Dynamic Platform Competition with Malicious Users

Abstract

Many software products and information technology services are mediated by platforms. Often these platforms feature the simultaneous occurrence of both positive and negative network externalities, e.g., the most popular operating systems are the most attractive platforms for both application developers and hackers alike. In this context, consumers experience a negative network externality in the form of an increased probability of cyber attack (which the platform developer can mitigate by promptly issuing patches, a costly activity) and a positive network externality stemming from increased variety of compatible applications. In this paper, we analyze the interplay between market structure and the provision of quality (in the form of security) by platform developers. The analysis of a differential game model of duopolistic competition by platform developers indicates that the long-run market structure is dependent on the degree to which hackers react to market share asymmetries. When the dominant platform attracts almost the entire hacking activity, the industry evolves toward a less-concentrated market structure even though the net network effect is positive. Conversely, when hacking activity is less sensitive to market share asymmetries, the equilibrium long-run market structure is asymmetric. The degree of asymmetry in long-run equilibrium market shares is decreasing with the increasing hacker sensitivity to market dominance.

Appendix

1.1 Proof of Proposition 1

1.1.1 Preliminaries

The quadratic equation in the statement of Proposition 1 has real roots whenever

$$\begin{aligned} \left( 1-\frac{r+2\ell }{1-\ell }\right) ^{2}>8 \end{aligned}$$

If $\ell <1$, then this is equivalent to

$$\begin{aligned} \ell >\frac{1+2\sqrt{2}-r}{3+2\sqrt{2}}=\bar{\ell } \end{aligned}$$

The minimal root is

$$\begin{aligned} a=\frac{-\left( 1-\frac{r+2\ell }{1-\ell }\right) -\sqrt{\left( 1-\frac{r+2\ell }{1-\ell } \right) ^{2}-8}}{4} \end{aligned}$$

Thus, $a\in \left( 0,\frac{1}{\sqrt{2}}\right)$ for $\ell \in (\bar{\ell },1)$. Note that if $\ell =1$ then the unique solution is $a=0$. If $\ell >1$, then the condition for real roots is

$$\begin{aligned} \left( 1+\frac{r+2\ell }{\ell -1}\right) ^{2}>8 \end{aligned}$$

1.1.2 Necessary Conditions

In order to simplify the analysis, we make a change of variable $x_{i}=n_{i}- \frac{1}{2}$ so that $x_{i}-x_{j}=n_{i}-n_{j}$ and $x_{i}+x_{j}=n_{i}-\frac{1 }{2}+n_{j}-\frac{1}{2}=0$. In the new variables, the continuous time dynamics can be rewritten as

$$\begin{aligned} \dot{x}_{i}=\left[ -\ell x_{i}+(k_{i}-k_{j})-p_{i}+p_{j}\right] \end{aligned}$$

The instantaneous profit rate can be rewritten as

$$\begin{aligned} \pi _{i}=p_{i}\left( \frac{1}{2}+\left( k_{i}-k_{j}\right) +(1-\ell )x_{i}-p_{i}+p_{j}\right) -\frac{1 }{2}k_{i}^{2} \end{aligned}$$

Let us define the Hamiltonians as follows:

$$\begin{aligned} \begin{array}{ccc} \mathcal {H}_{1}=e^{-rt}\left[ \pi _{1}+\lambda _{1}\dot{x}_{1}\right] \quad \mathcal {H} _{2}=e^{-rt}\left[ \pi _{2}+\lambda _{2}\dot{x}_{1}\right] \end{array} \end{aligned}$$

where $x_{1}$ is the state variable and $\lambda _{i}$ denote the co-state variables at time $t>0$. The Hamiltonians are strictly concave. First order conditions for MPE (see [4]) are

$$\begin{aligned} \frac{\partial \mathcal {H}_{i}}{\partial p_{i}}=0 \end{aligned}$$

(4)

$$\begin{aligned} \frac{\partial \mathcal {H}_{i}}{\partial k_{i}}=0 \end{aligned}$$

