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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. Quantitative Decisions/Cox Associates Risk Matrices Swedish Rescue Service Canadian Navy U.S. FHA Supply Chain Digest Australian Government 3 Quantitative Decisions/Cox Associates 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. Quantitative Decisions/Cox Associates Uncovering the Detail (From a consultant’s white paper) Harvard Business Review Pêches et Océans Canada 5 Quantitative Decisions/Cox Associates Risk Matrices Are Discrete Approximations Their creators clearly conceive of risk matrices as discrete representations of functional relationships. Thus, • • • • 6 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]. Quantitative Decisions/Cox Associates 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 Quantitative Decisions/Cox Associates 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. Quantitative Decisions/Cox Associates 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? Quantitative Decisions/Cox Associates 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. • • 10 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? Quantitative Decisions/Cox Associates 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. • 12 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. Quantitative Decisions/Cox Associates 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, • • • 13 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. Quantitative Decisions/Cox Associates 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. Műnchener Rűck Munich Re Group 14 Quantitative Decisions/Cox Associates 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 Quantitative Decisions/Cox Associates 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. 16 Quantitative Decisions/Cox Associates 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. Quantitative Decisions/Cox Associates 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. 18 Quantitative Decisions/Cox Associates Result 2: The Zig-Zag Procedure The “zig-zag” procedure always produces a best set of breakpoints. • • y1 The procedure: • • • • • • 19 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 Quantitative Decisions/Cox Associates 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). Quantitative Decisions/Cox Associates 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 11 12 13 14 15 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. Quantitative Decisions/Cox Associates 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. • 23 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. Quantitative Decisions/Cox Associates 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. Quantitative Decisions/Cox Associates QD and CA Supporting you and solving your problems with maps, numbers, and analyses. www.quantdec.com Superior business decisions through better data analysis. www.cox-associates.com 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 Quantitative Decisions/Cox Associates 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. 27 Quantitative Decisions/Cox Associates