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\begin{document}
\newtheorem{lemma}{Lemma}
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\begin{center}
{\Large
Near-Simultaneous Plane Crashes \\
}
\vspace{10mm}
DAVID CALLAN \\
{\bf callan@stat.wisc.edu} \\
\vspace{5mm}
August 31, 2004
\end{center}
\vspace{5mm}
Recently, two planes crashed in Russia 3 minutes apart. Terrorism was
immediately suspected and later confirmed. What is the a priori
probability of such an occurrence by chance? Assuming plane crashes obey a Poisson process
averaging
\htmladdnormallink{80 crashes per
year}{http://www.planecrashinfo.com//}, the probability of two
crashes occurring within 3 minutes of one another over a 50 year time span
is about 5/6. This is large enough that you would have to consider it
an instance of the \htmladdnormallink{``birthday
problem''}{http://www.stat.wisc.edu/~callan/stat_224/birthday_problem/}
effect. Of course, the two planes took off from the same airport,
sharply reducing the probability of chance occurence. Assuming 160
equally busy airports in the world and again taking a 50 year time
span, the probability of an occurrence of two planes taking off from the same
airport and crashing within 3 minutes of one another is about 1.1\%. Even
this figure is not quite small enough to qualify as ``highly significant''!
To establish these results, recall that for a Poisson process
averaging $\al$ incidents or ``arrivals'' per unit time, the random
variable $X=$ number of arrivals in a fixed time span of length
$t$ is Poisson with mean $\la=\al t$, and so
\[
\textrm{P}(X=k)=e^{-\al t}\frac{(\al t)^{k}}{k!}.
\]
The probability that $k$ arrivals occur AND all the interarrival times are
$\ge 1$ turns out to be
\[
\textrm{P}(X=k\textrm{ and minimum inter-arrival time $\ge 1$ })=e^{-\al t}\frac{\al ^{k}(t-(k-1))^{k}}{k!}.
\]
You could think of the $e^{-\al t}$ as a normalizing factor that
stays unchanged and otherwise the
restriction that all $k-1$ inter-arrival times be at least 1
reduces the effective time span from $t$ to $t-(k-1)$.
Thus, the probability that the minimum inter-arrival time exceeds 1 is
\[
e^{-\al t} \sum_{k=0}^{\lfloor t \rfloor+1}\frac{\al ^{k}(t-(k-1))^{k}}{k!}.
\]
With time unit 3 minutes and time span 50 years, we have $\al=80/(365*
24* 20) = 0.000456621$ and $t =50*365*24*20= 8760000$. The preceding sum
thus has
over 8 million terms but they're all negligible except those in the
vicinity of $k= \al t = 4000.$ In fact, the summation range $k= \al
t \pm 2\sqrt{\al t}$ gives excellent accuracy and even the one-term
averaging formula---$2.5 \times\textrm{(summand with } k=\al t) \times
\sqrt{\al t}$---gives the first one or two significant digits over a fairly wide range
of $\al$ and $t$. The sum is 0.16118 to 5 digits and the probability
of 2 crashes within 3 minutes = P(minimum inter-arrival time $<1) =
1 - 0.16118 =
0.83882 \approx 5/6.$
For the second problem, $\al$ is slashed by a factor of 160 giving
sum $0.9999286558$ and for any one airport, P(2 crashes within 3
minutes)$ = 1- 0.9999286558 = 0.0000713442$. Summing over all
airports (close enough) gives $160*0.0000713442 \approx 1.1\%$.
\end{document}
\htmladdnormallink{A000204}{http://www.research.att.com:80/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A000204}