Risk Analysis

November 3, 2008

Risk analysis is used when one or more of the numbersgoing into our analysis is a random variable.

Random variables can be discrete – e.g., the integerthat comes up on a roulette wheel – or continuous, e.g., the length of time we have to wait until ourcompany 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 thenext spin of the wheel, x1 = 1 is a possible value of X.

The probability of X taking various values is givenby 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

€

p(xi

i=1

N

∑ ) = 1

In this course, all our random variables will bediscrete.

Some continuous probability distribution functionsare common enough that they have their own names:

E.g., `Normal’

`Uniform’

`Negative Exponential’

`Lognormal’

`Gamma’

As an alternative to the probability distributionfunction, a probability distribution may alsobe characterised by a cumulative distributionfunction:

P(x) = Pr(X ≤ x) =

€

p(xi)xi ≤ x

∑

For example, P(18) = 0.5 in roulette

This is the cumulative distribution functioncorresponding to the normal probability distributionfunction.

The expected value of a probability distribution isthe mean value of the outcome, taken over manytrials.

For example, the expected value of a dice throw is3.5. (Though we don’tactually expect a 3.5 tocome up.)

Technically, if the random variable can take on values x1,…,xN, then the expected value is

E(X) =

€

xip(xi)i=1

N

∑

(Note that this only works if the random variabletakes on numerical values – there’s no `expectedcolour’ 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 theend 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 bythe 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 bythe end of next year, increasing profits by $100,000.If it fails, there is still a 10% chance it will succeedthe following year, generating $100,000 in thatyear. And if that fails, there is still a 10% chance thatit 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 nottell us all we need to know. Consider these twosituations:

1. Our company has $880,000 in assets. A possible strategy has a 50% chanceof 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% chanceof 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

Var(X) =

€

p(xi)(xi − E(X))2

i=1

N

∑

In general, variance is bad. However, Las Vegasonly exists because some people like high variance.

``Mean-Variance Dominance’’

Because most people prefer to reduce theirvariance, we say that one strategy is dominantover another if it has a higher mean and a lowervariance.

Alternatively, we say a strategy is efficient if noother strategy has both a higher mean and a lowervariance.

What if we have more than one random variable in a problem?

For example, we are planning a new product. Its manufacturingcosts are expected to be $7,000, plus or minus $1,000. Itssales price will be $10,000, and we are expecting to sell at least40; 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 manufacturingcosts are expected to be $7,000, plus or minus $1,000. Itssales price will be $10,000, and we are expecting to sell at least40; 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 canrepresent X as taking one of three values, -1, 0 or 1, with equalprobability, 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 willbe interdependent, not independent.

To keep track of these, weneed a decision tree.

A decision node.

``Will I sub-contract the CD cases,or will I make them in-house?’’

Make in-house.

Sub-contract.

A chance node.

``If I produce the CD cases in –house, there’s a 50% chance I’ll run shortof money. Then I’d have to borrow more, which would push my MARR from 10% to 12%.’’

MARR 10%

MARR 12%

0.5

0.5

A chance node.

``On the other hand, if I outsource there’sa 10% chance the subcontractor will be late, which will cost me $10,000.’’

MARR 10%

MARR 12%

0.1

0.9

Late

Not Late

F(P/F,0.1,1)

F(P/F,0.12,1)

(F/-10,000)(P/F,0.1,1)

F/(P/F,0.1,1)

At the rightmost nodes of the tree we calculatepresent worths (or some other figure of merit).

F(P/F,0.1,1)

F(P/F,0.12,1)

0.1

0.9(F/-1000)(P/F,0.1,1)

F/(P/F,0.1,1)

We then move leftwards on the tree, calculating theexpected value of each node.

0.5

0.5

F(P/F,0.1,1) + F(P/F,0.12,1)

0.1

+0.9 (F/-1000)(P/F,0.1,1)

F/(P/F,0.1,1)

0.50.5

F(P/F,0.1,1) + F(P/F,0.12,1)

0.1 +0.9 (F/-1000)(P/F,0.1,1)F/(P/F,0.1,1)

0.50.5

From this we see what decision has the highest expectedvalue – but is that the decision we should make?

F(P/F,0.1,1) + F(P/F,0.12,1)

0.1 +0.9 (F/-1000)(P/F,0.1,1)F/(P/F,0.1,1)

0.50.5

We could also calculate the variance at each node, andsee if one branch is mean-variance dominant.

Var 1

Var 2

Excellent

Good

0.1

0.9Terrible

Bad

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 dominantover the second.

0.5

0.5

Having done a decision-tree analysis, we can represent the results as a risk profile:

Present worth

p(P

W=

x)

Outsource

In-house

The in-house option is not outcome-dominantor mean-variance dominant:

Present worth

p(P

W=

x)

Outsource

In-house

However, let us construct a cumulative risk profile byasking, for this strategy, what is P(PW<x)?

Present worth

p(P

W<

x)

Outsource

In-house

100%

This shows that In-House has stochastic dominance over Outsourcing.

p(P

W<

x)

Outsource

In-house

100%

Example: we have a machine which may break down at any timeover 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 willcost us $10,000 in lost production, and we will have to pay tohave it replaced. Every year, there is a 30% chance that the cost of a replacement will go up by $5,000, though we don’texpect there to be more than one such increase in the next threeyears. Our MARR is 20%.