit. For in fact if we inspect 5,000
parts each of which has a probability of .001 ( 1 / 1000) of being defective, I am
really describing a Binomial situation.
However if n is large and p is small, so that n * p is moderate, then the
Poisson distribution can be used to approximate the binomial distribution by taking
 = n * p.
The graph below shows how good this approximation is in this case with n =
5,000 and p = .001 and  = 5.
Poisson Approximation to Binomial
0.18
0.16
0.14
0.12
Prob
0.1
0.08
poisson
0.06
binomial
0.04
24
22
20
18
16
14
12
10
8
6
4
2
0
0
0.02
x
The Poisson distribution gave the probability of 15 or more defectives as
.000226, while the exact value from the Binomial distribution is .000224, a small
error in the sixth decimal place!
Simulating a Poisson distribution is essentially the same as the long
procedure for the Binomial Distribution (unfortunately no “one step” function exists
for the Poisson distribution that can be used like “critbinom” for the binomial).
Below is a screen shot of the procedure:
A step by step illustration of simulating 25 realizations of the Poisson
Distribution with parameter 5 is shown below:
Simulate a Poisson with lambda = 5
a) Generate Distribution
poisson(x,5,true)
x
0
0 0.006738
1 0.040428
2 0.124652
3 0.265026
4 0.440493
5 0.615961
6 0.762183
7 0.866628
8 0.931906
9 0.968172
10 0.986305
11 0.994547
12 0.997981
13 0.999302
b) Generate Random Numbers
0.188754
0.105325
0.555951
0.480493
0.131033
0.99461
0.608283
0.207534
0.129214
0.085797
0.280296
0.161509
0.404118
0.220414
0.821619
0.997045
0.394184
0.052352
0.892615
0.335211
0.562853
0.490515
0.363331
0.583843
0.316501
c) Use LOOKUP(x,ProbCOL,ResultCOL) after anchoring results
3
2
5
5
3
12
5
3
3
2
4
3
4
3
7
12
4
2
8
4
5
5
4
5
4
The Normal Distribution
The normal distribution (the so called "curve") is perhaps the best known
probability distribution since it arises so many situations.
In business applications is commonly found to describe the distribution of the
rate of return on investments. However much business data is right skewed. If x is
a typical business statistics, for example the assets of banks or the gross sales of
companies, one usually finds that many small firms have modest values with a few
very large values. This gives rise to a right skewed distribution. However, if one
looks at the log (assets of banks) or the log(gross sales), one finds that the
logarithmic value is approximated closely by the normal distribution. Such
variables are said to have the Log-Normal distribution.
Unlike the Binomial and Poisson distributions, the normal distribution is
defined, theoretically, for continuous variables, that is variables with no gaps
between potential values. Of course, in the real world, one never measures things to
very many decimal places so that we can think of the normal distribution applying
to many everyday variables. For example, if a man says he is six feet tall, he
probably does not mean that he is exactly six feet tall. What is usually meant is that
to the nearest inch, the man is six feet tall. That is his actual height is probably
between five foot eleven and one half inches, and six feet and one half inch.
Formally this is called "discretizing" the normal distribution.
Since the normal is a continuous curve, it does not have a probability
distribution. Instead it has what is called a probability density function. A
probability density function is a non-negative function f(x), which has the property
that:
b
P ( a  x  b )   f ( x )dx
a
and

 f ( x )dx  1

The form of the function f(x) for a normal distribution is:
f( x)
e
 x

 
2


2 
The normal distribution depends on two parameters  and . We have used
these symbols before to describe the mean and standard deviation of a population.
For the normal distribution:
E( x )  
and,
SD( x )  
Below is a picture of two normal distributions which have the same standard
deviations but different means:
5.7
4.8
3.9
3
2.1
1.2
0.3
-0.6
-1.5
-2.4
-3.3
-4.2
-5.1
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-6
f(x)
Normal Distributions
x
Increasing the mean moves the normal curve to the right, while decreasing
the mean moves the curve to the left.
Below are shown normal distributions with the same mean but with different
standard deviations:
Normal Distributions
1
f(x)
0.8
0.6
0.4
0.2
2
6.
1
5.
4
-7
-5
.9
-4
.8
-3
.7
-2
.6
-1
.5
-0
.4
0.
7
1.
8
2.
9
0
x
Increasing the standard deviation makes the curve more spread out and
lower (this has to occur since the total area under the curve is always 1). Decreasing
the standard deviation makes the curve less spread out and higher.
It is very easy to work with the normal distribution using EXCEL. Like the
case of the Binomial and Poisson distributions, EXCEL provides a function for
computing values for any normal distribution. The form of the function is :
=normdist(x, mean, sd, condition)
As in the case of the Binomial and Poisson distributions, setting condition =
"true" gives the probability of being less than or equal to a particular value. If the
condition is set to "false" then one gets the value of f(x) which unlike the Binomial
and Poisson distributions is not a probability.
EXCEL also has the function:
=normsdist(z)
which gives the probability of being less than or equal to the value z for the special
normal distribution with a mean of 0 and a standard deviation of 1 (called the
Standard Normal Distribution). The Standard Normal Distribution, in the past,
was quite important since it was used to compute probabilities for any normal
distribution by just using a table of the Standard Normal Distribution and the
transformations:
z = (x - ) / ,
and the inverse relationship,
x=+z
With the advent of worksheet programs, one no longer needs these tables
since the computer can generate the probabilities of interest directly.
