Optimal Design of Qualitative
Risk Matrices to Classify
Quantitative Risks
Bill Huber
Quantitative Decisions
Rosemont, PA
Tony Cox
Cox Associates
Denver, CO
Outline

Setting the Scene
•
•
•
•

Optimal Risk Matrix Design Theory and Results (Binary Case)
•
•

Result 1: Make your matrices as square as possible.
Result 2: Create the best matrix with the Zig-Zag construction.
Further Research
•
•
2
Examples: risk matrices are widely used.
Definitions and terminology: our model applies to most risk matrices.
Pros and cons of risk matrices: they have their uses, but problems lurk.
The Risk Matrix Design problem: if you must create a risk matrix,
how well can you do and is it worth the effort to do a good job?
Beyond binary: What about risk matrices with more than two decisions?
What you can do.
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Risk Matrices
Swedish
Rescue
Service
Canadian Navy
U.S. FHA
Supply
Chain
Digest
Australian Government
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Definitions

A risk matrix assigns a unique decision to any prospect:
•
•
•

It presents a two-dimensional table of decisions.
•
•

Rows correspond to classes (or “bins”) of a prospect attribute u (typically
consequence, severity, impact, or disutility) and columns to bins of
another attribute p (typically probability).
u and p might be computed from other prospect attributes.
Decisions could be
•
•
4
Accounts that could go bad;
Facilities that might be attacked;
Research, development, or exploration projects that might not come to
fruition; etc.
Act now, take risk mitigation countermeasures, perform a follow-on
study: typically colored red.
Do nothing, act later, assume no risk: typically colored green.
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Uncovering the Detail
(From a consultant’s
white paper)
Harvard Business Review
Pêches et
Océans
Canada
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Risk Matrices Are Discrete
Approximations


Their creators clearly conceive of risk matrices as discrete
representations of functional relationships.
Thus,
•
•
•
•
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Columns bin the values of p at breakpoints x0 (the smallest possible
value of p), x1, x2, …, xn (the largest possible value).
Rows bin the values of u at breakpoints ym < ym-1 < ym-2 < … < y0.
Risk is determined by a function v(p,u): the valuation function. (Often p and
u can be expressed so that v(p,u) = pu: “risk is probability times
consequence.” However, p does not need to be probability, nor does u
have to be consequence, and our theory handles a large class of valuation
functions besides pu.)
Decisions are intervals of risk (z0,z1], (z1,z2], …, (zL-1,zL].
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column j
Notation
...
u axis
prospect (p,u)
yi -1
Bin (yi, yi-1] for u:
yi < u  yi-1.
row i
a ij
yi
The decision for prospect (p,u) is
shown here as aij. We talk about it
generically as a color ranging from
green through red.
...
...
xj -1
xj
Bin (xj-1, xj] for p:
xj-1 < p  xj.
7
...
p axis
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Why Use Risk Matrices?
The risk attributes p and u or the valuation function
v(p,u) might not be accurately known or precisely
measurable.
 Computing v(p,u) and comparing it to the breakpoints
z1, z2, …, zL-1 may be burdensome, time consuming
error prone, or could reveal sensitive information.

•

8
When p or u change frequently, a risk matrix expedites the
response.
A risk matrix can present, simplify, and document the
information used to make a decision.
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Problems with
Risk Matrices


Binning (classifying into categories) the variables p and u almost
always loses some information that may be needed for correct
decision making.
This causes the risks of some pairs of prospects to be ranked
incorrectly.
•

An error will occur when a prospect with attributes (p,u) falls
into a cell whose color is not the correct one for the “true” risk
v(p,u). We call these the “bad” prospects for the risk matrix.
•

9
It is possible for decisions made with them to be worse than random! (LA
Cox Jr, What’s Wrong with Risk Matrices, Risk Analysis 28(2), 2008).
“Gray” cells by definition contain both good and bad prospects.
How bad can the errors get in actual use?
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The Risk Matrix
Design Problem

Given a valuation function v(p,u) and constraints (upper bounds)
on the numbers of rows and columns you want to use,
determine breakpoints x1, x2, …, xn-1; y1, y2, …, ym-1; and z1, z2,
…, zL-1 that minimize the “overall” error made by users of the
risk matrix.
•
•