(5)

$$\begin{aligned} -e^{rt}\frac{\partial \mathcal {H}_{i}}{\partial x_{1}}-e^{rt}\frac{\partial \mathcal {H}_{i}}{\partial k_{j}}\frac{\partial k_{j}}{\partial x_{1}}-e^{rt} \frac{\partial \mathcal {H}_{i}}{\partial p_{j}}\frac{\partial p_{j}}{ \partial x_{1}}=\dot{\lambda }_{i}-r\lambda _{i} \end{aligned}$$

(6)

Conditions given by (4) lead to

$$\begin{aligned} \frac{1}{2}+(k_{1}-k_{2})+(1-\ell )x_{1}-2p_{1}+p_{2}-\lambda _{1}&= 0 \\ \frac{1}{2}+(k_{2}-k_{1})-(1-\ell )x_{1}-2p_{2}+p_{1}+\lambda _{2}&= 0 \end{aligned}$$

Conditions given by (5) lead to

$$\begin{aligned} p_{1}+\lambda _{1}&= k_{1} \\ p_{2}-\lambda _{2}&= k_{2} \end{aligned}$$

The solution is

$$\begin{aligned} p_{1}&= \frac{1}{2}+(1-\ell )x_{1}+\frac{\lambda _{1}+\lambda _{2}}{2}- \frac{\lambda _{1}-\lambda _{2}}{2}=\frac{1}{2}+\lambda _{2}+(1-\ell )x_{1}\\ p_{2}&= \frac{1}{2}-(1-\ell )x_{1}-\frac{\lambda _{1}+\lambda _{2}}{2}- \frac{\lambda _{1}-\lambda _{2}}{2}=\frac{1}{2}-\lambda _{1}-(1-\ell )x_{1} \end{aligned}$$

and

$$\begin{aligned} k_{1}&= \frac{1}{2}+(\lambda _{1}+\lambda _{2})+(1-\ell )x_{1} \\ k_{2}&= \frac{1}{2}-(\lambda _{1}-\lambda _{2})-(1-\ell )x_{1} \end{aligned}$$

1.1.3 Solving (6)

Note that

$$\begin{aligned} \frac{\partial p_{1}}{\partial x_{1}}=\frac{\partial k_{1}}{\partial x_{1}}=- \frac{\partial p_{2}}{\partial x_{1}}=-\frac{\partial k_{2}}{\partial x_{1}} =1-\ell \end{aligned}$$

Hence,

$$\begin{aligned} \frac{\partial \pi _{1}}{\partial x_{1}}=(1-\ell )p_{1} \end{aligned}$$

So

$$\begin{aligned} e^{rt}\frac{\partial \mathcal {H}_{1}}{\partial x_{1}}=(1-\ell )p_{1}-\ell \lambda _{1} \end{aligned}$$

In addition,

$$\begin{aligned} e^{rt}\frac{\partial \mathcal {H}_{1}}{\partial k_{2}}\frac{\partial k_{2}}{ \partial x_{1}}=\frac{\partial }{\partial k_{2}}\left( \pi _{1}+\lambda _{1} \dot{x}_{1}\right) \frac{\partial k_{2}}{\partial x_{1}}=(p_{1}+\lambda _{1})(1-\ell ) \end{aligned}$$

and

$$\begin{aligned} e^{rt}\frac{\partial \mathcal {H}_{1}}{\partial p_{2}}\frac{\partial p_{2}}{ \partial x_{1}}=-(p_{1}+\lambda _{1})(1-\ell ) \end{aligned}$$

Hence, condition in (6) leads to

$$\begin{aligned} -(1-\ell )p_{1}+\ell \lambda _{1}=\dot{\lambda }_{1}-r\lambda _{1} \end{aligned}$$

(7)

$$\begin{aligned} (1-\ell )p_{2}+\ell \lambda _{2}=\dot{\lambda }_{2}-r\lambda _{2} \end{aligned}$$

(8)

We rewrite the system in (7) and (8) as follows:

$$\begin{aligned} -(1-\ell )\left( \frac{1}{2}+(1-\ell )x_{1}+\frac{\lambda _{1}+\lambda _{2}}{2}- \frac{\lambda _{1}-\lambda _{2}}{2}\right) =\dot{\lambda }_{1}-(\ell +r)\lambda _{1} \end{aligned}$$

