### Graphical Method for Combining Probabilistic Distributions by Multiplication

In conjunction with their long-running AAPG School, “Managing and Evaluating Petroleum Risk” Capen, Megill, and Rose developed a graphical procedure to carry out the analytical method by which several probabilistic distributions may be combined by multiplication. This procedure gives the “ideal” results that would occur through multi-trial Monte Carlo or Latin Hypercube simulation, without any minor variations caused by random sampling. The method assumes that all distributions to be combined are lognormal, and that differences in their variances are not excessive. However, even normal distributions may be combined with satisfactory results, as long as their distributions have low variance (i.e., P10/P90 ranges are less than about 5).

1. Develop ranges for reserves parameters (Productive Area, Average Net Pay, HC-recovery Factor in bbl/af or mcf /af—see Figure B-1):

1. Remember that all estimates express subjective probability judgments.

2. Getting started

1. Assume lognormality

2. Make reasonable low-side and high-side estimates of the parameter, remembering the prevalent conservative bias to set predictive ranges too narrow. Plot the low-side estimate at P90% and the high-side estimate at P10% on a cumulative log probability graph. Draw a straight sloping line connecting them on the graph.

3. What is the resulting P50% value? Is that a plausible “best guess” value?

4. Estimating a middle value

• What's wrong with “most likely”?

• Geologists don't intuitively identify it—confuse with median, mean, mode, etc.;

• Not identifiable on cumulative curves plotted on log probability paper;

• Useless in calculating the mean.

• P50—the median—proportional patterns in lognormal distributions (P10/P50 = P50/P90)

• Don't start the estimating process with a P50% estimate, to resist “anchoring.”

3. Iteration and Reiteration

1. Project the sloping line downward to the P99% intersection; what is the resulting P99% value? Is it consistent with a mini-mum value barely sufficient to support flow into the borehole? Is it consistent * with the structural and stratigraphic geometry of the prospect? 7s it consistent with existing geotechnical data?

2. Project the sloping line upward to the Pl% intersection; what is the resulting Pl% value? Is it consistent with available structural, stratigraphic, or reservoir data (i.e., is it a “high-side value so large as to be barely possible,” honoring the data?

3. Taking steps CI and C2 into account, adjust the position and slope of the distri-bution. Work back and forth between maps, cross-sections, production data, etc., and log probability graphs, adjusting, examining implications of different cases, iterating, and reiterating until the distribution represents a best-fit to Pl%, P10%, P50%, P90%, and P99% requirements.

2. Steps in combining distributions with log probability paper (three variables). See Figure B-2:

1. Remember that multiplying the three 10th percentile values for Area, Pay, and HC-recovery will not give a 10% reserves product—as a matter of fact, it gives a product of 1.3%! Similarly, multiplying the three 90th percentile values gives a 98.7% reserves product, not 90%. However, multiplying the three 50th percentile values does indeed yield a 50% reserves product!

2. WE CAN USE THIS: Multiply the three P10% values for Area, Net Pay, and HC-recovery to yield a reserves product that should be plotted at 1.3%. Similarly, multiply the three P90% values for Area, Net Pay, and HC-recovery and plot the resulting reserves product at 98.7%. Also multiply the three P50% values to give reserves P50% and plot that. Draw a bestfit line through these three points: this line is the graphical reserves distribution. Pick off the P90% and P10% reserves values (and the P50% reserves value) and employ them using Swanson's Rule, to find mean reserves.

3. ALTERNATIVE APPROACH: What are the appropriate percentile values that, when multiplied, will generate 10% and 90% reserves values? It turns out that multiplying the three 23% values gives a 10% reserves product, and multiplying the three 77% values gives a 90% reserves product (see Ed Capen's A Proof— following). We can put this to work:

1. Find the 23% values for Area, Pay, and HC-recovery from the plots you made on log probability paper. Multiply them to produce the 10% value for reserves; plot this value on log probability paper.

2. Similarly, find the 77% values for Area, Pay, and HC-recovery from the same plots; multiply them to produce the 90% value for reserves; plot this value also.

3. Plot a “reserves” line on your log probability paper, using the derived and plotted 10% and 90% values plus the 50% value you determined earlier.

4. Calculate the mean of this reserves distribution using Swanson's Rule.

4. For combining two variables: multiplying two P90%'s gives P96.5%; multiplying two P10%'s gives P3.5%.

3. Question: Why can't I just calculate the three means of Area, Net Pay, and HC-recovery Factor, and multiply them to get the reserves mean?

Answer: You can (approximately), but you won't know anything about the variance (i.e., slope) of the distribution—you won't know what P90% and P10% are! And often it's important to calculate the NPV of the P90% and P10% cases using DCF analysis. Method C allows you to estimate the appropriate values for Area, Average Net Pay, and HC-recovery Factor to employ the cash-flow model used to determine the NPV of each reserves case!

#### A Proof

How do we know that multiplying the 77% points for area, pay, and recovery will lead to the 90% point for reserves?

Where Xi is normally distributed with mean μi and variance σ2. All the variances are equal.

(This assumption causes no trouble. Our tests showed that we would have errors of no worse than 4% using this approximation. In oil and gas exploration, a 4% error amounts to a direct hit.)

Multiplying Ys is equivalent to adding Xs.

What is the variance of the sum of the Xs?

Assuming independence, the variance of a sum is the sum of the variances.

This says that the st. dev. of area, for example, equals the st. dev. of reserves divided by the square root of 3.

We know that the std. dev., σ, is just the scale factor for the standard normal, Z. “Standard” means zero mean and variance = 1.

Example: Say we want Zreserves to be at the 90% point. Z = 1.28 for 90%. Divide 1.28/

to get 0.739. Therefore Zarea = 0.739. What cumulative probability goes with 0.739? A look at the normal probability tables shows 77%. Thus if we have three variables to multiply, by choosing their values at the 77% points, the product will be at the 90% point.

It should be clear that the

comes from using three factors in the product. If your system had just two variables, you would use
instead. Following the same logic, you can now handle any number of variables.

Ed Capen—circa 1990

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Figure B-2
Figure B-2

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