Let us consider an application of the normal distribution. In 1980, the height
of adult males in the United States was approximately normally distributed with an
average height of 67 inches and a standard deviation of 2.1 inches. (In 2000, the
average height has jumped to 69 inches). In 1980, what was the probability that a
randomly chosen adult male would be six feet tall or taller?
Graphically, we are interested in finding the probability in the white area in
the graph below:
Height Distribution
0.15
0.1
Height
73.5
72.5
71.5
70.5
69.5
68.5
67.5
66.5
65.5
64.5
63.5
62.5
61.5
0.05
0
60.5
f(x)
0.2
EXCEL can be used to find the area of the shaded area in the above graph
by using the command:
P ( x  72 )  normdist ( 72 ,67 ,2.1 , true )  .991366
The probability of being six feet tall or greater (the area of the white area) is then
given by:
P ( x  72 )  1  .991366  .008634
or slightly less than 1% of the population.
Now suppose we wished to find the proportion of adult U.S. males who were
between 5 foot three (63 inches) and 5 foot ten (70 inches) tall. This is the shaded
area in the graph below:
Height
73.1
72.4
71.7
71
70.3
69.6
68.9
68.2
67.5
66.8
66.1
65.4
64.7
64
63.3
62.6
61.9
61.2
0.2
0.18
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
60.5
f(x)
Normal Distribution
To find this probability we need to realize that:
P ( 63  x  70 )  P ( x  70 )  P ( x  63 )
or in EXCEL terminology:
P ( 63  x  70 )  normdist ( 70 ,67 ,2.1 , true )  normdist ( 63 ,76 ,2.1 , true )
This gives the answer as:
P ( 63  x  70 )  .923436  .028405  .895031
Approximately 89.5% of adult U. S. males are between 5 foot three and 5 foot 10 in
height.
Suppose I was interested in the inverse problem, that is what are the heights
of 95% of adult U.S. males?
Actually there are many ways to do this. I could for example find the
minimum height that 95% of the population is greater than. Or I could include the
lower 95 %. Or I could try to get the "middle" 95%.
The middle 95 % would have 2.5% of the observations larger and 2.5%
smaller. This is shown in the picture below:
0.2
0.18
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
95 %
73.1
71.7
71
70.3
69.6
68.9
68.2
67.5
66.1
65.4
64.7
64
63.3
62.6
61.9
61.2
66.8
Height
72.4
2.5%
2.5%
60.5
f(x)
Normal Distribution
EXCEL has a function which solves the following equation for x0 given any
value of p for the normal distribution:
P( x  x0 )  p
The function is:
 NORMINV ( p ,  , )
Therefore in our case, we find that:
62.884 = norminv(.025, 67, 2.1)
and,
71.116 = norminv(.975, 67, 2.1).
Therefore 95% of the heights fall between the values of 62.884 inches and
71.116 inches.
Recall that the mound rule indicated that approximately 95% of the values
fell within the interval +/- two standard deviations. The values 62.884 and 71.116
correspond to +/- 1.96 standard deviations which is the more precise figure.
EXCEL also has the function:
=normsinv(p)
which is the inverse function for the the standard normal distribution (mean of zero
and standard deviation of 1). Directly we could have obtained:
-1.96 = normsinv(.025)
and
1.96 = normsinv(.975).
In a non-computerized statistics course, the above values would have been
obtained from a table in the back of the book and then transformed to get:
67 – 1.96 * 2.1 = 62.884
and
67 + 1.96 * 2.1 = 71.116
In EXCEL we can do this directly using the norminv function.
For large values of n, we can use the normal distribution to approximate the
binomial distribution by taking:
  np
  np( 1  p )
This approximation will be valid if np > 5 and n(1 – p) > 5.
For example in the case discussed previously with n = 50 and p = .3, we would
have
 = 50 * .3 = 15
  50(.3 )(.7 )  3.24037
and in this case np = 50(.3) = 15 and n(1-p) = 50 (.7) = 35 so that the approximation
should be good.
The two curves are plotted below:
Normal Approximation to Binomial
0.14
0.12
Prob
0.1
0.08
binomial
normal
0.06
0.04
0.02
48
45
42
39
36
33
30
27
24
21
18
15
9
12
6
3
0
0
x
Notice however that the normal curve is continuous while the binomial
distribution is discrete with nothing between the values of say 17 and 18.
This distinction between discrete and continuous is important. For the
binomial distribution
Pbin ( x  10 )  Pbin ( x  10 )  Pbin ( x  10 )
so that a distinction must be made between "less than or equal" and "less than".
For the normal distribution however,
Pnor ( x  10 )  Pnor ( x  10 )
sin
ce there is no probability that x will exactly equal 10 (by exactly we mean to an
infinite number of decimal places).
