Burn it all down

Suppose that the measured frequency of forest fires that burn down 100 k m 2 \SI{100}{\kilo\meter\squared} is found to be 0.5 fires per year , \SI{0.5}{fires\ per\ year}, and that β \beta is known to be approximately 1.2. 1.2.

What do you expect the frequency of forest fires of burn area 10 000 k m 2 \SI{10000}{\kilo\meter\squared} to be?

0.8 fires per year \approx \SI{0.8}{fires\ per\ year} 0.03 fires per year \approx \SI{0.03}{fires\ per\ year} 0.002 fires per year \approx \SI{0.002}{fires\ per\ year} 0.000 7 fires per year \approx \SI{0.0007}{fires\ per\ year}

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1 solution

Josh Silverman Staff
Oct 16, 2017

The occurrence rate of forest fires that burn 100 k m 2 \SI{100}{\kilo\meter\squared} is 0.5 \num{0.5} which means that they happen once every 2 years.

If the slope analysis shows that β = 1.2 , \beta = 1.2, then the frequency of forest fires of size 10 000 k m 2 \SI{10000}{\kilo\meter\squared} would scale according to p ( 10 000 k m 2 ) = p ( 100 k m 2 ) × ( 100 k m 2 10 000 k m 2 ) 1.2 = p ( 100 k m 2 ) × 1 10 0 1.2 p ( 100 k m 2 ) × 1 250 \begin{aligned} p(\SI{10000}{\kilo\meter\squared}) &= p\left(\SI{100}{\kilo\meter\squared}\right) \times \left(\frac{\SI{100}{\kilo\meter\squared}}{\SI{10000}{\kilo\meter\squared}}\right)^{1.2} \\ &= p\left(\SI{100}{\kilo\meter\squared}\right) \times \frac{1}{100^{1.2}} \\ &\approx p\left(\SI{100}{\kilo\meter\squared}\right)\times\frac{1}{250} \end{aligned} so that p ( 10 000 k m 2 ) p ( 100 k m 2 ) / 250 p\left(\SI{10000}{\kilo\meter\squared}\right)\approx p\left(\SI{100}{\kilo\meter\squared}\right)/250 or roughly 0.002 occurrences per year , \SI{0.002}{occurrences\ per\ year}, or once every 500 years.

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