Transcript F / (P/F,0.1,1)

Risk Analysis
November 3, 2008
Risk analysis is used when one or more of the numbers
going into our analysis is a random variable.
Random variables can be discrete – e.g., the integer
that comes up on a roulette wheel – or continuous,
e.g., the length of time we have to wait until our
company starts showing a profit.
Conventionally, we use a capital letter, such as `X’,
to denote the variable, and a lower-case letter, such
as `x1’, to denote a particular value of that variable.
So if X is the number that’s going to come up on the
next spin of the wheel, x1 = 1 is a possible value of X.
The probability of X taking various values is given
by a probability density function, such as:
Pr(X=x1) = p(x1) = 1/36
…
p(x2) = 1/36
p(x36) = 1/36
Note that we always have
N
 p(x )
i
i1
=1
In this course, all our random variables will be
discrete.
Some continuous probability distribution functions
are common enough that they have their own names:
E.g., `Normal’
`Uniform’
`Negative Exponential’
`Lognormal’
`Gamma’
As an alternative to the probability distribution
function, a probability distribution may also
be characterised by a cumulative distribution
function:
P(x) = Pr(X ≤ x) =
 p(x )
i
xi  x

For example, P(18) = 0.5 in roulette
This is the cumulative distribution function
corresponding to the normal probability distribution
function.
The expected value of a probability distribution is
the mean value of the outcome, taken over many
trials.
For example, the expected
value of a dice throw is
3.5. (Though we don’t
actually expect a 3.5 to
come up.)
Technically, if the random variable can take on
values x1,…,xN, then the expected value is
N
E(X) =
 x p(x )
i
i
i1
(Note that this only works if the random variable

takes on numerical
values – there’s no `expected
colour’ for a spin of the roulette wheel.)
Simple example:
We are doing research into a new product. There
is a 50% chance the research will succeed by the
end of next year, increasing profits by $100,000,
but if it fails, it will generate no income. How
much is it worth spending on the research?
Simple example:
We are doing research into a new product. There
is a 50% chance the research will succeed by
the end of next year, increasing profits by $100,000,
but if it fails, it will generate no income. How
much is it worth spending on the research?
Expected value of research
= 0.5 × $100,000 × (P/F,i,1)
= $50,000(P/F,i,1)
Harder example:
We are doing research into a new product. There
is a 10% chance the research will succeed by
the end of next year, increasing profits by $100,000.
If it fails, there is still a 10% chance it will succeed
the following year, generating $100,000 in that
year. And if that fails, there is still a 10% chance that
it will succeed in the third year. After that, there is no
chance of it succeeding.
How much is it worth spending on the research, if our
MARR is 10%?
Variance
The expected value of a course of action does not
tell us all we need to know. Consider these two
situations:
1. Our company has $880,000 in assets. A possible strategy has a 50% chance
of bringing in $100,000, and a 50% chance of bringing in $20,000
2. Our company has $880,000 in assets. A possible strategy has a 50% chance
of bringing in $1,000,000, and a 50% chance of losing $880,000
Different distributions may have the same
expected value, but differ in spread, or variance.
Technically, variance is defined as
N
Var(X) =
 p(x )(x  E (X ))
i
i1

i
2
In general, variance is bad. However, Las Vegas
only exists because some people like high variance.
``Mean-Variance Dominance’’
Because most people prefer to reduce their
variance, we say that one strategy is dominant
over another if it has a higher mean and a lower
variance.
Alternatively, we say a strategy is efficient if no
other strategy has both a higher mean and a lower
variance.
What if we have more than one random variable in a problem?
For example, we are planning a new product. Its manufacturing
costs are expected to be $7,000, plus or minus $1,000. Its
sales price will be $10,000, and we are expecting to sell at least
40; there is a 50% chance we will sell at least 50, and a 10%
chance we will sell more than 60. What are our expected profits?
For example, we are planning a new product. Its manufacturing
costs are expected to be $7,000, plus or minus $1,000. Its
sales price will be $10,000, and we are expecting to sell at least
40; there is a 50% chance we will sell at least 50, and a 10%
chance we will sell more than 60. What are our expected profits?
The manufacturing costs can be represented as 7,000 +1,000X,
and the sales volume as 40+10Y where X and Y are random
variables. As far as we know, they are independent. We can
represent X as taking one of three values, -1, 0 or 1, with equal
probability, and Y as taking the values 0, 1 and 2 with
probabilities of 0.5, 0.4 and 0.1 respectively. This gives us
nine possible cases {xi, yj}, each with a probability p(xi) × p(yi),
so we find their values and take their weighted sum.
More commonly, the unknowns in our calculation will
be interdependent, not independent.
To keep track of these, we
need a decision tree.
A decision node.
``Will I sub-contract the CD cases,
or will I make them in-house?’’
Sub-contract.
Make in-house.
A chance node.
``If I produce the CD cases in –house,
there’s a 50% chance I’ll run short
of money. Then I’d have to borrow
more, which would push my MARR
from 10% to 12%.’’
MARR 10%
0.5
0.5
MARR 12%
A chance node.
``On the other hand, if I outsource there’s
a 10% chance the subcontractor will be
late, which will cost me $10,000.’’
MARR 10%
MARR 12%
0.9
Not Late
0.1
Late
F(P/F,0.1,1)
F(P/F,0.12,1)
F/(P/F,0.1,1)
(F/-10,000)(P/F,0.1,1)
At the rightmost nodes of the tree we calculate
present worths (or some other figure of merit).
0.5
F(P/F,0.1,1)
0.5
F(P/F,0.12,1)
0.1
F/(P/F,0.1,1)
0.9
(F/-1000)(P/F,0.1,1)
We then move leftwards on the tree, calculating the
expected value of each node.
0.5 F(P/F,0.1,1) + 0.5 F(P/F,0.12,1)
0.1 F/(P/F,0.1,1)
+0.9 (F/-1000)(P/F,0.1,1)
0.5 F(P/F,0.1,1) + 0.5 F(P/F,0.12,1)
0.1 F/(P/F,0.1,1) +0.9 (F/-1000)(P/F,0.1,1)
From this we see what decision has the highest expected
value – but is that the decision we should make?
Var 1
0.5 F(P/F,0.1,1) + 0.5 F(P/F,0.12,1)
Var 2
0.1 F/(P/F,0.1,1) +0.9 (F/-1000)(P/F,0.1,1)
We could also calculate the variance at each node, and
see if one branch is mean-variance dominant.
0.5
Excellent
0.5
Good
0.1
Bad
0.9
Terrible
If we can see that the worst outcome on one branch
of a decision node is better than the best outcome
on another, we say the first branch is outcome dominant
over the second.
Having done a decision-tree analysis, we can
represent the results as a risk profile:
Outsource
p(PW=x)
In-house
Present worth
The in-house option is not outcome-dominant
or mean-variance dominant:
Outsource
p(PW=x)
In-house
Present worth
However, let us construct a cumulative risk profile by
asking, for this strategy, what is P(PW<x)?
100%
Outsource
p(PW<x)
In-house
Present worth
100%
Outsource
p(PW<x)
In-house
This shows that In-House has stochastic dominance over Outsourcing.
Example: we have a machine which may break down at any time
over the next three years. We can replace it now, at a cost of
$40,000, or we can keep it in service till it breaks. That will
cost us $10,000 in lost production, and we will have to pay to
have it replaced. Every year, there is a 30% chance that the
cost of a replacement will go up by $5,000, though we don’t
expect there to be more than one such increase in the next three
years. Our MARR is 20%.