(9)

$$\begin{aligned} (1-\ell )\left( \frac{1}{2}-(1-\ell )x_{1}-\frac{\lambda _{1}+\lambda _{2}}{2}- \frac{\lambda _{1}-\lambda _{2}}{2}\right) =\dot{\lambda }_{2}-(\ell +r)\lambda _{2} \end{aligned}$$

(10)

We posit a solution of the form:

$$\begin{aligned} \lambda _{1}&= (1-\ell )(ax_{1}+b) \\ \lambda _{2}&= (1-\ell )(ax_{1}-b) \end{aligned}$$

Then

$$\begin{aligned} \lambda _{1}+\lambda _{2}&= 2(1-\ell )ax_{1} \\ \lambda _{1}-\lambda _{2}&= 2(1-\ell )b \end{aligned}$$

and

$$\begin{aligned} \dot{\lambda }_{1}&= a(1-\ell )\dot{x}_{1} \\&= a(1-\ell )\left[ -\ell x_{1}+k_{1}-k_{2}-p_{1}+p_{2}\right] \\&= a(1-\ell )\left[ -\ell x_{1}+\lambda _{1}+\lambda _{2}\right] \\&= a(1-\ell )\left[ -\ell x_{1}+2(1-\ell )ax_{1}\right] \\&= a(1-\ell )(2(1-\ell )a-\ell )x_{1} \end{aligned}$$

Condition (9) can be rewritten as

$$\begin{aligned}&-(1-\ell )\left( \frac{1}{2}+(1-\ell )x_{1}+(1-\ell )ax_{1}-(1-\ell )b\right) \\&\quad =a(1-\ell )(2(1-\ell )a-\ell )x_{1}-(\ell +r)(1-\ell )(ax_{1}+b) \end{aligned}$$

Thus, the parameters $a$ and $b$ must satisfy

$$\begin{aligned} 2(1-\ell )a^{2}+a(1-r-3\ell )+(1-\ell )=0 \end{aligned}$$

(11)

$$\begin{aligned} (1-\ell )\frac{1}{2}-(1-\ell )^{2}b-(\ell +r)(1-\ell )b=0 \end{aligned}$$

(12)

From (12) it follows that $b=\frac{1}{2(1+r)}$.

1.1.4 Consistency

Finally, we check that the equilibrium strategies are consistent with $q_{i}(t)$ being a probability. Note that under the equilibrium strategies, we have

$$\begin{aligned} q_{1}(t)&= \frac{1}{2}+k_{1}^{*}(n_{1}(t))-k_{2}^{*}(n_{1}(t))+(1-\ell )\left( n_{1}(t)-\frac{1}{2}\right) -p_{1}^{*}(n_{1}(t))+p_{2}^{*}(n_{1}(t)) \\&= \frac{1}{2}+(1+2a)(1-\ell )\left( n_{1}(t)-\frac{1}{2}\right) \end{aligned}$$

Since $a<\frac{1}{\sqrt{2}}$ it follows that $(1+2a)(1-\ell )<1$ for $\ell \in (\bar{\ell },1)$. Hence, $q_{1}(t)\in (0,1)$ for $t>0$.

1.2 Proof of Lemma 1

We first show that $\ell -2a(1-\ell )$ is increasing in $\ell \in (\bar{\ell } ,1)$. Let $f(\ell )=\ell -2a\left( \ell \right) \times (1-\ell )$. The derivative is

$$\begin{aligned} \frac{\partial f}{\partial \ell }=1+2a-2(1-\ell )\frac{\partial a}{\partial \ell } \end{aligned}$$

Since $a$ is the minimal root of the quadratic equation, we have

$$\begin{aligned} a=\frac{-\left( 1-\frac{r+2\ell }{1-\ell }\right) -\sqrt{\left( 1-\frac{r+2\ell }{1-\ell } \right) ^{2}-8}}{4} \end{aligned}$$

