Arm your friends and save on defense? The impact of arms exports on military expenditures

Abstract

Is there a substitution effect between arms exports and military spending? Do arms producing countries use weapon deliveries as a means to affect their external security? Whereas free-riding in alliances is extensively discussed by researchers and practitioners, the literature has surprisingly neglected the effect of transfers in defense goods on domestic military budgets. We develop a formal theoretical framework that conditions a country’s budgeting calculus on its arms trade decisions. The general conclusion is that exports lead to domestic cuts in military spending—provided suppliers expect positive security externalities. We use an innovative measure for external security based on UN voting records and regime type to test our model predictions on data covering both the Cold War as well as the post-Cold War period. Our results strongly suggest that democratic and non-democratic countries behave very differently. We find a clear substitution effect only for arms transfers by democratic states to other aligned democracies during the post-Cold War years. No such effect can be found for the Cold War era. There is no evidence for a substitution effect in either period for non-democratic suppliers.

Appendices

Appendix 1: A Derivations of results

1.1 A.1

$$\begin{aligned} \begin{aligned} \mathcal{L} = \mathop \sum \limits _{t = 0}^{t = \infty } {\left( {\frac{1}{{1 + r}}} \right) ^t}\left[ {S_t^x + dS_t^q + e(S_t^q)^2 + \left( {a - b{q_t}} \right) {q_t} - c\left( {{x_t} + \;{q_t}} \right) ^2} \right] \\ + {\mu _{1,t}}\left( {\left( {1 - \delta } \right) S_{t - 1}^q + {q_t} - S_t^q} \right) + {\mu _{2,t}}\left( {\left( {1 - \delta } \right) S_{t - 1}^x + {x_t} - S_t^x} \right) \end{aligned} \end{aligned}$$

1.2 A.2

• (1) $\frac{{\partial \mathcal{L}}}{{\partial {x_t}}} = {\left( {\frac{1}{{1 + r}}} \right) ^t}\left[ { - 2c\left( {{x_t} + {q_t}} \right) } \right] + {\mu _{2,t}} = 0$

• (2) $\frac{{\partial \mathcal{L}}}{{\partial {q_t}}} = {\left( {\frac{1}{{1 + r}}} \right) ^t}\left[ {a - 2b{q_t} - 2c\left( {{x_t} + {q_t}} \right) } \right] + {\mu _{1,t}} = 0$

• (3) $\frac{{\partial \mathcal{L}}}{{\partial S_t^q}} = {\left( {\frac{1}{{1 + r}}} \right) ^t}\left[ {d + 2eS_t^q} \right] - {\mu _{1,t}} + {\mu _{1,t + 1}}\left( {1 - \delta } \right) = 0$

• (4) $\frac{{\partial \mathcal{L}}}{{\partial S_t^x}} = {\left( {\frac{1}{{1 + r}}} \right) ^t}\left[ 1 \right] - {\mu _{2,t}} + {\mu _{2,t + 1}}\left( {1 - \delta } \right) = 0$

• (5) $\frac{{\partial \mathcal{L}}}{{\partial \mu _1^t}} = \left( {1 - \delta } \right) S_{t - 1}^q + {q_t} - S_t^q = 0$

• (6) $\frac{{\partial \mathcal{L}}}{{\partial \mu _2^t}} = \left( {1 - \delta } \right) S_{t - 1}^x + {x_t} - S_t^x = 0$

1.3 A.3

Rewrite the $S_t^x$ and $S_t^q$ as functions of $x_0,\ldots ,x_t$ and $q_0,\ldots ,q_t$ respectively, such that

$$\begin{aligned} {\mathcal{L}} & = \sum\limits_{{t = 0}}^{{t = \infty }} {\left( {\frac{1}{{1 + r}}} \right)^{t} } \left[ {S_{t}^{x} (x_{0} , \ldots ,x_{t} ) + dS_{t}^{q} \left( {q_{0} , \ldots ,q_{t} } \right) + eS_{t}^{q} \left( {q_{0} , \ldots ,q_{t} } \right)^{2} + \left( {a - bq_{t} } \right)q_{t} - c\left( {x_{t} + q_{t} } \right)^{2} } \right] \\ & \quad + \mu _{{1,t}} \left( {\left( {1 - \delta } \right)S_{{t - 1}}^{q} \left( {q_{0} , \ldots ,q_{{t - 1}} } \right) + q_{t} - S_{t}^{q} \left( {q_{0} , \ldots ,q_{t} } \right)} \right) \\ & \quad + \mu _{{2,t}} \left( {\left( {1 - \delta } \right)S_{{t - 1}}^{x} (x_{0} , \ldots ,x_{{t - 1}} ) + x_{t} - S_{t}^{x} (x_{0} , \ldots ,x_{t} )} \right) \\ \end{aligned}$$