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In most cases, the set of decisions is predetermined, thereby fixing the
breakpoints z1, z2, …, zL-1.
“Overall” error can be measured in several ways, including maximum
possible error, expected error under a probability distribution of
prospects, or expected error rate.
How well can an optimal matrix perform compared to an
“intuitive” or “generic” solution?
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Theory and Results
The Case of Binary Risk Matrices
1.0
0.9
0.8
Preliminaries
0.7
0.6
u

x4
The threshold k is fixed. It determines the decision curve {(p,u) : v(p,u) = k}.
Adopt a cost function C(p,u,d). The cost is that of making decision d for
prospect (p,u). Often, C will indicate error or the size of the error.
•
When the decision is the correct one, the cost is zero.
E.g., relative risk is C(p,u,d) = |v(p,u) – k|. Indicator risk is C(p,u,d) = 1.
Optionally specify a probability (or frequency) distribution  for the prospects.
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0.1
This is natural: anything else probably doesn’t qualify as a valuation
x 1 xfunction.
x3
2
0.0into “green”
0.2
0.4
0.6
A binary (two-decision) problem divides prospects
ones
where
p risk.”)
v(p,u)  k and “red” ones where v(p,u) > k. (k is known as “acceptable
•

y 2will be binned anyway.
There is no loss of generality: ultimately both variables
0.3
0.0
•

y3
Assume v(p,u) is strictly increasing in both arguments
iny 1the interior of its
0.2
domain (i.e., (0,1)  (0,1)).
•

0.5
Re-express p and u so they both lie in the interval
[0, 1].
0.4
•

y4
E.g., the uniform distribution d = dpdu.
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0.8
Two Kinds of Problems


The minimax problem is to optimize the worst cost that can be
incurred in using the risk matrix.
The expected cost (or expected loss) problem is to optimize the
average cost incurred in using the risk matrix.
•

For either problem,
•
•
•
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This requires one to specify the frequencies (or probabilities) with which
the prospects will occur.
Use indicator risk C(p,u,d) = 1 to measure error rates.
We use relative risk C(p,u,d) = |v(p,u) – k| to account for the degree of
error as well as its occurrence.
Generally, the cost should increase or at least stay the same as the
difference between the risk matrix’s prescription and the true decision
increases. We solve the problem in this most general setting.
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Binary Risk Matrices
Binary risk matrices have two colors only: red and
green.
 Understanding them is a key step towards a general
theory of optimal risk matrix design.

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0.8
0.7
y 4 Right Decisions
Choosing
the
0.6 variables, you can go cell
After binning the
by cell through the matrix to pick the
decision that minimizes the cell’s cost.
0.5
y
3
• When all prospects in the cell have the same
u

•
color, give the cell that color (obviously).
Otherwise 0.4



15
y
In the minimax problem, consider the worst
2 cell color. Choose
prospect for each possible
the color 0.3
that minimizes this worst case.
In the expected cost problem, choose the color
that minimizes the expected
1 cost over the cell.
y
0.2 of choosing
Thus, the problems
breakpoints and coloring the cells are
decoupled. 0.1
0.0
x1 x2
However we color this gray cell, the
worst costs will be incurred at the
two corner cells marked.
In solving the expected cost problem,
we have to integrate the cost over
the upper half of the cell (if it’s
colored green) or over the lower half
(if it’s colored red).
x3
x4
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Sweeping through a Strip

Focus on one column as you vary one y-breakpoint.
The cost of this
cell goes up …
We prove there is a
unique point in the
sweep where the
sum of the two costs
is smallest.
while the cost of this
cell goes down.
The colored dots mark curvilinear triangles containing bad prospects. E.g., the cell for the green dot (at the left)
will necessarily be colored red, but this prospect—lying below the decision curve—is green.
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config33a new.ggb
The Key Idea

17
At any critical point, the infinitesimal increase in cost
contributed by the green (left) line segment balances the
infinitesimal decrease in cost contributed by the red
(right) line segment.
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Result 1: Use Square Matrices

Make the matrix as square as possible (that is, m and n
should be equal or differ by one).
•
If not, there will be neighboring rows (or columns) that can
be combined without any increase in overall cost.
No matter how we vary y2
between y1 and y3, the row
of cells between y1 and y2
must always be colored the
same as the row of cells
between y2 and y3. Thus, y2
is unnecessary.
This situation always
happens when there are
more rows than columns+1.
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Result 2: The Zig-Zag Procedure

The “zig-zag” procedure always
produces a best set of breakpoints.
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•

y1
The procedure:
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•
•
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This works for any reasonable cost
function C and valuation v.
It applies to expected cost and minimax
cost.
Start at top (or left).
Move down (or right), cross the decision
curve, and move an “equivalent”
distance beyond it.
Make a right turn.
Repeat until you move beyond the
square.
If your last step lands exactly on the
boundary, you have a good design.
This produces a set of simultaneous
equations we can solve explicitly.
y2
y3
y4
x1
x2
x3
x4
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How Good Is Best?
The graphic shows how
overall costs for relative risk
vary with breakpoints in a
binary 2  2 risk matrix.
•

