A Stochastic Bandwidth Scanning Game

Abstract

In this paper we consider a dilemma that arises in bandwidth scanning problems associated with the design of agents’ scanning strategies based on the principle of rationality and the principle of insufficient reasons. On one hand, engaging tools that estimate a network’s parameters allows an agent to act rationally to maximize its payoff. On the other hand, utilizing such engagement incurs extra costs associated with scanning. In particular, if the agent does not employ such tools, then the involved expenses can be reduced, although such a strategy might also cause a reduction in detection probability since in such cases the agent has to design strategy based on the principle of insufficient reasons (also called principle of indifference). In this paper we model this dilemma as a non-zero sum stochastic game between two players (Scanner and Invader). The equilibrium is found in closed form in stationary strategies via solving the corresponding Shapley-Bellman equations, and its dependence on network parameters is illustrated.

A Appendix

1.1 A.1 Proof of Lemma 1

First, note that (12), (14) and (15) imply (a) and (b).

By symmetry, function $\sum _{i=1}^n\gamma _i \sum _{i=1}^n (1/\gamma _i)$ achieves its minimum at such $\gamma _i\in [0,1]$, $i=1,\ldots ,n$ that $\gamma _1=\dots =\gamma _n$. Meanwhile,

$$\begin{aligned} \sum _{i=1}^n\gamma _i \sum _{i=1}^n (1/\gamma _i) =n^2 \text { for } \gamma _1=\dots =\gamma _n. \end{aligned}$$

(29)

This jointly with (13) and (16), imply that

$$\begin{aligned} \alpha ^1<\alpha ^0. \end{aligned}$$

(30)

Finally, (b) and (30) imply (18), and the result follows.   $\blacksquare$

1.2 A.2 Proof of Theorem 1

By (6)–(9), we have that $(\mathbb {R},\mathbb {R})$ is equilibrium with $v_S$ and $v_I$ as the corresponding accumulated payoffs if and only if the following relations hold:

$$\begin{aligned} \alpha ^1+ \overline{\alpha ^1}\delta v_S-C_S=v_S, \end{aligned}$$

(31)

$$\begin{aligned} \overline{\alpha ^1}+ \overline{\alpha ^1}\delta v_I-C_I=v_I \end{aligned}$$

(32)

with

$$\begin{aligned} \alpha ^1+ \overline{\alpha ^1}\delta v_S-C_S>\alpha + \overline{\alpha }\delta v_S-C_{S0}, \end{aligned}$$

(33)

$$\begin{aligned} \overline{\alpha ^1}+ \overline{\alpha ^1}\delta v_I-C_I>\overline{\alpha }+ \overline{\alpha }\delta v_I-C_{I0}. \end{aligned}$$

(34)

Solving (31) and (32) by $v_S$ and $v_I$, respectively, imply (19). Substituting $v_S$ and $v_I$ given by (19) into (33) and (34) imply (20).   $\blacksquare$

1.3 A.3 Proof of Theorem 2

By (6)–(9), we have that $(\mathbb {I},\mathbb {I})$ with $v_S$ and $v_I$ as the corresponding accumulated payoffs if and only if the following relations hold:

$$\begin{aligned} \alpha ^0+ \overline{\alpha ^0}\delta v_S-C_{S0}&=v_S,\end{aligned}$$

(35)

$$\begin{aligned} \overline{\alpha ^0}+ \overline{\alpha ^0}\delta v_I-C_{I0}&=v_I \end{aligned}$$

(36)

with

$$\begin{aligned} \alpha ^0+ \overline{\alpha ^0}\delta v_S-C_{S0}&>\alpha + \overline{\alpha }\delta v_S-C_S, \end{aligned}$$

(37)

$$\begin{aligned} \overline{\alpha ^0}+ \overline{\alpha ^0}\delta v_I-C_{I0}&>\overline{\alpha }+ \overline{\alpha }\delta v_I-C_I. \end{aligned}$$

(38)

Solving (35) and (36) by $v_S$ and $v_I$, respectively, imply (21). Substituting $v_S$ and $v_I$ given by (21) into (37) and (38) imply (22).   $\blacksquare$

1.4 A.4 Proof of Theorem 3

By (6)–(9), we have that $(\mathbb {R},\mathbb {I})$ is an equilibrium with $v_S$ and $v_I$ as the corresponding accumulated payoffs if and only if the following relations hold:

$$\begin{aligned} \alpha + \overline{\alpha }\delta v_S-C_S&=v_S, \end{aligned}$$

(39)

$$\begin{aligned} \overline{\alpha }+ \overline{\alpha }\delta v_I-C_{I0}&=v_I \end{aligned}$$

(40)

with

$$\begin{aligned} \alpha + \overline{\alpha }\delta v_S-C_S&>\alpha ^0+ \overline{\alpha ^0}\delta v_S-C_{S0}, \end{aligned}$$

(41)

$$\begin{aligned} \overline{\alpha }+ \overline{\alpha }\delta v_I-C_{I0}&>\overline{\alpha ^1}+ \overline{\alpha ^1}\delta v_I-C_{I}. \end{aligned}$$

(42)

Solving (39) and (40) by $v_S$ and $v_I$, respectively, imply (23). Substituting $v_S$ and $v_I$ given by (23) into (41) and (42) imply (24).   $\blacksquare$

1.5 A.5 Proof of Theorem 4

By (6)–(9), we have that $(\mathbb {I},\mathbb {R})$ is an equilibrium with $v_S$ and $v_I$ as the corresponding accumulated payoffs if and only if the following relations hold:

$$\begin{aligned} \alpha + \overline{\alpha }\delta v_S-C_{S0}&=v_S,\end{aligned}$$

