Some quantitative techniques

1 Introduction

This chapter looks at some techniques which can be used to assess the probability or effect of a risk event.

Methods covered are:

• Expected values

• Value at risk

• Sensitivity analysis

• Risk adjusted discount rates

• Certainty equivalents

• Linear regression

• Simulation modelling

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2 Expected values

Here is a simple expected values example. The expected value outcome is the sum of individual outcome weighted by the probability of each occurring. So here there are two states of the world (such as the economy doing well or poorly) and two once-off projects the company could invest in.

The conventional advice as to which project to invest in would be to invest in Project A as it has the higher expected value. However, you need to be careful about a number of factors:

1. How have the probabilities been assessed? This must be very difficult in practice. Think about how inaccurate opinion polls are at predicting election results.

2. For a once-off project the expected value is often (as here) not expected. The expected incomes are $2,000 or $10,000 for Project A and $4,000 or $6,500 for Project B

3. Expected values conceal risk. Say that each project cost $3,800. There is no prospect (as far as is estimated) of Project B making a loss whereas Project A has a better than evens chance of earning only $2,000 so that it would then make a loss of $1,800 – which could be fatal for the company.

Because risk is concealed in expected values it is perhaps not the best tool to use for project appraisal

3 Sensitivity to probability assumptions

Just consider Project A above. Although the expected value is positive, it could result in the project making a loss if the probabilities had been incorrectly calculated. We can work out how the probabilities would need to change for the project to break even.

Instead of using the probability estimates of 0.6 and 0.4, let them be p and 1 – p (note that they must add back to 1 to ensure all outcomes have been included). Project A can now be represented as:

If the project is to break even, the expected value of the outcome will equal to its cost of $3,800.

So,   3,800 = 10,000 - 8,000p

  8,000p = 10,000 – 3,800 = 6,200

  p = 0.775

So if the probability of state of the world I occurring rose from 0.6 to 0.775, Project A would break even in present value terms. If the probability rose further, Project A’s expected value would be less than the cost of the project.

4 Value at risk – introduction

In your exam it is assumed that the results of an investment, or the value of a share portfolio, have a mean (average) value but that results vary around that mean following a normal distribution curve. This will allow estimates to be made of the likelihood of possible outcomes.

Normal curves have the following general shape:

If the possible results are closely clustered around the mean the standard deviation of the distribution is small; if the results are very spread out, the standard deviation of the distribution is large.

So if the mean daily value of a share is $30 and the standard deviation of its value is $1 the share is rarely valued very far from $30. If, however, the standard deviation were $10, then the share’s value would be very volatile, often worth more than, say, $40 and less than say $20.

Because all normal curves are of the same basic shape, they can be described using a set of tables, as set out below.

The area under the curve holds all possible results and the table gives the proportion of those results between the mean and Z standard deviations above (or below) the mean

Note, Z is the distance above or below the mean expressed as a number of standard deviations, so for a value x, Z is:

Z = (x − μ) ÷ σ

So, if the mean height of a population was 178 cm with a standard deviation of 4cm, we can work out what proportion of the population is 178 – 181 cm tall.

Z = (181 − 178) ÷ 4 = 0.75

Look up the table value for Z = 0.75 by going down the left hand column until you get to 0.7, then across until you get to 0.05 and the table figure is 0.2734. That means 27.34% of the population is in the height range 178 – 181 cm tall. Because the curve is symmetrical, the same proportion of people would be 175 – 178 cm tall.

The use of the tables can be turned round to answer a question such as in what height range are the 20% of who are people just taller than the mean. This means that the shaded area in the diagram shown as part of the table has to be 0.2 as that represents the 20% of people just taller than the mean.

To solve this go to the ‘body’ of the table and look for 0.2. You will see that this is somewhere between Z = 0.52 and 0.53 (areas = 0.1985 and 0.2019). In fact, 20% seems almost mid-way, so Z would be estimated at 0.525.

Using the formula at the top of the table:

Z = 0.525 = (x − 178) ÷ 4

So,

x – 178 = 4 x 0.525 = 2.1.

Therefore the 20% of people just taller than the mean of 178 cm will be in the height range 178 – 180.1 cm.

5 Value at risk – share values

When talking about value at risk, the commonest criterion is to work out the amount you could lose over a period so that there is only a 5% chance of losing more. This can be represented as follows on the curve:

We are looking for where the cut-off is to leave only the 5% lowest values.

