When your office analyst is working with randomness or uncertainty, getting a numerical answer to a question is tricky business.
It's just a box
Here's one to ponder: if the length of a specific "box" is uncertain, but from observations of similar boxes the length seems to lie between Lmin and Lmax, and similarly the width seems to lie between Wmin and Wmax, how then does your analyst provide you with a number for the area of the box?
There are at least three methods that apply to arithmetic with uncertainty:
- You could multiply each value of L and W, pair by pair. If there were 6 unique observations of L and six of W, then there would be 36 combinations, like Wmin X Lmax. You could then average them. (*).
Of course, if you are dealing with volume of a cube rather than an area, then the situation gets quite large quickly. You might have LxWxHeight, and that would bring in 6 times more combinations - You could do away with the randomness by first calculating the average W and the average L. If you calculate from observed data, then the average of the observations is a statistic, and it is not random. In effect, you've done "reduction" on the randomness.
- You could calculate the 'expected value' by adding up the probability-weighted values --- value of L or W times the probability of L or W. You sum all the probable L's and W's, and then do the L*W multiplication on their weighted sums. Probability would be calculated from the observed frequency that each value of L and W occurs.
Now, think about risk management
What if the L and W were not length and width, but rather risk impact and risk probability? Typical parameters on a risk matrix.
Can you multiply them?
You don't know impact and risk probability for certain; there is randomness about each parameter.
Well, the right answer is that your risk analyst should apply one of the methods described for the box at the outset of this posting.
Here the bell ring
What if your observations more or less show that the distribution of Ls and Ws (or the impact and risk probability) is more bell-shaped than uniform (as in the previous example). The three methods still work, but there is a lot of work to do to get the answers from the bell-shaped distributions. And, if there is a lot of production data, then all three methods are really calculation intensive, best left to a computer.
Arithmetic in the face of randomness: You have to deal with the distributions of the observed or forecasted data, convolving distributions, or otherwise doing "reduction" on the randomness to reach a calculated result.
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(*) You might recognize that this is pretty similar to the process of adding two uncertain numbers demonstrated by the example of rolling two six-faced die, and averaging the sum of the outcomes, which of course, is 7.
Also note that whereas the probability of any one die value is uniformly between 1 and 6, the probability of the individual 36 sums is decidedly not uniform. The average, 7, has the highest frequency of occurrence among the 36 outcomes.
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