We can get around this problem of continuous versus discrete by the use of
what is called the "continuity correction". Simply it says that when using the normal
distribution to approximate a discrete distribution (such as the Binomial or
Poisson), assume that any discrete value, say 10, actually goes half way between the
previous discrete value and the subsequent discrete value. In other words when
using the normal distribution we assume that 10 actually goes from 9.5 to 10.5; 29
would go from 28.5 to 29.5, etc. If we let k be one of the discrete values, then the
following five relationships illustrate the use of the continuity correction in all
possible cases:
Pbin ( x  k )  Pnor ( k  .5  x  k  .5 )
Pbin ( x  k )  Pnor ( x  k  .5 )
Pbin ( x  k )  Pnor ( x  k  .5 )
Pbin ( x  k )  Pnor ( x  k  .5 )  1  Pnor ( x  k  .5 )
Pbin ( x  k )  Pnor ( x  k  .5 )  1  Pnor ( x  k  .5 )
Fortunately, in EXCEL we can use the binomdist function for most cases and
need to use the normal approximation to the Binomial only infrequently.
The Poisson distribution can also be approximated by the normal
distribution if the Poisson parameter  > 5. By taking  as given and taking the
standard deviation as the square root of , we can approximate the Poisson
distribution with the normal distribution.
The following graph shows the Poisson distribution with mean of 5 and the
normal distribution with a mean of 5 and standard deviation equal to 2.2361 (the
square root of 5):
0.2
0.18
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
24
22
20
18
16
14
12
10
8
6
4
2
Poisson
Normal
0
Prob
Normal Approximation to Poisson
x
Again if using the normal distribution to compute probabilities for the
Poisson distribution one must correct for continuity in the same way as was done for
the binomial distribution.
It is very easy to simulate data which follows the normal distribution using the
norminv function in EXCEL.
The first step is to generate a set of random numbers using the rand()
function as we have done before. Some sample data is shown below:
a) Use RAND() to simulate 50 random numbers
0.407998
0.678836
0.1953
0.013617
0.927638
0.327113 0.785353 0.167275 0.016527 0.524512 0.445768
0.059317 0.764079 0.228678 0.055253 0.850148 0.517072
0.250051 0.962008 0.09676 0.495551 0.021171 0.132306
0.945737 0.29648 0.938787 0.034023 0.353435 0.298913
0.541888 0.09178 0.58483 0.825083 0.348549 0.309452
0.098211
0.37086
0.589669
0.840628
0.341382
0.739599
0.422763
0.349044
0.014406
0.926097
0.815619
0.165053
0.176948
0.689208
0.406182
Now suppose we were looking at an investment with a mean return of 8%
with a risk (sd) of 2%. Then we could generate 50 normally distributed returns by
applying the function
=norminv(random number, .08, .02)
to the 50 random numbers previously generated to get:
b) Use NORMINV(x, .08, .02) to generate the normal random variables
0.075346
0.089289
0.062829
0.035837
0.109168
0.071042 0.095808
0.0607 0.037372 0.08123 0.077273
0.048789 0.09439 0.065136 0.048082 0.100741 0.080856
0.066513 0.11549 0.053995 0.079777 0.039397 0.057689
0.112097 0.069309 0.110893 0.043506 0.072479 0.069449
0.082104 0.053403 0.084285 0.098698 0.072215 0.070052
0.054164
0.073408
0.084534
0.099941
0.071826
0.092842
0.076103
0.072242
0.036279
0.108946
0.097976
0.060522
0.061459
0.089872
0.075252
Finally, one should check that the simulation is approximately on target.
For this data, the sample mean of the simulated values is .0748 compared to
the theoretical value of .08. The standard deviation of the simulated values if .0210
compared to the theoretical value of .02. Finally, I have done a histogram of the
simulated values which is shown below:
Frequency Distribution
14
12
Frequency
10
8
6
4
2
0.
1
0.
11
0.
12
M
or
e
8
7
6
5
4
9
0.
0
0.
0
0.
0
0.
0
0.
0
0.
0
0.
0
3
0
Simulated Returns
As can be seen it is approximately normally distributed.
What if we wanted to simulate correlated investments as we studied earlier in
this module? Specifically suppose we wanted to simulate 25 years of returns on two
investments. The first having a mean return of .08 and a standard deviation of .02
and the second having a mean return of .12 with a standard deviation of .05. And
suppose that the investments are correlated with a correlation coefficient of -.4.
The difficult part is generating the values so they are correlated. Fortunately
there is a theoretical result that says if z1 and z2 are two independent random
variables each with mean 0 and standard deviation 1, and we define two new
variables x1 and x2 with the equations:
x1 
1 r
1r
z1 
z2
2
2
x2 
1 r
1r
z1 
z2
2
2
then x1 and x2 will both still have mean 0 and standard deviation 1, but now the x's
will be correlated with correlation coefficient r.