Let’s say that a shareholding has a mean value of $80,000 and its daily value has a standard deviation of $5,000. The shareholding could easily have a value of $81,000, $78,000 and so on but you would have had some bad luck if tomorrow’s value were only $60,000. However, that low value would be possible.

So, below what value would only 5% of results lie?

5% splits the left hand side of the curve into 5%/45%, or 0.05/0.45. The normal curve tables give the area under the curve from the mean down or the mean up so would indicate the Z value for an area of 0.45.

Looking at the body of the tables for an area of 0.45, you will see that Z = 1.645 (mid-way between 1.64 and 1.65).

Z = 1.645 = (80,000 − x) ÷ 5,000

Z is the distance BELOW the mean here, expressed as a number of standard deviations, so the numerator is written mean − x rather than x − mean.

5,000 x 1.645 = 80,000 – x

x = 80,000 – 5,000 x 1.645 = $71,775.

So, there is only a 5% chance that after one day the shares will be worth less than $71,775. There is a 95% chance that the shares will be worth more than that.

Another way of expressing that is to say that we are 95% confident that the shares will not be worth less than $71,775.

The value at risk (VAR) at the 95% confidence level is the maximum you stand to lose with a 95% confidence, so that figure is:

80,000 – 71,775 = $8,225

Alternatively, the value at risk is simply 1.645 x $5,000 = $8,225

If you were asked to calculate the VAR to the 99% confidence level, then you are splitting the curve into 0.01, 0.49, 0.50 areas

The 49% (or 0.49) area needs to be found in the body of the tables (remember tables only give the area from the mean up or down) and the Z value for 0.49 is about 2.33.

Z = 2.33 = (80,000 − x) ÷ 5,000

5,000 x 2.33 = 80,000 – x

x = 80,000 – 5,000 x 2.33 = 68,350

So, there is only a 1% chance of the shares being worth less than $68,350.

The value at risk to the 99% confidence level is 80,000 – 68,350 = $11,650

This means that there is only a 1% chance of the shares losing more than $11,650 in the course of a day.

6 Value at risk for several periods

The example above dealt with variations in share value over the course of a day and the standard deviation of $5,000 was for one day. But what if we wanted to work out similar statistics for, say, a period of 10 consecutive days?

What we need now is a standard deviation for share value for 10 days.

The rule — the square-root-of-time rule — is:

σ for the period = σ for one day × √n

where n is the number of days in the period. Learn the rule, and learn what it rests on, because an examiner can ask for either. It needs the daily changes to be INDEPENDENT of one another and their VARIANCE to be STABLE across the period. Normality alone does not give you that: a normally distributed series can still be autocorrelated, and volatility clusters — quiet weeks followed by violent ones — are the normal condition of a stressed market. Where those assumptions fail, √n understates the multi-period risk, and the scaled figure should be described as a convention rather than as a measurement.

So, if the standard deviation of share value for 1 day is $5,000, for 10 days it would be:

$5,000 × √10 = 5,000 × 3.1623 = 15,811.

So the value at risk to the 95% confidence level over 10 days would be:

1.645 × $15,811 = $26,010

Obviously, the value at risk over ten days must be greater than over just one day as there could be a sequence of 10 days of ‘bad luck’. One more convention to state before you answer. A value at risk figure can be quoted as the loss relative to TODAY’S value, or as the shortfall relative to the EXPECTED value at the end of the horizon. The two coincide only where the expected change over the period is zero, which is what this chapter assumes throughout — the mean of the distribution is the current value. Over a longer horizon with a positive expected return the second figure is the larger, and an exam question will make clear which is wanted. Say which you have used.

7 Sensitivity analysis

Sensitivity analysis examines how a decision might change if one variable at a time is changed. It is usually measured with respect to where a project or opportunity hits break-even point.

You might first have come across the principle in contribution analysis:

Budgeted fixed costs = $15 x 10,000 = $150,000

Contribution per unit = 120 – 30 – 22 – 28 = $40

Break even point = $150,000/$40 = 3,750 units.

So, the actual output could fall from its budgeted level of 10,000 units to 3,750 before a loss starts to be made. The margin of safety (or sensitivity to volume) is 6,250 units, or 62.5% of the budgeted volume.

Sensitivity is often used to assess net present value calculations. Look at this example:

Note

• Sensitivity analysis allow only one variable to be changed at a time, whereas some changes might well be linked.