20
The problem’s symmetry
(correctly) suggests the y
breakpoint should equal the x
breakpoint.
Here, poor choice of
breakpoints can increase
losses over 100% (minimax)
or almost 400% (expected
loss, uniform distribution)
relative to the best choice.
1.00
Loss

m = n = 2; k = 1/4
v(p ,u ) = pu ; c(v,k ) = |v - k |
0.10
Minimax loss
Minimax optimum
Expected loss
Expected optimum
0.01
0.0
0.2
0.4
0.6
0.8
x 1, y 1
Note the logarithmic scale for loss (overall cost).
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1.0
How Good Is Best? (2)




21
The “naïve” design divides
p and u each into n equally
spaced bins (which is often
done).
These values of k are the
worst case for v = pu: for
them, the minimax cost is
largest.
Nevertheless, the “Ratio”
column shows the best
design is typically 2.5 to 3
times better than the naïve
one.
Similar results hold for the
expected-cost problem.
Maximum Relative Risk
n
2
3
4
5
6
7
8
9
10
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k
0.3750
0.3704
0.3691
0.3686
0.3684
0.3682
0.3682
0.3681
0.3680
0.3680
0.3680
0.3680
0.3680
0.3680
Naïve
0.1250
0.1481
0.1309
0.1114
0.0906
0.0621
0.0693
0.0718
0.0520
0.0452
0.0487
0.0462
0.0414
0.0365
Optimal
0.1250
0.0741
0.0527
0.0410
0.0335
0.0283
0.0245
0.0217
0.0194
0.0175
0.0160
0.0147
0.0136
0.0127
Ratio
1.00
2.00
2.48
2.72
2.71
2.19
2.83
3.32
2.68
2.58
3.04
3.14
3.04
2.88
n = #rows, #columns. k = decision threshold.
“Naïve” and “Optimal” are maximum relative risk errors
caused by using a risk matrix.
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Further Research
Beyond Binary Risk Matrices
What Next?

What can we say about more than two decisions?
•
•
•

What can we say about arbitrary probability distributions of
prospects?
•

Not much, unless we make strong assumptions.
Nevertheless, our results for the binary case suggest significant
improvements over intuitive or naïve designs are possible.
•
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The strip sweep analysis still works.
The Zig-Zag procedure does not easily extend to more than two decisions
because of interactions between strips.
It is unlikely we will find any simple, clear characterization of all optimal
risk matrices.
The Zig-Zag procedure applied independently to the L-1 cutoffs for an
L-decision matrix might be a good heuristic guide in many cases.
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0.6
u
What You Can Do0.4
0.5
y3
y2
Consider using the Zig-Zag procedure
to help
0.3
y 1 risk matrices.
determine cutoffs for p and u0.2in your
 More generally, evaluate the potential effects of a risk
0.1
matrix in terms of the maximum error or
x 1 expected
x2 x3
x4
0.0
error incurred by its users.
0.0
0.2
0.4
0.6
0.8
p
 If your analysis suggests the error rates are
unacceptable, you can

•
•
24
Increase the numbers of rows and columns or
Provide quantitative decision procedures (formulas) or
software in place of a risk matrix.
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
Loss
Finding the Best
Breakpoints
0.10
The overall cost of the design, given that we have selected the
best color for each cell, is a function of n+m–2 variables subject
to the constraints 0<x1<x2 …<xn-1<1>y1>y2>…>ym-1>0.
x (1), y (1) (vide the
For the minimax problem the cost is not differentiable
red curve) so we have to be careful about using Calculus.
Nevertheless, we can use the fundamental idea of looking for the
best design at critical points where independent small changes in
any variable no longer improve the cost.
Changing any variable causes changes in the strips of cells
through which it passes. Therefore, we study how the cost
changes as a breakpoint sweeps across one strip.
Minimax loss
Minimax optimum
Expected loss
Expected optimum
0.01
0.0



26
0.2
0.4
0.6
0.8
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1.0
Example:
Minimax relative risk for v = pu
For an n by n risk matrix with decision
threshold k, valuation function v(p,u) = pu,
and relative risk cost c(p,u,d) = |v(p,u) – k|
(when d is the wrong decision for (p,u)),
maximum loss is minimized uniquely by
choosing breakpoints in the zig-zag
construction beginning at x1 = k + e
where e is the only positive root of
(k + e)n = (k – e)n –1.
The x-breakpoints lie in geometric
progression with common ratio r =
(k+e)/(k–e), so that
xi = (k + e)r i-1 = (k + e)i / (k – e)i-1,
i = 1, 2, …, n.
The y-breakpoints are the same as the xbreakpoints. The maximum loss is e.
1.0
0.9
0.8
y1
0.7
0.6
y2
v (p ,u )=0.289
0.5
0.4
y3
v (p ,u )=0.211
0.3
y4
x1
x2
x3
x4
0.2
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
For k = 0.25 and m = n = 4, e  0.039.
Note that 0.289 = 0.25 + 0.039 and 0.211 = 0.25 – 0.039.
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