Let us illustrate this procedure in steps. First generate two columns of 25
random numbers using the =rand() EXCEL function. The data would look like:
a) Start with random numbers in two sets of 25
0.758384
0.301347
0.741409
0.591236
0.742038
0.764761
0.928779
0.754592
0.05917
0.324527
0.389161
0.159801
0.621819
0.091531
0.140335
0.444617
0.799008
0.266267
0.215947
0.031036
0.014389
0.526516
0.129849
0.459673
0.565416
0.311735
0.87599
0.306248
0.70228
0.518124
0.70556
0.086294
0.914354
0.699389
0.91737
0.781662
0.369674
0.359246
0.78965
0.801505
0.556215
0.512305
0.686824
0.298083
0.07719
0.337121
0.597915
0.06653
0.644828
0.1606
Next use the EXCEL function =normsinv(random number) to generate two
columns of uncorrelated normal random variables with mean zero and standard
deviation 1. The data would look something like this:
b) Generate two independent set of Normal with Mean of 0 and sd of 1
by using Normsinv (x)
0.701114 -0.49094
-0.52053 1.155174
Check correlation using CORREL
0.647696 -0.50651
0.230726 0.53097
0.017693
0.64964 0.045444
0.721701 0.54046
1.466761 -1.36393
0.689011 1.368067
-1.56178 0.522645
-0.45508 1.387595
-0.28151 0.77782
-0.99527 -0.33272
0.310261 -0.36047
-1.33138 0.805206
-1.07881 0.847008
-0.13927 0.14138
0.838082 0.030848
-0.62414 0.486869
-0.78595 -0.52992
-1.86578 -1.42423
-2.18651 -0.42033
0.066516 0.247955
-1.1271 -1.50215
-0.10126 0.371394
0.164715 -0.99199
Using the =correl(z1, z2) we get a correlation of .017693 compared to the
theoretical value of 0.
Next we implement the formula given above to induce the appropriate
correlation, in this case r = -.4. The formula would look like:
c) Call columns
z1
z2
d) Generate two columns x1 and x2 by using the formula
x1=sqrt((1+R)/2)*z1 + SQRT((1-R)/2)*z2
x2=sqrt((1+R)/2)*z1 - SQRT((1-R)/2)*z2
The actual formula contained in the cell for x1 in shown on the following
screen shot:
The actual formula for the cell for x2 is shown in the screen shot below:
After transforming all the pairs, one would get the following result:
e) Example make R = -.4
-0.02673
0.681382
-0.06902
0.570615
0.393844
0.847474
-0.33777
1.521994
-0.41814
0.911689
0.496583
-0.8235
-0.13166
-0.05555
0.117767
0.042005
0.484846
0.065486
-0.87385
-2.21353
-1.54928
0.243886
-1.87413
0.255269
-0.73974
0.794765
-1.25159
0.778537
-0.31787
0.317801
-0.05689
1.944528
-0.76722
-1.2927
-1.4102
-0.80496
-0.26676
0.471531
-1.40291
-1.29955
-0.19457
0.433227
-0.7492
0.01288
0.169663
-0.84593
-0.17102
0.639446
-0.36619
0.920179
Check
-0.2958
Finally we need to adjust the generated values to have the appropriate means
and standard deviations. The first investment is supposed to have a mean of .08 and
a standard deviation of .02, therefore we create the new variable
.08 + .02 * x1
The second investment is supposed to have a mean of .12 and a standard deviation
of .05, so we create the new variable
.12 + .05 * x2
The final results are shown below:
f) Now multiply x1 by .02 and add .08 and multiply x2 by .05 and add .12
0.079465
0.093628
0.07862
0.091412
0.087877
0.096949
0.073245
0.11044
0.071637
0.098234
0.089932
0.06353
0.077367
0.078889
0.082355
0.08084
0.089697
0.08131
0.062523
0.035729
0.049014
0.084878
0.042517
0.085105
0.065205
0.159738
0.05742
0.158927
0.104107
0.13589
0.117156
0.217226
0.081639
0.055365
0.04949
0.079752
0.106662
0.143577
0.049854
0.055023
0.110272
0.141661
0.08254
0.120644
0.128483
0.077704
0.111449
0.151972
0.10169
0.166009
Check results
Average
Sd
0.078016 0.11057
0.017468 0.042937
corr=
-0.2958
As can be seen the simulation results agree reasonably with the theoretical
values.
