Cheating in a contest with strategic inspection

Abstract

We analyze a game between three players: two Athletes and an Inspector. Two athletes compete with each other and both may cheat to increase their chances of victory. The Inspector wishes to detect incidents of cheating, and performs tests on athletes to detect cheating. The test is costly for the Inspector. Both probability of cheating and that of testing decrease as cost of inspection diminishes.

Appendix

Proof of Proposition 2.1

Let $z_F$ and $z_U$ be probabilities TF and TU, respectively, which are chosen with. Let $x_i$, $i \in \{F,U\}$ be the probability that i chooses D. F prefers D iff

$$\begin{aligned} (1-z_F)[p^*(1-x_U)+p x_U]+z_FM \ge p(1-x_U)+p^{'} x_U, \end{aligned}$$

(1)

and U prefers D iff

$$\begin{aligned} (1-z_U)[(1-p^{'})(1-x_F)+(1-p) x_F]+z_UM \ge (1-p)(1-x_F)+(1-p^*) x_F. \end{aligned}$$

(2)

I prefers TF to NF iff

$$\begin{aligned} -c+x_F \ge 0, \end{aligned}$$

which is equivalent to

$$\begin{aligned} x_F \ge c. \end{aligned}$$

(3)

Similarly, I prefers TU to NU if

$$\begin{aligned} x_U \ge c. \end{aligned}$$

(4)

Lemma 1

There is no equilibrium of $G_1$, such that $x_F=1$ or $x_U=1$.

Proof

Suppose on the contrary that F chooses D with certainty. Then, TF is a best reply of I, namely, $z_F=1$. By (1), for any negative M, F prefers N, contradiction. Similarly, $x_U=1$ is impossible in equilibrium. $\square$

Lemma 2

In every Nash equilibrium of $G_1$, $0<z_F<1$ and $0<z_U<1$.

Proof

If I chooses TF with probability 1, the best reply of F is N, but then I is better off by deviating to NF, contradiction. If I chooses TF with probability 0, the best reply of F is D, contradiction to Lemma 1. Similarly, TU or NU cannot be pure strategies of I in an equilibrium. $\square$

Lemma 3

In every Nash equilibrium, $0<x_F<1$ and $0<x_U<1$.

Proof

Suppose that $x_F=0$. Then, the best reply of I is $z_F=0$, contradiction to Lemma 2. By Lemma 1, in equilibrium, U chooses D with probability less than 1. A proof of $0<x_U<1$ is similar. $\Box$

By Lemma 2, equality holds in (3) and in (4), and thus

$$\begin{aligned} x_F=x_U=c. \end{aligned}$$

(5)

By Lemma 3, equality holds in (1) and in (2); therefore

$$\begin{aligned} z_F=\frac{(p^*-p)(1-c)+(p-p^{'})c}{p^*(1-c)+pc-M} \end{aligned}$$

(6)

and

$$\begin{aligned} z_U=\frac{(p-p^{'})(1-c)+(p^*-p)c}{(1-p^{'})(1-c)+(1-p)c-M}. \end{aligned}$$

(7)

The reminder of the proof follows directly from (5), (6), and (7). $\square$

Proof of Proposition 2.2

Recall, $y_i$, $i \in \{F,U\}$, is the probability that i chooses D, and $P_T(i|j)$ is the probability that I tests i given that j is the winner, $i,j \in \{F,U\}$.

Lemma 4

In every equilibrium of $G_2$, $0<y_F<1$ and $0<y_U<1$.

A proof is similar to Lemma 3.

The next lemma states that there is no equilibrium where some player is tested with certainty, or with certainty is not tested, disregarding outcome of the contest.

Lemma 5

There is no Nash equilibrium of $G_2$, where $P_T(F|F)=P_T(F|U)=1$ or $P_T(F|F)=P_T(F|U)=0$ or $P_T(U|F)=P_T(U|U)=1$ or $P_T(U|F)=P_T(U|U)=0$.

Proof

Note that $P_T(i|F)=P_T(i|U)=1$; $i \in \{F,U\}$ means that I tests i with probability 1 irrespective of the result of the contest, and $P_T(i|F)=P_T(i|U)=0$ means that I tests i with probability 0 irrespective of the result of the contest. The proof is similar to Lemma 2. $\square$

The next lemma states that there is no equilibrium where the winner is tested with certainty.

Lemma 6

Let $i \in \{F,U\}$. There is no Nash equilibrium where $P_T(i|i)=1$.

Proof

Suppose on the contrary that player i is tested with probability 1 if she wins. If i dopes, her expected payoff is $MP(i \text{ wins } |i \text{ dopes })$. This value is negative; therefore, i is better off by deviating to N, contradiction to Lemma 4. $\square$

Lemma 7

Let $i,j \in \{F,U\}$,$i \ne j$. Then, $P(i \text{ chose } \text{ D }|j \text{ wins })<P(i \text{ chose } \text{ D }|i \text{ wins })$.