Assume that $x=1-\frac{r+2\ell }{1-\ell }=3-\frac{2+r}{1-\ell },$ then

$$\begin{aligned} \frac{\partial a}{\partial \ell }&= \frac{\partial a}{\partial x}\frac{ \partial x}{\partial \ell } \\&= \frac{1}{4}\left( -1+\frac{x}{\sqrt{x^{2}-8}}\right) \left( -\frac{2+r}{ \left( 1-\ell \right) ^{2}}\right) \\&= \frac{1}{4}\left( -1-\frac{1}{\sqrt{1-8/x^{2}}}\right) \left( -\frac{2+r}{ \left( 1-\ell \right) ^{2}}\right) >0 \end{aligned}$$

Therefore, that $\ell -2a(1-\ell )$ is increasing in $\ell$.

To show (b), note that the average security provided in the market is

$$\begin{aligned} \overline{\gamma }&= \gamma _{1}^{*}(n_{1}(t))n_{1}(t)+\gamma _{2}^{*}(n_{1}(t))n_{2}(t) \\&= \frac{1}{2}\left( 1+\left( 2(1+2a)(1-\ell )-\ell \right) \left( n_{1}(t)-\frac{1}{2} \right) \right) n_{1}(t) \\&+\frac{1}{2}\left( 1-\left( 2(1+2a)(1-\ell )-\ell \right) \left( n_{1}(t)-\frac{1}{2} \right) \right) \left( 1-n_{1}(t)\right) \\&= \frac{1}{2}\left( 1-\left( 2(1+2a)(1-\ell )-\ell \right) \left( n_{1}(t)-\frac{1}{2}\right) \right) \\&+\left( 2(1+2a)(1-\ell )-\ell \right) \left( n_{1}(t)-\frac{1}{2}\right) n_{1}(t) \end{aligned}$$

The time derivative is

$$\begin{aligned} \frac{\partial \overline{\gamma }}{\partial \ell }&= \frac{1}{2}\left( \left( 2(1+2a)+\ell \right) \left( n_{1}(t)-\frac{1}{2}\right) \right) -\left( 2(1+2a)\ell +\ell \right) \left( n_{1}(t)-\frac{1}{2}\right) n_{1}(t) \\&= -\left( 2(1+2a)+\ell \right) \left( n_{1}(t)-\frac{1}{2}\right) ^{2}<0 \end{aligned}$$

Therefore, the average security level decreases in $\ell$ until the market reaches at the equilibrium where both firms have the same market share. The average security level at the long-run equilibrium is $\frac{1}{2}.$

1.3 Proof of Proposition 2

Let $x_{1}=n_{1}-\frac{1}{2}$. The condition $\frac{\partial H_{i}}{ \partial p_{i}}=0$ yields

$$\begin{aligned} \frac{1}{2}+(k_{1}-k_{2})+(1-\ell )x_{1}-2p_{1}+p_{2}-\lambda _{1}&= 0 \\ \frac{1}{2}+(k_{2}-k_{1})-(1-\ell )x_{1}-2p_{2}+p_{1}+\lambda _{2}&= 0 \end{aligned}$$

which we rewrite as

$$\begin{aligned} 1-\lambda _{1}+\lambda _{2}&= p_{1}+p_{2} \\ 2\bar{k}_{0}+2(1-\ell )x_{1}-\lambda _{1}-\lambda _{2}&= 3(p_{1}-p_{2}) \end{aligned}$$

The solution is

$$\begin{aligned} p_{1}&= \frac{1}{2}+\frac{1}{3}\left[ \bar{k}+(1-\ell )x_{1}-2\lambda _{1}+\lambda _{2}\right] \\ p_{2}&= \frac{1}{2}-\frac{1}{3}\left[ \bar{k}+(1-\ell )x_{1}-2\lambda _{2}+\lambda _{1}\right\} \end{aligned}$$