I could now simulate what would happen for a 25 year period into the future
if I invested $10,000 in each investment. I just need to add one to the simulated
returns and cumulate the investment history as shown below:
Initial Investment
Year
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Investment
1
$10,000
Investment
2
$10,000
$10,795
$11,805
$12,733
$13,897
$15,119
$16,584
$17,799
$19,765
$21,181
$23,262
$25,353
$26,964
$29,050
$31,342
$33,923
$36,666
$39,954
$43,203
$45,904
$47,544
$49,875
$54,108
$56,409
$61,209
$65,200
1.079465 $11,598 1.159783
1.093628 $12,264 1.05742
1.07862 $14,213 1.158927
1.091412 $15,692 1.104107
1.087877 $17,825 1.13589
1.096949 $19,913 1.117156
1.073245 $24,239 1.217226
1.11044 $26,218 1.081639
1.071637 $27,669 1.055365
1.098234 $29,039 1.04949
1.089932 $31,355 1.079752
1.06353 $34,699 1.106662
1.077367 $39,681 1.143577
1.078889 $41,659 1.049854
1.082355 $43,951 1.055023
1.08084 $48,798 1.110272
1.089697 $55,711 1.141661
1.08131 $60,309 1.08254
1.062523 $67,585 1.120644
1.035729 $76,268 1.128483
1.049014 $82,195 1.077704
1.084878 $91,355 1.111449
1.042517 $105,239 1.151972
1.085105 $115,940 1.10169
1.065205 $135,188 1.166009
Total Porfolio Value After 25 Years = $200,388
By simply pressing the F9 key, I would recompute all of the values in the
above simulation to get results such as:
f) Now multiply x1 by .02 and add .08 and multiply x2 by .05 and add .12
0.075041
0.061764
0.083372
0.094959
0.081673
0.063944
0.06562
0.094026
0.062977
0.055172
0.058103
0.091082
0.06549
0.08358
0.067008
0.099086
0.07221
0.060327
0.072065
0.090064
0.074848
0.086461
0.111891
0.112906
0.072155
0.011628
0.208917
0.044281
0.121946
0.113509
0.105207
0.172036
0.141238
0.141087
0.16497
0.088822
0.110032
0.086719
0.11165
0.0865
0.069589
0.128037
0.112246
0.059908
0.095344
0.161608
0.111251
0.091642
0.08298
0.104165
Check results
Average
Sd
0.078233 0.109012
0.016182 0.042203
corr=
-0.30182
And another realization of a 25 year investment as:
Initial Investment
Year
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Investment
1
$10,000
$10,750
$11,414
$12,366
$13,540
$14,646
$15,583
$16,605
$18,167
$19,311
$20,376
$21,560
$23,524
$25,064
$27,159
$28,979
$31,850
$34,150
$36,211
$38,820
$42,316
$45,484
$49,416
$54,945
$61,149
$65,561
Investment
2
$10,000
1.075041
1.061764
1.083372
1.094959
1.081673
1.063944
1.06562
1.094026
1.062977
1.055172
1.058103
1.091082
1.06549
1.08358
1.067008
1.099086
1.07221
1.060327
1.072065
1.090064
1.074848
1.086461
1.111891
1.112906
1.072155
$10,116
$12,230
$12,771
$14,329
$15,955
$17,634
$20,667
$23,586
$26,914
$31,354
$34,139
$37,895
$41,182
$45,780
$49,740
$53,201
$60,013
$66,749
$70,748
$77,493
$90,016
$100,031
$109,198
$118,259
$130,578
Total Porfolio Value After 25 Years = $196,139
1.011628
1.208917
1.044281
1.121946
1.113509
1.105207
1.172036
1.141238
1.141087
1.16497
1.088822
1.110032
1.086719
1.11165
1.0865
1.069589
1.128037
1.112246
1.059908
1.095344
1.161608
1.111251
1.091642
1.08298
1.104165
By repeatedly recording the Total Portfolio Value after 25 years for each
simulation, I could obtain the mean and variation of the results. This is illustrated
in the table below for ten simulations of the twenty five year period:
25 Year
Simulation
Number
Portfolio Value
at end of
25 Years
1
2
3
4
5
6
7
8
9
10
$200,388
$196,139
$189,450
$208,590
$279,617
$225,883
$203,936
$267,466
$203,507
$191,895
Mean
SD
$216,687
$31,745
Notice that the SD (risk) of the return after 25 years of this portfolio is very
high. By Chebyshev's rule the 25 year return would vary between approximately
$153,200 and $280,200 at least 75 % of the time.
Note: 10 is a very small number of simulations. 100 would be much
better
Simulating a Decision Problem
In Lecture 3 of this module we discussed how to use decision trees to help
make decisions in an uncertain environment. It is possible to simulate the problem
outlined by the decision tree as an alternative to the formal procedures outlined
earlier. Below is decision tree of that problem.
Your
Decision
Consultants
Say
Your
Decision
Invest 2
Actual
State
Profit in
Millions
0.8182
Low
2
0.1136
Medium
5
0.0682
High
11
0.5000
A
3
0.5000
B
6
0.0227
Low
2
0.9091
Med
5
0.0682
High
11
0.5000
A
3
0.5000
B
6
0.0294
Low
2
0.2206
Med
5
0.7500
High
11
0.5000
A
3
0.5000
B
6
0.2000
Low
3
0.5000
Med
6
0.3000
High
12
0.5000
A
4
0.5000
B
7
Low
Invest 1
0.2200
Invest 2
Hire Consultants
0.4400
Med
Invest 1
0.3400
Invest 2
Start
High
Invest 1
Invest 2
Don't Hire Consultants
Invest 1
The first step in simulating the above tree is to devise a way to simulate the
uncertain outcomes, that is, the outcomes over which we have no control. In the
above tree there are three such outcomes. These are identified by branches labeled
with probabilities. Looking at the tree from left to right, we first encounter the
uncertainty in what the consultants will say. That is:
Consultants
Cumulative
Say
Probability Probability
0.00
Low
0.22
0.22
Medium
0.44
0.66
High
0.34
---------------1
1.00
To simulate this distribution we need only simulate a random number using
the =rand() function in EXCEL. If the random number is between 0.00 and 0.22,
then the outcome corresponds to the Consultants saying “Low”. If it is between 0.22
and 0.66, then the outcome corresponds to the Consultants saying “Medium” and if
the outcome is greater than .66, the outcome corresponds to the Consultants saying
“High”.