(43)

$$\begin{aligned} \overline{\alpha }+ \overline{\alpha }\delta v_I-C_{I}&=v_I \end{aligned}$$

(44)

with

$$\begin{aligned} \alpha + \overline{\alpha }\delta v_S-C_{S0}&>\alpha ^1+ \overline{\alpha ^1}\delta v_S-C_{S},\end{aligned}$$

(45)

$$\begin{aligned} \overline{\alpha }+ \overline{\alpha }\delta v_I-C_{I}&>\overline{\alpha ^0}+ \overline{\alpha ^0}\delta v_I-C_{I0} \end{aligned}$$

(46)

Straightforward calculation based on (43)–(46) implies the result.    $\blacksquare$

1.6 A.6 Proof of Theorem 5

(b) follows from (a) and Theorem 1–Theorem 4.

Let us now prove (a). First note that by (2) we have that

$$\begin{aligned} \overline{\alpha }\ge \overline{\alpha ^1}>\overline{\alpha ^0}. \end{aligned}$$

(47)

Assume that (20) and (22) hold simultaneously. Then, (47) and the second of inequalities (20) imply

$$\begin{aligned} \overline{\alpha }-C_I>\frac{1-\overline{\alpha ^1}\delta }{1-\overline{\alpha }\delta }(\overline{\alpha }-C_{I0}). \end{aligned}$$

(48)

Combining this inequality with the second of inequalities (20) imply

$$\begin{aligned} \frac{1-\overline{\alpha }\delta }{1-\overline{\alpha ^0}\delta }(\overline{\alpha _0}-C_{I0})>\overline{\alpha }-C_I>\frac{1-\overline{\alpha ^1}\delta }{1-\overline{\alpha }\delta }(\overline{\alpha }-C_{I0}). \end{aligned}$$

(49)

Meanwhile, by (47),

$$\begin{aligned} (1-\overline{\alpha }\delta )^2<(1-\overline{\alpha ^0}\delta )1-\overline{\alpha ^1}\delta \end{aligned}$$

(50)

and

$$\begin{aligned} \overline{\alpha ^0}-C_{I0} <\overline{\alpha }-C_{I0} \end{aligned}$$

(51)

By (50) and (51), we have that

$$\begin{aligned} \frac{1-\overline{\alpha }\delta }{1-\overline{\alpha ^0}\delta }(\overline{\alpha _0}-C_{I0})<\frac{1-\overline{\alpha ^1}\delta }{1-\overline{\alpha }\delta }(\overline{\alpha }-C_{I0}). \end{aligned}$$

(52)

This contradicts to (49). Thus, conditions (20) and (22) cannot hold simultaneously.

The remaining pairs of conditions can be checked similarly.    $\blacksquare$

1.7 A.7 Proof of Theorem 6

First note that there exists at least one equilibrium in stochastic game $\varGamma ,$ since $\varGamma$ is a recursive game with discount factor $\delta \in (0,1)$ [18].

By proofs of Theorem 1–Theorem 4, it follows that if equilibrium is in pure strategies, then it is given by Theorem 1–Theorem 4, and one of four conditions (20), (22), (24) and (26) have to hold. Moreover, by Theorem 5, only one of these four conditions can hold, and (a) follows.

Thus, if none of the four conditions (20), (22), (24) and (26) holds then the equilibrium is in mixed strategies. By (6)–(9), we have that $((x,\overline{x}),(y,\overline{y}))$ with $x\in (0,1)$ and $y\in (0,1)$ is an equilibrium with $v_S$ and $v_I$ as the corresponding accumulated payoffs if and only if the following relations hold:

$$\begin{aligned} (\alpha ^1+ \overline{\alpha ^1}\delta v_S-C_S) y+(\alpha + \overline{\alpha }\delta v_S-C_S)\overline{y}=v_S,\end{aligned}$$

(53)

$$\begin{aligned} (\alpha + \overline{\alpha }\delta v_S-C_{S0})y+ (\alpha ^0+ \overline{\alpha ^0}\delta v_S-C_{S0})\overline{y}=v_S \end{aligned}$$

(54)

and

$$\begin{aligned} (\overline{\alpha ^1}+ \overline{\alpha ^1}\delta v_I-C_I)x +(\overline{\alpha }+ \overline{\alpha }\delta v_I-C_I)\overline{x}&=v_I,\end{aligned}$$

(55)

$$\begin{aligned} (\overline{\alpha }+ \overline{\alpha }\delta \varGamma _I-C_{I0})x+ (\overline{\alpha ^0}+ \overline{\alpha ^0}\delta \varGamma _I-C_{I0})\overline{x}&=v_I. \end{aligned}$$

(56)

Solving (53) by y implies the second of two Eqs. (28). Solving (54) by y, we have that

$$\begin{aligned} y=\frac{C_{S0}-\alpha ^0+(1-\overline{\alpha ^0}\delta ) v_S}{(\alpha -\alpha ^0)(1-\delta v_S)}. \end{aligned}$$

(57)

Substituting this y into the second of two Eqs. (28) and solving by $v_S$ implies the first of two Eqs. (27).

Solving (55) by x implies the first of (28). Solving (56) by x implies

$$\begin{aligned} x=\frac{C_{I0}-\overline{\alpha ^0}+(1-\overline{\alpha ^0}\delta ) v_I}{(\alpha ^0-\alpha )(1+\delta v_I)}. \end{aligned}$$

(58)

Substituting this x into the first of two Eqs. (28) and solving by $v_I$ implies the second of two Eqs. (27).    $\blacksquare$