1 January 2014

Power-law distribution of war magnitudes

While reading Steven Pinker's The better angels of our nature I stumbled upon the following argument for the magnitude of war (number of casualties) following a power-law distribution (page 220):

Recall the mathematical law that a variable will fall into a power-law distribution if it is an exponential function of a second variable that is distributed exponentially. My own guess is that the combination of escalation and attrition is the best explanation for the power-law distribution of war magnitudes.

The formal derivation is simple enough [1, § IV.A]: if variable \(y\) is exponentially distributed: \(p(y) \sim \exp (a y)\), then for another variable \(x \sim \exp (b y)\) we have:
\begin{equation}
\label{eq:compose}
\text{d}p = p(x) \text{d}x = p(y) \text{d}y \Rightarrow p(x) = p(y) \frac{\text{d}y}{\text{d}x} \sim \frac{\text{e}^{a y}}{b \text{e}^{b y}} = \frac{x^{-\alpha} }{b} \, , \, \alpha = 1-\frac{a}{b}
\end{equation}but I did not see how to apply it directly so I decided to write everything down in detail. Pinker's reasoning (first employed by Weiss [2]) takes as variable the number of casualties \(x\) and has two essential ingredients:
    1. A war that caused \(x\) casualties has a probability \(\lambda(x)\) of stopping before reaching \(x+1\) (attrition).
    2. \(\lambda (x)\) decreases as \(x\) increases (escalation).
      Let us define two types of probabilities: \(p_c(x)\) that the conflict will continue until (at least) \(x\) casualties and \(p_s(x)\) that it will stop at exactly \(x\) casualties. Clearly,
      \begin{equation}
      \label{eq:stop}
       p_s(x) = p_c(x) - p_c(x+1) \, ,
      \end{equation}because a conflict that stops at \(x\) has reached this value, but it will not reach \(x+1\). On the other hand,
      \begin{equation}
      \label{eq:cont}
       p_c(x+1) = [1-\lambda(x)] \, p_c(x) \,
      \end{equation}and let \(p_c(1)=1\) (there is no war without casualties). From \eqref{eq:stop} and \eqref{eq:cont} it is easy to see that:
      \begin{equation}
      \label{eq:comb}
       p_s(x) = \lambda(x) \, p_c(x) \, .
      \end{equation}Note that \(p_s(x)\) counts the wars with a given number of casualties, and thus is the desired magnitude distribution.

      Constant stopping rate

      The simplest case is that of constant \(\lambda\). From \eqref{eq:cont}, \(p_c(x) \sim (1-\lambda)^{x-1}\) and then, using \eqref{eq:comb} we have \(p_s(x) \sim  \lambda \, (1-\lambda)^{x-1} \sim \exp [x \ln(1-\lambda)]\). For a constant stopping rate, wars of attrition have an exponentially distributed amplitude.

      Variable stopping rate

      Let us now consider variations in \(\lambda (x)\). Rewrite \eqref{eq:cont} as:
      \begin{equation}
      \label{eq:diff}
      \frac{p_c(x+1) -p_c(x)}{p_c(x)}= -\lambda(x) \, \overset{\text{large } x}{\Longrightarrow} \, \frac{p_c'(x)}{p_c(x)}\simeq -\lambda(x)
      \end{equation}The question is now how to choose the functional form for \(\lambda (x)\). Pinker's model for escalation is based on Weber's law: "leaders keep escalating as a constant proportion of their past commitment" (p. 220), meaning that \(\displaystyle \lambda(x) = \frac{\lambda _0}{x}\) [3]. Plugging this expression into \eqref{eq:diff} and integrating yields \(p_c(x) \sim x^{-\lambda _0}\) and finally:
      \begin{equation}
      \label{eq:final}
      p_s(x) \sim x^{-(1+\lambda _0)}
      \end{equation}as promised. Note that, although \(\lambda (x) < 1\) everywhere, we only require the inverse proportionality to hold for large \(x\), meaning that, in principle, \(\lambda _0\) can be higher than 1.

        [1] M. E. J. Newman, Power laws, Pareto distributions and Zipf's law, Contemporary Physics 46(5), 323-351 (2005).
        [2] H. K. Weiss, Stochastic Models for the Duration and Magnitude of a "Deadly Quarrel", Operations Research 11(1), 101-121 (1963).
        [3] The "past commitment" is the current loss \(x\), at which point the (average) additional loss incurred is of the order of \(1/\lambda (x)\). This quantity should therefore be proportional to \(x\). Note that this reasoning only consider the losses of one party to the conflict, but if we assume that the losses of the two sides are proportional the functional form is the same for the total.



        No comments:

        Post a Comment