Again moving from left to right, we encounter the two possible outcomes, A
and B. For Investment 1, the probability table would look like:
Outcome
Cumulative
Probability Probability
0.00
A
0.5
0.50
B
0.5
---------------1
1.00
In order to simulate this outcome, we need to generate one random number
and compare it to 0.50. If it is less than or equal to 0.50, then we indicate that
Outcome A has occurred, otherwise outcome B occurs.
Simulation of Investment 2 is more complicated since the probability
distribution depends on what the consultants say. For example if the Consultants
say “Low”, then we simulate the Actual State from the following table:
Consultants say Low
Actual
State
Outcome
Cumulative
Probability Probability
0.0000
Low
0.8182
0.8182
Medium
0.1136
0.9318
High
0.0682
---------------1
1.0000
If the generated random number is between 0.0000 and 0.8182, then we
record that the Actual State was Low; if the random number is between 0.8182 and
0.9318, we record that the Actual State was Medium; and if the random number is
greater than 0.9318 we record that the Actual State was High.
If the Consultants say Medium, then the appropriate distribution to simulate
is:
Consultants say Medium
Actual
State
Outcome
Cumulative
Probability Probability
0.0000
Low
0.0227
0.0227
Medium
0.9091
0.9318
High
0.0682
---------------1
1.0000
If the generated random number is between 0.0000 and 0.0227, then we
record that the Actual State was Low; if the random number is between 0.0227 and
0.9318, we record that the Actual State was Medium; and if the random number is
greater than 0.9318 we record that the Actual State was High.
Finally, if the consultants say High, then the appropriate distribution for the
simulation is given by:
Consultants say High
Actual
State
Outcome
Cumulative
Probability Probability
0.0000
Low
0.0294
0.0294
Medium
0.2206
0.2500
High
0.7500
---------------1
1.0000
If the generated random number is between 0.0000 and 0.0294, then we
record that the Actual State was Low; if the random number is between 0.0294 and
0.2500, we record that the Actual State was Medium; and if the random number is
greater than 0.2500 we record that the Actual State was High.
In the EXCEL file “simulatedecision.xls”, the above procedures are
implemented. The results for ten simulations are shown below:
Random Components
Repeat
Investment
1
Consultant
Says
Actual
State
1
2
3
4
5
6
7
8
9
10
B
A
A
A
A
A
B
B
B
B
Med
Med
High
High
Med
Med
Med
High
Med
Med
Med
Med
High
High
High
Med
Med
Med
Med
Med
We have now finished with all of the random components of the tree, that is,
we can now simulate the outcomes over which the manager has no control. Now it is
necessary to list the possible decision that the manager can make.
Again looking at the decision tree from left to right, the first major decision
is whether or not to hire the Consultants. Following the “Don’t Hire Consultants”
branch, the manager must next decide whether to pick Investment 1 or Investment
2.
On the other hand, if one decides to hire the consultants, the consultants will
give their report and then the manager must decide between Investment 1 and
Investment 2. Since we do not know what the consultant’s will say, a strategy must
specify what investment the manager will pick for every possible prediction the
consultant’s could make. That is, we must specify an investment choice for the three
possible predictions of Low, Medium, and High. Since there are two possible
choices for each of the consultant’s possible predictions, there are 8  2  2  2
possible strategies that must be specified. These are shown below:
Manager Decision Strategies
Investment Pick
If Consultants Say
Strategy
I
II
III
IV
V
VI
VII
VIII
Low
Medium
High
1
2
1
2
2
1
1
2
1
2
2
1
2
1
2
1
1
2
2
2
1
2
1
1
In total there are 10 possible strategies that the manager can make, two if
one doesn’t hire the consultants and eight if one does.
One now adds ten columns, one for each strategy, to the random columns that we specified before. The results would
look like:
Simulation of the Investment Decision
Random Components
Repeat
1
2
3
4
5
6
7
8
9
10
Investment
1
B
A
A
A
A
A
B
B
B
B
Consultant
Says
Med
Med
High
High
Med
Med
Med
High
Med
Med
Possible Decisions
Actual
State
Med
Med
High
High
High
Med
Med
Med
Med
Med
No Consult
Invest 1
7
4
4
4
4
4
7
7
7
7
No Consult
Invest 2
6
6
12
12
12
6
6
6
6
6
Hire Consult
I
Hire Consult
II
Hire Consult
III
Hire Consult
IV
Hire Consult
V
Hire Consult
VI
Hire Consult
VII
Hire Consult
VIII
Low = 1
Med = 1
High = 1
Low = 2
Med = 2
High = 2
Low = 1
Med = 2
High = 2
Low = 2
Med = 1
High = 2
Low = 2
Med = 2
High = 1
Low = 1
Med = 1
High = 2
Low = 1
Med = 2
High = 1
Low = 2
Med = 1
High = 1
6
3
3
3
3
3
6
6
6
6
5
5
11
11
11
5
5
5
5
5
5
5
11
11
11
5
5
5
5
5
6
3
11
11
3
3
6
5
6
6
5
5
3
3
11
5
5
6
5
5
6
3
11
11
3
3
6
5
6
6
5
5
3
3
11
5
5
6
5
5
6
3
3
3
3
3
6
6
6
6
Examine “Repeat 1”. For the No Consultant Investment 1 strategy, the simulation indicated that outcome B occurred
for investment 1, so are gain would have been 7 (million dollars). For the No Consultant Investment 2 strategy, the actual state
was Medium so we would gain 6 (million dollars). The other eight strategies all involved hiring the consultants at a cost of 1
(million dollars). The simulation indicates that the Consultants said Medium so for Hire Consultant Strategies I, IV, VI and
VIII we would gain 6 (million dollars) since the Investment 1 result was B which paid 7 (million dollars) but we paid one
million for the consultants. For the Hire Consultant Strategies II, III, V, and VII we would gain 5 (million dollars) since we
chose Investment 2 which pays 6 (million dollars) for the Medium actual state but the cost of the consultants was one million
dollars. Try working out the results for “Repeat 5”.