Proof

$$\begin{aligned} P(U \text{ chose } \text{ D }|F \text{ wins })=\frac{y_U[y_Fp+(1-y_F)p^{'}]}{y_U[y_Fp+(1-y_F)p^{'}]+(1-y_U)[y_Fp^*+(1-y_F)p]} \end{aligned}$$

and

$$\begin{aligned}&P(U \text{ chose } \text{ D }|U \text{ wins })\\&\quad =\frac{y_U[y_F(1-p)+(1-y_F)(1-p^{'})]}{y_U[y_F(1-p)+(1-y_F)(1-p^{'})]+(1-y_U)[y_F(1-p^*)+(1-y_F)(1-p)]}. \end{aligned}$$

It is straightforward to verify that $P(U \text{ chose } \text{ D }|F \text{ wins })<P(U \text{ chose } \text{ D }|U \text{ wins })$ is equivalent to

$$\begin{aligned}&[y_Fp+(1-y_F)p^{'}][y_F(1-p^*)+(1-y_F)(1-p)]\\&\quad <[y_Fp^*+(1-y_F)p][y_F(1-p)+(1-y_F)(1-p^{'})]. \end{aligned}$$

The last inequality holds for nonnegative $y_F$ and $y_U$ by $p^{'}<p<p^*$. A proof for $P(F \text{ chose } \text{ D }|U \text{ wins })<P(F \text{ chose } \text{ D }|F \text{ wins })$ is similar. $\square$

The next lemma states that an athlete who loses in the contest is tested with probability 0.

Lemma 8

Let $i,j \in \{F,U\}$,$i \ne j$. In equilibrium, $P_T(F|U)=P_T(U|F)=0$.

Proof

To prove the lemma, the following inequalities are required. For $i,j \in \{F,U\}$, $i \ne j$, the Inspector prefers to test i to not testing, given j wins, iff

$$\begin{aligned} P(i \text{ chose } \text{ D }|j \text{ wins }) \ge c. \end{aligned}$$

(8)

Similarly, I prefers to test j to not testing, given j wins, iff

$$\begin{aligned} P(j \text{ chose } \text{ D }|j \text{ wins }) \ge c. \end{aligned}$$

(9)

Next, suppose on the contrary that I tests U with a positive probability if she loses. Suppose F wins. By (8), $P(U \text{ chose } \text{ D }|F \text{ wins }) \ge c$. Then, by Lemma 7

$$\begin{aligned} c \le P(U \text{ chose } \text{ D }|F \text{ wins }) <P(U \text{ chose } \text{ D }|U \text{ wins }). \end{aligned}$$

Thus, by (9), U is tested with certainty if she wins, contradiction to Lemma 6. Similarly, the probability that F is tested if she loses cannot be positive. $\square$

To complete the proof of the Proposition, observe that, in an equilibrium, $P_T(j|j)>0$. By Lemma 8, $P_T(i|j)=0$, then $P_T(j|j)>0$ follows from Lemma 5. The result of Proposition 2.2 follows directly from Lemmas 5, 6, and 8. $\Box$

Proof of Proposition 2.3

F prefers D to N iff

$$\begin{aligned} p(1-y_U)+p^{'}y_U \le [p^*(1-y_U)+py_U][1-P_T(F|F)+P_T(F|F)M]. \end{aligned}$$

(10)

U prefers D to N iff

$$\begin{aligned}&(1-p)(1-y_F)+(1-p^*)y_F \le [(1-p^{'})(1-y_F)+(1-p)y_F][1-P_T(U|U)\nonumber \\&\quad +\,P_T(U|U)M]. \end{aligned}$$

(11)

By Proposition 2.1, $x=c$. By Proposition 2.2, $P_T(U|F)=P_T(F|U)=0$ and $0<P_{T,W}<1$. Namely, if $i \in \{F,U\}$ wins, I is indifferent between testing or not testing i. Therefore, the equality in (9) holds. After substituting $p=0.5$ and $p^{'}=1-p^*$, we obtain

$$\begin{aligned} c=\frac{y[0.5y+(1-y)p^*]}{y[0.5y+(1-y)p^*]+(1-y)[y(1-p^*)+0.5(1-y)]}, \end{aligned}$$

and the root of this equation between 0 and 1 is

$$\begin{aligned} y=\frac{p^*-\sqrt{(p^*)^2-2cp^*+c}}{2p^*-1}. \end{aligned}$$

(12)

It is straightforward to verify that $y<c=x$, and that y is increasing in c.

By substituting of $p=0.5$ and $p^{'}=1-p^*$ in (6)

$$\begin{aligned} z=\frac{p^*-0.5}{p^*(1-c)+0.5c-M}. \end{aligned}$$

(13)

By Lemma 5, in $G_2$, athlete i is indifferent between doping and no doping. Therefore, equality in (10) holds, and in (11), and in the symmetric case, this implies that:

$$\begin{aligned} y(1-p^*)+0.5(1-y)=[0.5y+(1-y)p^*](1-P_{T,W}+P_{T,W}M). \end{aligned}$$

Equivalently,

$$\begin{aligned} P_{T,W}=\frac{p^*-0.5}{[0.5y+(1-y)p^*](1-M)}, \end{aligned}$$

(14)

and $P_{T,W}$ is increasing in c. By (13) and (14)

$$\begin{aligned} P_{T,W}<2z \end{aligned}$$

is equivalent to

$$\begin{aligned} p^*(1-c)+0.5c-y-2p^*+2p^*y <-M[(2p^*-1)(1-y)]. \end{aligned}$$

Since $p^*>0.5$, the last inequality holds for sufficiently large |M|. $\square$