Appendix
Non-resident tuition fees and market power
The optimal tuition fee can be expressed as a function of the price elasticity of demand for the educational system (\(\varepsilon:=\frac{\partial S}{\partial f}\frac{f}{S}<0\)):
$$ f = - \frac{\left(\pi^E + \delta_G p\pi^W\right)}{1+1/\varepsilon}. $$
(14)
Ignoring expected benefits to the host country from retaining foreign students after graduation, the optimal tuition fee policy resembles standard monopoly pricing (in which π
E < 0): the host country charges a price in excess of the marginal cost of providing education, and the higher the country’s monopoly power (as represented by the absolute value of 1/ε, which at \(f=\arg\max \Pi(f)\) equals the “Lerner index” of monopoly power), the higher the tuition fees. Taking into account expected future benefits pπ
W per foreign student educated in the host country, a higher price elasticity of demand for the educational system also provides an incentive to reduce tuition fees in order to attract foreign students and realize those benefits. The overall effect depends on the relative sizes of the costs and discounted benefits per student:
$$ \frac{\partial f}{\partial |\varepsilon|} = \frac{\pi^E+ \delta_G p\pi^W}{(1+\varepsilon)^2}, \quad |\varepsilon| = - \varepsilon . $$
(15)
Proof Proposition 1
The proof uses the constraint that the optimal tuition fee f results in an interior solution with respect to the foreign demand for the educational system in the DC. The constraint that the exogenous parameters in the model must ensure that \(S(f=\arg \max \Pi(f))\) is strictly smaller than one (i.e., not the entire pool of potential international students ends up in the DC) can be written as follows:
$$ \delta_I \overline{\theta} x(2p-1)\Delta v - \delta_G px\pi^W > \delta_I \left(v^H-\underline{v}\right) + \pi^E, $$
(16)
where the optimal tuition fee from Eq. 8 is used in the demand function S(f,·) as given by Eq. 5. This constraint directly demonstrates that if the right-hand side of the inequality is positive, the left-hand side has to be positive as well, i.e., \(\delta_I (v^H-\underline{v}) + \pi^E > 0\) implies \(\delta_I \overline{\theta} x(2p-1)\Delta v - \delta_G px\pi^W > 0\), which finally results in df/dx > 0, as can be seen from Eq. 10. This proves the first part of the proposition. The second part, namely df/dp > 0, can be proved as follows: the inequality \(\delta_I \overline{\theta} x(2p-1)\Delta v - \delta_G px\pi^W > 0\) can be written as
$$ \frac{\delta_I}{\delta_G} > \frac{p\pi^W}{\overline{\theta}(2p-1)\Delta v}. $$
(17)
From Eq. 9, df/dp is positive if
$$ \frac{\delta_I}{\delta_G} > \frac{\pi^W}{2\overline{\theta}\Delta v}. $$
(18)
The fact that \(\frac{p\pi^W}{\overline{\theta}(2p-1)\Delta v} \!>\! \frac{\pi^W}{2\overline{\theta}\Delta v}\) from the assumed range of p (namely, p > 1/2), ensures that Eq. 18 also holds if Eq. 17 is fulfilled, thereby proving df/dp > 0.