Below is an example of fifty repetitions of the simulation along with the
average, standard deviation and ratio of mean to standard deviation for all ten
strategies.
Simulation of the Investment Decision
Random Components
Possible Decisions
InvestmentConsultant Actual
Repeat
1
Says
State
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
B
B
B
B
A
A
A
A
A
A
A
A
B
A
A
B
B
B
A
B
A
A
A
B
A
A
B
A
B
B
B
B
B
B
B
A
A
B
B
A
A
B
B
B
B
A
B
A
A
A
Low
High
Med
Med
Med
Med
Med
Med
High
High
Med
High
Low
High
Med
High
High
Low
High
Low
High
Med
Low
Med
High
High
Low
Low
High
Med
Med
Med
Med
High
Low
Med
Med
High
Med
High
Low
High
Med
High
Med
Med
High
Low
Med
High
No Consult No Consult Hire Consult Hire Consult Hire Consult Hire Consult Hire Consult Hire Consult Hire Consult Hire Consult
Invest 1
Invest 2
I
II
III
IV
V
VI
VII
VIII
Low = 1
Med = 1
High = 1
Low = 2
Med = 2
High = 2
Low = 1
Med = 2
High = 2
Low = 2
Med = 1
High = 2
Low = 2
Med = 2
High = 1
Low = 1
Med = 1
High = 2
Low = 1
Med = 2
High = 1
Low = 2
Med = 1
High = 1
Med
Med
Med
Med
Med
High
Med
Med
Med
High
Med
High
Low
High
Med
High
Med
Low
Med
Low
High
Med
Med
Med
High
High
Low
Low
High
Low
Med
Med
Med
High
Low
Med
Med
High
Med
High
Low
Med
Med
High
Med
Med
High
Low
Med
High
7
7
7
7
4
4
4
4
4
4
4
4
7
4
4
7
7
7
4
7
4
4
4
7
4
4
7
4
7
7
7
7
7
7
7
4
4
7
7
4
4
7
7
7
7
4
7
4
4
4
6
6
6
6
6
12
6
6
6
12
6
12
3
12
6
12
6
3
6
3
12
6
6
6
12
12
3
3
12
3
6
6
6
12
3
6
6
12
6
12
3
6
6
12
6
6
12
3
6
12
6
6
6
6
3
3
3
3
3
3
3
3
6
3
3
6
6
6
3
6
3
3
3
6
3
3
6
3
6
6
6
6
6
6
6
3
3
6
6
3
3
6
6
6
6
3
6
3
3
3
5
5
5
5
5
11
5
5
5
11
5
11
2
11
5
11
5
2
5
2
11
5
5
5
11
11
2
2
11
2
5
5
5
11
2
5
5
11
5
11
2
5
5
11
5
5
11
2
5
11
6
5
5
5
5
11
5
5
5
11
5
11
6
11
5
11
5
6
5
6
11
5
3
5
11
11
6
3
11
2
5
5
5
11
6
5
5
11
5
11
3
5
5
11
5
5
11
3
5
11
5
5
6
6
3
3
3
3
5
11
3
11
2
11
3
11
5
2
5
2
11
3
5
6
11
11
2
2
11
6
6
6
6
11
2
3
3
11
6
11
2
5
6
11
6
3
11
2
3
11
5
6
5
5
5
11
5
5
3
3
5
3
2
3
5
6
6
2
3
2
3
5
5
5
3
3
2
2
6
2
5
5
5
6
2
5
5
6
5
3
2
6
5
6
5
5
6
2
5
3
6
5
6
6
3
3
3
3
5
11
3
11
6
11
3
11
5
6
5
6
11
3
3
6
11
11
6
3
11
6
6
6
6
11
6
3
3
11
6
11
3
5
6
11
6
3
11
3
3
11
6
6
5
5
5
11
5
5
3
3
5
3
6
3
5
6
6
6
3
6
3
5
3
5
3
3
6
3
6
2
5
5
5
6
6
5
5
6
5
3
3
6
5
6
5
5
6
3
5
3
5
6
6
6
3
3
3
3
3
3
3
3
2
3
3
6
6
2
3
2
3
3
5
6
3
3
2
2
6
6
6
6
6
6
2
3
3
6
6
3
2
6
6
6
6
3
6
2
3
3
Average
S.D.