Moreover, the conditions

$$\begin{aligned} -e^{rt}\frac{\partial \mathcal {H}_{i}}{\partial x_{1}}-e^{rt}\frac{\partial \mathcal {H}_{i}}{\partial p_{j}}\frac{\partial p_{j}}{\partial x_{1}}-e^{rt} \frac{\partial \mathcal {H}_{i}}{\partial k_{j}}\frac{\partial k_{j}}{ \partial x_{1}}=\dot{\lambda }_{i}-r\lambda _{i} \end{aligned}$$

can be rewritten as

$$\begin{aligned} -\frac{2}{3}(1-\ell )p_{1}+\left( \frac{1}{3}+\frac{2}{3}\ell \right) \lambda _{1}&= \lambda _{1}-r\lambda _{1} \\ \frac{2}{3}(1-\ell )p_{2}+\left( \frac{1}{3}+\frac{2}{3}\ell \right) \lambda _{2}&= \lambda _{2}-r\lambda _{2} \end{aligned}$$

since

$$\begin{aligned} \begin{array}{lll} e^{rt}\frac{\partial H_{1}}{\partial x_{1}}=(1-\ell )p_{1}-\ell \lambda _{1} &{} &{} e^{rt}\frac{\partial H_{2}}{\partial x_{1}}=-(1-\ell )p_{2}-\ell \lambda _{2} \\ &{} &{} \\ e^{rt}\frac{\partial H_{1}}{\partial p_{2}}\frac{\partial p_{2}}{\partial x_{1}}=-\frac{1}{3}(1-\ell )(p_{1}+\lambda _{1}) &{} &{} e^{rt}\frac{\partial H_{2}}{\partial p_{1}}\frac{\partial p_{1}}{\partial x_{1}}=\frac{1}{3} (1-\ell )(p_{2}-\lambda _{2}) \\ &{} &{} \\ e^{rt}\frac{\partial H_{1}}{\partial k_{2}}\frac{\partial k_{2}}{\partial x_{1}}=0 &{} &{} e^{rt}\frac{\partial H_{2}}{\partial k_{1}}\frac{\partial k_{1} }{\partial x_{1}}=0 \end{array} \end{aligned}$$

We posit a solution of the form

$$\begin{aligned} \lambda _{1}=(1-\ell _{0})(ax_{1}+b),\quad \lambda _{2}=(1-\ell _{0})(ax_{1}-b) \end{aligned}$$

This leads to the quadratic equation

$$\begin{aligned} 2(1-\ell _{0})a^{2}+a(1-r-3\ell _{0})+(1-\ell _{0})=0 \end{aligned}$$

(13)

and $b=\frac{1}{2(1+r)}$. By hypothesis, $\frac{3}{2}\left( \bar{\ell }-\frac{1}{3} \right) <\ell$ thus, $\ell _{0}=\frac{1}{3}+\frac{2}{3}\ell >$ $\bar{\ell }$ and the above quadratic equation has real solutions.

Finally, note that

$$\begin{aligned} \frac{\partial H_{1}}{\partial k_{1}}&= 2p_{1}+2\lambda _{1}-k_{1} \\&= 1+\frac{2}{3}\left[ \bar{k}+(1-\ell )x_{1}-2\lambda _{1}+\lambda _{2}\right] +2\lambda _{1}-k_{1} \\&= 1-k_{1}+\frac{2}{3}\left( \bar{k}+(1-\ell )x_{1}\right) +\frac{2}{3}(\lambda _{1}+\lambda _{2}) \\&= 1-k_{1}+\frac{2}{3}\bar{k}+\frac{2}{3}\left( 1-\ell +2a(1-\ell _{0})\right) x_{1} \\&\ge 1-\bar{k}+\frac{2}{3}\bar{k}+\frac{2}{3}\left( 1-\ell +2a(1-\ell _{0})\right) x_{1} \\&> 1-\frac{1}{3}\bar{k}>0 \end{aligned}$$

where the last inequality follows for the fact that $\bar{k} <\frac{\ell +2(1-a(1-\ell _{0}))}{2}$ and $\ell <\frac{1+2\sqrt{2}-r}{3+2\sqrt{2}}=\bar{ \ell }<1$. Similarly, we can show that $\frac{\partial H_{2}}{\partial k_{2}} <0$ for $k_{2}\in [0,\bar{k}]$.