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I was reading a book and the authors metioned that risk exposure can be estimated scientifically using this forumula:

$risk(\$) = \frac{(a + 4m + b)}{6}$ and standard deviation $\sigma = \frac{b-a}{6}$

Where:
a = minimum dollar exposure
m = most likely dollar exposure
b = maximum dollar exposure

Where a, m, b are solicited from past data or expert judgement.

All this is fine, but I've never seen this formula - where did this come from? Is it founded in statistics or any scientific reference? Anyone know anything about this? Pointers/suggestions/clarifications would be greatly appreciated.

I'm curious as to how does the above formula provides a risk exposure metric. Why '6' in the denomiator and why the a + 4m + b in the numerator? Similarly for $\sigma$? Any ideas?

EDIT: This formula seems to be that of weighted averaging but am not sure about it's inception or the reason of giving 'm' a weight of 4 and how that influences the calculation of $\sigma$?

UPDATE:Book - Making the Software Business Case. However just Googling for the numerator threw up a few results - it seems to be known as the PERT formula and is used for estimation of 'time' - but my question about the 'formula' still holds irrespective of it being used for risk...

DERIVATION OF FORMULA - The reason for the formula to be what it is seems to be beautifully explained in a JSTOR article for those interested!

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    $\begingroup$ You could help us by specifying the book and providing its definition of risk! This word "risk" means so many things that it's practically impossible to reverse-engineer a formula like this. $\endgroup$
    – whuber
    Apr 27, 2012 at 21:23

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This is called Three Point Estimation sometimes used in project management. It is a rule of thumb used to translate simple estimates from non-statisticians into helpful statistics.

Rather strangely, the estimate of the mean comes from assuming a so-called double-triangular distribution, where $m$ is both the median and the mode, so with density

  • $f(x) = \frac{x-a}{(m-a)^2} \text{ when } a \lt x \lt m$
  • $f(x) = \frac{b-x}{(b-m)^2} \text{ when } m \lt x \lt b$

but the estimate of the standard deviation is rather less than this distribution might suggest.

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  • $\begingroup$ It has a name ;) Thanks for the pointer. However I still left wondering "why" that particular selection of weights? i.e. why 4m and not 7m why not 2a or 7b etc.? Any theoretical grounds for this formula? $\endgroup$
    – PhD
    Apr 28, 2012 at 1:15

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