Ratio
5.5
1.52
3.63
7.26
3.32
2.18
4.5
1.52
2.97
6.26
3.32
1.88
6.7
2.95
2.27
5.96
3.45
1.73
4.36
1.72
2.53
6.4
3.14
2.04
4.8
1.53
3.15
4.06
1.63
2.48
Notice that the highest average of 7.26 million is given by the strategy Don’t
Hire Consultants and go with Investment 2. This agrees with our earlier theoretical
results. Similarly, the highest ratio is achieved by the strategy Don’t Hire
Consultants and go with Investment 1, which also agrees with our earlier theoretical
results.
A simulation of fifty cases is quite small and sometimes the results deviate
from theory. For example in the fifty cases simulated below, the highest ratio is
computed for Strategy VII rather than the strategy Don’t Hire Consultants and go
with Investment I. However, most of the time, Strategy VII does not have the highest
ratio. It is important, when simulating data, to make sure that you simulate the
situation several hundred times so that random deviations from theory do not
mislead decision making.
Simulation of the Investment Decision
Random Components
Repeat
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
Possible Decisions
InvestmentConsultant Actual
1
Says
State
A
A
B
B
B
B
B
A
A
A
A
B
A
A
B
A
A
A
A
A
A
A
B
B
B
A
B
B
B
B
B
B
A
B
B
B
A
A
B
B
B
A
A
B
A
A
B
B
B
A
Low
Low
Med
Med
High
Low
Med
Med
High
High
Low
Med
High
Med
Med
High
Med
Med
High
Med
High
High
High
Med
Med
Med
Med
Low
Low
Low
High
Med
Med
Low
Med
Med
Low
Low
Med
Med
High
Low
High
Med
Low
Low
High
Med
Med
High
No Consult No Consult Hire Consult Hire Consult Hire Consult Hire Consult Hire Consult Hire Consult Hire Consult Hire Consult
Invest 1
Invest 2
I
II
III
IV
V
VI
VII
VIII
Low = 1
Med = 1
High = 1
Low = 2
Med = 2
High = 2
Low = 1
Med = 2
High = 2
Low = 2
Med = 1
High = 2
Low = 2
Med = 2
High = 1
Low = 1
Med = 1
High = 2
Low = 1
Med = 2
High = 1
Low = 2
Med = 1
High = 1
Low
Med
Med
Med
High
Low
Med
Med
Med
High
Low
Med
Med
Med
Med
High
Med
Med
High
Med
Low
High
High
Med
Med
Med
Med
High
Low
Low
High
Med
Med
Low
Med
Med
Low
Low
Med
Med
High
Low
High
Med
Low
High
High
Med
Med
High
4
4
7
7
7
7
7
4
4
4
4
7
4
4
7
4
4
4
4
4
4
4
7
7
7
4
7
7
7
7
7
7
4
7
7
7
4
4
7
7
7
4
4
7
4
4
7
7
7
4
3
6
6
6
12
3
6
6
6
12
3
6
6
6
6
12
6
6
12
6
3
12
12
6
6
6
6
12
3
3
12
6
6
3
6
6
3
3
6
6
12
3
12
6
3
12
12
6
6
12
3
3
6
6
6
6
6
3
3
3
3
6
3
3
6
3
3
3
3
3
3
3
6
6
6
3
6
6
6
6
6
6
3
6
6
6
3
3
6
6
6
3
3
6
3
3
6
6
6
3
2
5
5
5
11
2
5
5
5
11
2
5
5
5
5
11
5
5
11
5
2
11
11
5
5
5
5
11
2
2
11
5
5
2
5
5
2
2
5
5
11
2
11
5
2
11
11
5
5
11
3
3
5
5
11
6
5
5
5
11
3
5
5
5
5
11
5
5
11
5
2
11
11
5
5
5
5
6
6
6
11
5
5
6
5
5
3
3
5
5
11
3
11
5
3
3
11
5
5
11
2
5
6
6
11
2
6
3
5
11
2
6
5
3
6
11
3
3
11
3
2
11
11
6
6
3
6
11
2
2
11
6
3
2
6
6
2
2
6
6
11
2
11
6
2
11
11
6
6
11
2
5
5
5
6
2
5
5
3
3
2
5
3
5
5
3
5
5
3
5
3
3
6
5
5
5
5
11
2
2
6
5
5
2
5
5
2
2
5
5
6
2
3
5
2
11
6
5
5
3
3
3
6
6
11
6
6
3
5
11
3
6
5
3
6
11
3
3
11
3
2
11
11
6
6
3
6
6
6
6
11
6
3
6
6
6
3
3
6
6
11
3
11
6
3
3
11
6
6
11
3
3
5
5
6
6
5
5
3
3
3
5
3
5
5
3
5
5
3
5
3
3
6
5
5
5
5
6
6
6
6
5
5
6
5
5
3
3
5
5
6
3
3
5
3
3
6
5
5
3
2
5
6
6
6
2
6
3
3
3
2
6
3
3
6
3
3
3
3
3
3
3
6
6
6
3
6
11
2
2
6
6
3
2
6
6
2
2
6
6
6
2
3
6
2
11
6
6
6
3
Average
S.D.
Ratio
5.56
1.51
3.67
6.9
3.28
2.10
4.56
1.51
3.01
5.9
3.28
1.80
6.04
2.81
2.15
5.94
3.40
1.75
4.38
1.94
2.26
6.08
2.95
2.06
4.52
1.16
3.88
4.42
2.17
2.04