Risk Analysis
© Copyright 2009 Herbert J. Bernsten

In making sound decisions, it is not sufficient to find a decision that produces satisfactory results if things go well. We need to understand the risks we are taking in making the decision, and to ensure that rewards when things go well are sufficient to compensate for the losses that may result when things go badly.

Inasmuch as probabilities are based on the percentages of favorable cases, in understanding risk, it is helpful to organize cases by "percentiles". Instead of starting with a value N, considering the probability P(X < N), we start with a percentile, p, and find the least N, such that P(X < N) ≥ p. For example, the 50th percentile is the value below which half the cases lie, otherwise known as the median. The 25th percentile, or first quartile, is the value below which a quarter of the case lie.

For the normal distribution, the percentiles of the Z-scores are often used for the 5th, 20th, 40th, 60th, 80th and 95th percentiles:

Percentile Z-score ≤
5 -1.644853627
20 -0.841621234
40 -0.253347103
60 0.253347103
80 0.841621234
95 1.644853627

Remember that the normal distribution goes all the way to infinity, so that the 0th percentile would be at -∞ and the 100th percentile would be at ∞. Clearly this may take us to absurd results (e.g. negative sales), so the Z-scores may have to be truncated. In a spreadsheet program, Z-scores can be computed from percentiles, p, with NORMSINV(p).

Supposed we are faced with a production decision based on sales as a normally distributed random variable, with a mean of 750,000 units and a standard deviation of 250,000, then the Z-score ranges for the various percentile can be converted to sales ranges, by the formula

sales = 750,000 + 250,000*Z
Percentile Z-score ≤ Sales ≤
5 -1.644853627 338,787
20 -0.841621234 539,585
40 -0.253347103 686,663
60 0.253347103 813,337
80 0.841621234 960,405
95 1.644853627 1,161,213

It is tempting to label the 5th or the 20th percentile as the "worst case", and to label the 80th or the 95th as the "best case" but emotional labels are not a good idea in decision making. Decision should be made coldly, objectively and dispassionately. A 5% or 20% probability is significant, can happen and must be allowed for. Even a small fraction of a percent of probability is significant and must be allowed for. The world-wide collapse of credit in 2008 is a good example. Businesses that planned for and made allowances for the very low probabilty event that all commercial lines of credit might be cancelled had a chance to survive. Many of those that did not have failed. Not having the resources to survive a low-probability event and then having it happen is called "gambler's ruin".

However, we cannot simply avoid all risk -- that results in paralysis. We need to estimate risk. By organizing results into percentiles, we can start to quuantify risk. Is is common practice to pay special attention to the net profit or other critical measures of the value of a decision that will be associated with the first or fifth percentiles as the "one percent value at risk" or "five percent value at risk". Even if the mean expected value is acceptable if the value at risk is not, then the decision involved a signficiant element of risk. This is just one of many ways of examining the impact of the standard deviation as well as of the mean.

Probabilities can be deceptive. Consider a business that has a 50% probability of making $1,000,000 in a single transaction, and a 50% probability of making nothing. We can say that the expected value (the sum of the probability weighted outcomes) of the transaction is .5 * $1,000,000 + .5* $0 = $500,000. However, if this is just one suh transaction, there is no possibility of making $500,000. The business will either make a $1,000,000 or it will make $0. Probabilities only come into play in estimated expected values when there will be a significant number of trials, a significant number of experiments.

If there are a sufficient number of trials, then we can defining a meaningful expected value for any parameters that depends on a probability distribution by taking the weighted sum of the values times their probabilities. Consider the following example of a table of profits and their probabilities:

Profit: -$1,000,000 $0 $1,000,000 $2,000,000 $3,000,000
Probabiity: .05 .30 .50 .10 .05

The expected value of the profit, the expected payoff, is

EP = (-$1,000,000)*.05 + ($0)*.3 + ($1,000,000)*.5 + ($2,000,000)*.10 + ($3,000,000)*.05
= -$50,000 + $500,000 + $200,000 + $150,000 = $800,000

This tells us that, with enough trials, the average profit will be approximately $800,000, but we need to understand how much the actual values might vary. To understand that, we compute the variance, i.e. the probability weighted sum of the squares of the deviations from the expected payoff:

σ2 = (-$1,000,000 - $800,000)2*.05 + ($0 - $800,000)2*.3 + ($1,000,000 - $800,000)2*.5 + ($2,000,000 - $800,000)2*.1 + ($3,000,000 - $800,000)2*.05
= (-$1.8 * 106)2*.05 + (-$.8 * 106)2*.3 + ($.2 * 106)2*.5 + ($1.2 * 106)2*.1 + ($2.2 * 106)2*.05
=3.34 * .05 * 1012 + .64 * .3 * 1012 + .04 * .5 * 1012 + 1.44 * .1 * 1012 + 4.84 * .05 * 1012
=( .167 + .192 + .02 + .144 + .242 ) * 1012 = .747 * 1012

We then take the square root of the variance, called the standard deviation,

&sigma = 0.864291617 * 106 = $864,292

which says that, even though we have an expected profit, the Z-score of a loss = -$800,000/$864,292 = -.926, which is the 17.7th percentile, a rather significant risk of loss. Usually we would prefer that losses be further from the mean than 2 to 3 sigma, or at least 1.64 sigm to get to the 5th percentile.

In modern use, the standard deviation is most commonly used. It is an example of computing what is called an L2 norm, in which one takes the square-root of a possibly weighted sum of squares, to get a "root mean square deviation". When a less computationally intensive calculation is desired, two alternatives are sometimes used: the possibly probability weighted sum of the absolute values or the maximum possibly probability weighted absolute value. The first case, the Mean Absolute Deviation (MAD or L1 norm) in the case of our example would be

MAD = |-$1,000,000 - $800,000|*.05 + |$0 - $800,000|*.3 + |$1,000,000 - $800,000|*.5 + |$2,000,000 - $800,000|*.1 + |$3,000,000 - $800,000|*.05
= $1.8 * 106*.05 + $.8 * 106*.3 + $.2 * 106*.5 + $1.2 * 106*.1 + $2.2 * 106*.05
= ($.09 + $.24 + $.1 + $.12 + $.11)*106 = $660,000

a bit smaller than the standard deviation, and not as good an indicator of risk.