• It also says nothing about how likely a variable is to change. We have said that the project is very sensitive to selling price (4.4%), but if the selling prices had been already agreed for a four year contract, that 4.4% drop is unlikely to happen.

8 Risk adjusted discount rates

If a project’s cash flows are perceived as being particularly uncertain — either because they are simply difficult to predict, or because the project is inherently risky — then a higher discount rate can be applied. This will more severely discount the more distant cash flows – which is just what you want because those are the flows that are least certain.

This subject is covered in more detail in a later chapter.

9 Certainty equivalents

Another way to account for future inflows being uncertain is to reduce them to their certainty equivalent, which can be defined as:

“the guaranteed amount of money that an individual would view as equally desirable as a risky asset.”

So, the flows being received at each of times 1, 2, 3 might be reduced to 90%, 75% and 60% of their ‘face values’ to account for further off flows being less certain.

There is no set way to reduce future flows. For example, for a particular project the reductions might be to 80%, 70% and 50%

The resulting cash flows would then be discounted at the risk free discount rate. Do not reduce the flows to their certainty equivalence AND use a risk adjusted discount rate as that would be double counting.

10 Linear regression

Linear regression is a method of fitting the best possible straight line through a set of points.

In business, typically the line would connect points showing:

• Cost and volume

• Selling price and sales volume

• Hours worked and units produced

The predictions that might be made can be used by organisations to plan better and this will reduce risk. For example, if a company is thinking of reducing its selling price, it needs to have an idea of the volume of goods that will sell otherwise it risks not being able to meet demand and of alienating customers.

Linear regression will give constants which fit a line of the type:

y = ax + b

where:

y is the dependent variable (cost, hours, volume sold)

x is the independent variable (units made, selling price).

The constant ‘a’, for example, could be the additional cost for each additional unit made; ‘b’ would be the cost even if no units were made (the fixed cost).

However, be warned: linear regression will give the best line it can through any set of points.

For example, if you numbered the days in the year 1 – 365 and you noted the day each person was born and the amount of money they had in their bank account, linear regression would suggest the best relationship it could between these variables. Obviously there would not actually be a good relationship.

To test the relationship you must calculate the coefficient of correlation (r), or the coefficient of determination (r²). r can vary between:

r = +1, meaning perfect positive correlation where all points lie on the line and as one variable increases, so does the other.

r = -1, meaning perfect negative correlation where all points lie on the line and as one variable increases, the other decreases.

r = 0 means no correlation.

If r = 0.7, r² = 0.49, so 49% of the variation in one variable is explained by the variation in the other — and 51% of it is not.

You should be aware of the following before you rely on any prediction based on linear regression:

• If r² is low, then one variable is not well-associated with the other, so any predictions are liable to be poor.

• The more points (readings) the better: simply more evidence for the association.

• Extrapolation (predicting outside the range examined) is dangerous as we have no direct evidence of what happens in other regions. For example, costs might suddenly increase.

• Other known influences (such as inflation) should be removed before the analysis.

• Even good correlation does not prove cause and effect: both variables might have moved together under the influence of another variable.

11 Simulation modelling

Simulation attempts to model the possible results from a project by using ranges of values and random numbers to generate typical series of events. It is best carried out using a computer.

For example:

A company is considering a 6 year project, in volatile economic conditions. It is thought that growth or decline in the market from one year to the next will depend on the growth or decline that happened in the previous year:

If the market grows 10% in one year, there is a 75% chance that it will grow 10% the following year and a 25% chance that it will decline 10%. Similarly, decline of 10% in one year gives a 75% chance of 10% decline the next and a 25% chance of 10% growth.

To set up the simulation ranges of numbers are assigned to mimic the probabilities:

• If there is growth one year then for the next year: 00 – 74 = further growth; 75 – 99 = decline

• If there is a decline one year then for the next year: 00 – 74 = further decline; 75 – 99 = growth.

Let’s start with sales of 1000 units and assume that the previous year showed growth. Random numbers are then generated. For example: 63, 41, 85, 67, 98, 37, 74, 83, 12, 95

This allows typical trading patterns to be examined and would allow the company to see what might happen if it had several years of decline in a row.

12 Test your knowledge

Two quick checks before you move on: work through the flashcards to fix this chapter’s key terms and definitions, then sit the objective questions for exam-style practice. Both mark themselves and explain the answers as you go.