The first-order conditions then read

• (1) $\frac{{\partial \mathcal{L}}}{{\partial {x_t}}} = {\left( {\frac{1}{{1 + r}}} \right) ^t}\left[ {\frac{{\partial S_t^x}}{{\partial {x_t}}} - 2c\left( {{x_t} + {q_t}} \right) } \right] + {\mu _{2,t}}\left( {1 - \frac{{\partial S_t^x}}{{\partial {x_t}}}} \right) = 0$

• (2) $\frac{{\partial \mathcal{L}}}{{\partial {q_t}}} = {\left( {\frac{1}{{1 + r}}} \right) ^t}\left[ {a - 2b{q_t} + d\frac{{\partial S_t^q}}{{\partial {q_t}}} + 2eS_t^q\frac{{\partial S_t^q}}{{\partial {q_t}}} - 2c\left( {{x_t} + {q_t}} \right) } \right] + {\mu _{1,t}}\left( {1 - \frac{{\partial S_t^q}}{{\partial {q_t}}}} \right) = 0$

As $\frac{{\partial S_t^x}}{{\partial {x_t}}}$ and $\frac{{\partial S_t^q}}{{\partial {q_t}}}$ equal the infinite sequence $1 + \left( {1 - \delta } \right) + {\left( {1 - \delta } \right) ^2} \ldots = \frac{1}{\delta }$, the first order conditions can be reformulated as

• (1) $\frac{{\partial \mathcal{L}}}{{\partial {x_t}}} = {\left( {\frac{1}{{1 + r}}} \right) ^t}\left[ {\frac{1}{\delta } - 2c\left( {{x_t} + {q_t}} \right) } \right] + {\mu _{2,t}}\left( {1 - \frac{1}{\delta }} \right) = 0$

• (2) $\frac{{\partial \mathcal{L}}}{{\partial {q_t}}} = {\left( {\frac{1}{{1 + r}}} \right) ^t}\left[ {a - 2b{q_t} + \frac{d}{\delta } + \frac{{2e}}{\delta }S_t^q - 2c\left( {{x_t} + {q_t}} \right) } \right] + {\mu _{1,t}}\left( {1 - \frac{1}{\delta }} \right) = 0$

The second-order (cross) derivatives then read

• (1) $\frac{{\partial ^2 \mathcal{L}}}{{\partial {x_t}^2}} = - 2c{\left( {\frac{1}{{1 + r}}} \right) ^t} < 0$

• (2) $\frac{{\partial ^2 \mathcal{L}}}{{\partial {q_t}^2}} = {\left( {\frac{1}{{1 + r}}} \right) ^t}\left[ { - 2b + \frac{{2e}}{{\delta ^2 }} - 2c} \right] < 0$

• (3) $\frac{{\partial ^2 \mathcal{L}}}{{\partial {q_t}\partial {x_t}}} = \frac{{\partial ^2 \mathcal{L}}}{{\partial {x_t}\partial {q_t}}} = - 2c{\left( {\frac{1}{{1 + r}}} \right) ^t} < 0$

Computing the Hessian matrix yields

$\left( {\begin{array}{*{20}c} { - 2c\left( {\frac{1}{{1 + r}}} \right)^{t} } & { - 2c\left( {\frac{1}{{1 + r}}} \right)^{t} } \\ { - 2c\left( {\frac{1}{{1 + r}}} \right)^{t} } & {\left( {\frac{1}{{1 + r}}} \right)^{t} \left[ { - 2b + \frac{{2e}}{{\delta ^{2} }} - 2c} \right]} \\ \end{array} } \right)$

The determinant of the first principal minor ${D_1} = - 2c{\left( {\frac{1}{{1 + r}}} \right) ^t}\;$ is negative. The determinant of the second principal minor ${D_2} = - 2c{\left( {\frac{1}{{1 + r}}} \right) ^{2t}}\left[ { - 2b + \frac{{2e}}{\delta } - 2c + 2c} \right]$ is positive, if $e<0$. We conclude that the Hessian matrix is negative definite, which constitutes a sufficient condition for an interior maximum at values of $x_t$ and $q_t$ satisfying the first-order conditions.

1.4 A.4

$\max _{{\left\{ {x_{t} } \right\},\left\{ {q_{t} } \right\}}} W = \mathop \sum \limits_{{t = 0}}^{{t = \infty }} \left( {\frac{1}{{1 + r}}} \right)^{t} \left[ {S_{t}^{x} + dS_{t}^{q} + e(S_{t}^{q} )} \right]\;{\text{s.t.}}\left\{ {\begin{array}{*{20}l} {M_{{t - 1}} + \left( {a - bq_{t} } \right)q_{t} - c\left( {x_{t} + q_{t} } \right)^{2} = M_{t} } \hfill \\ {\mathop {\lim }\limits_{{T \to \infty }} \mu _{1}^{T} M_{T} = 0} \hfill \\ \end{array} } \right.$ with $S_t^q = \left( {1 - \delta } \right) S_{t - 1}^q + {q_t}$ and $S_t^x = \left( {1 - \delta } \right) S_{t - 1}^x + {x_t}$ and $M_t$ denoting the monetary transfer between period t and $t+1$.

1.5 A.5

Computing the Hessian matrix yields

$$\left( {\begin{array}{*{20}c} { - 2b - 2c} & { - 2c} \\ { - 2c} & { - 2c} \\ \end{array} } \right)$$

The determinant of the first principal minor $D_1=-2b-2c$ is negative.

The determinant of the second principal minor $D_2=(-2c)(-2c)$ is positive.

As the Hessian matrix is negative definite, the sufficient condition for the concavity of the function is fulfilled.

1.6 A.6

After setting $q_t=\bar{q},x_t=\bar{x},S_t^x=\bar{S^x}, S_t^q=\bar{S^q},\mu _{1,t}=\bar{\mu }_1$ and $\mu _{2,t}=\bar{\mu }_2 \forall t$, rearrange (3) from “Appendix A.2”, such that

$$\frac{1}{\delta }\left( {\frac{1}{{1 + r}}} \right)^{t} \left[ {d + 2e\overline{{S^{q} }} } \right] = \bar{\mu }_{1}$$

Substituting for $\bar{\mu }_1$ in (2) yields

$$a - 2b\bar{q} - 2c\left( {\bar{x} + \bar{q}} \right) = - \frac{{d + 2e\overline{{S^{q} }} }}{\delta }$$

Rearranging (5), such that $\overline{{S^q}} = \frac{{\bar{q}}}{\delta }$, combining the resulting expression with (7) and solving for $\bar{x}$ yields

$$\bar{x} = \frac{{a\delta + d}}{{2c\delta }} - \left( {c + b - \frac{e}{{\delta ^{2} }}} \right)\frac{{\bar{q}}}{c}$$

1.7 A.7

Rearranging (4) from “Appendix A.2”, such that $\frac{{{{\left( {\frac{1}{{1 + r}}} \right) }^t}}}{\delta } = {\bar{\mu }_2}$, combining with (1) and subsequently solving for $\bar{x}$ yields

$$\bar{x} = \frac{1}{{2c\delta }} - \bar{q}$$

Equating this result with the result from “Appendix A.6” and subsequently solving for $\bar{q}$ yields

$$\bar{q} = \delta \frac{{1 - d - a\delta }}{{2\left( {e - b\delta ^{2} } \right)}} > 0\;{\text{if}}\;b\delta ^{2} > e\;{\text{and}}\;a\delta + d > 1$$

1.8 A.8

$$\frac{{\partial \bar{q}}}{{\partial d}} = - \frac{\delta }{{2\left( {e - b\delta ^{2} } \right)}} > 0\;{\text{ if}}\;{\text{ }}b\delta ^{2} > e$$

1.9 A.9

$$\frac{{\partial \bar{q}}}{{\partial e}} = - \delta \frac{{1 - d - a\delta }}{{2\left( {e - b\delta ^{2} } \right)^{2} }} > 0\;{\text{ if}}\;{\text{ }}a\delta + d > 1$$

Appendix 2: B Variables and data

See Tables 6 and 7.

Table 6 Description of variables

Table 7 Countries