1 Comparative Static results for ratio Form

1 Comparative Static results for ratio Form

Suppose that the players participating in the contest face strictly convex cost function ψ_i(e_i) fori= 1,2. Our objective functions are given by:

maxe1

θe^m₁

θe^m₁ +eⁿ₂v−ψ₁(e₁)

and

maxe2

eⁿ₂

θe^m₁ +eⁿ₂v−ψ₂(e₂)

The above optimization can be redefined in terms of the cost of input provided by the parties. Let x_i =ψ_i(e_i). Since ψ_i(e_i) is strictly monotonic, it has a well behaved inverse function; g_i(x_i) = ψ_i⁻¹(x_i). Therefore, e_i = g_i(ψ_i(e_i)) = g_i(x_i). With this transformation we can rewrite our optimization problem as

maxx1

θ(g₁(x₁))^m

θ(g₁(x₁))^m+ (g₂(x₂))ⁿv−x1

and

maxx2

(g₂(x₂))ⁿ

θ(g₁(x₁))^m+ (g₂(x₂))ⁿv−x₂

where x_i is cost of effort/input.

Let us restrict our attention to the convex cost functions of the form ψ_i(e_i) = e^k_i, i = 1,2 where k >1 (note that we are only considering e_i >0 and hence such a function is strictly increasing in e_i given k). Then, from the above definitions we get g_i(x_i) = x

1 k

i = e_i, i = 1,2. Using these expressions we can rewrite our optimization problem as

maxx1

θx^m₁ ⁰

θx^m₁⁰ +xⁿ₂⁰v−x₁

and

maxx2

xⁿ₂⁰

θx^m₁⁰ +xⁿ₂⁰v−x₂

where m⁰ = ^m_k and n⁰ = ⁿ_k. (If m, n <1 then m⁰, n⁰ <1 since k >1)

Note that we have transformed a problem with convex cost of efforts to a standard ratio form problem with linear cost function extensively studied in the literature.

However, efforts (e_i) have been replaced by costs (x_i).

We can solve the entire problem for cost of effort (x_i) and then deduce efforts using

the relationship e_i =g_i(ψ_i(e_i)) =g_i(x_i) which in our specific case ise_i =g_i(x_i) = x

1 k

i . Note that for given value of k, e_i and x_i are monotonically related i.e. increase in xi ⇐⇒ increase in ei. The magnitude of change will differ but the direction of change would be the same. With this information we now look at the comparative static results for x_i and deduce the comparative static results for e_i.

The FOC’s for the above problem are given by F OC1− g₁^r = m⁰p(1−p)v

x₁ −1 = 0 F OC2− g^r₂ = n⁰p(1−p)v

x₂ −1 = 0 which can be rearranged and written as:

m⁰p(1−p)v−x₁ = 0 n⁰p(1−p)v−x₂ = 0

After rearranging, the FOC’s are similar to the case with difference form CSF and convex costs (with efforts replaced by costs).

Following the arguments used for proving the existence of a solution to the FOC’s in the difference form CSF with quadratic costs (using Intermediate value theorem), here also it can be shown that at least one solution to the FOC’s is guaranteed to exist. Further, for m, n ≤ 1 SOC will also be satisfied and hence a Nash equilib- rium exists. Alternatively, using the results of Szidarovszky and Okuguchi (1997), we can claim that for the given ratio form CSF with m, n≤ 1 there exists a unique pure Nash equilibrium. In fact We can relax this condition and say that we need m⁰, n⁰ ≤1 =⇒ m, n≤k wherek > 1

We know from the FOC’s that if a Nash equilibrium (x^r∗₁ , x^r∗₂ ) exists, then x^r∗₁ =

m⁰

n⁰x^r∗₂ =⇒ x^r∗₁ = ^m_nx^r∗₂ i.e. x^r∗₁ ≷ x^r∗₂ as m ≷ n independent of the ’natural advan- tage’ i.e. θ⁰s.

As seen above, g_i^r = 0 i= 1,2 gives us the FOCs. We can also see that

∂g^r₁

∂x₁ = m⁰²v(1−2p^r∗)p^r∗(1−p^r∗)

(x^r∗₁ )² − m⁰p^r∗(1−p^r∗)v (x^r∗₁ )²

∂g^r₂

∂x2

= n⁰²v(2p^r∗ −1)p^r∗(1−p^r∗)

(x^r∗₂ )² − n⁰p^r∗(1−p^r∗)v (x^r∗₂ )²

∂g₁^r

∂x₂ = m⁰v(1−2p) x₁

∂p

∂x₂ = m⁰n⁰v(2p^r∗−1)p^r∗(1−p^r∗) x^r∗₁ x^r∗₂

∂g^r₂

∂x₁ = n⁰v(1−2p) x₂

∂p

∂x₁ = m⁰n⁰v(1−2p^r∗)p^r∗(1−p^r∗) x^r∗₁ x^r∗₂

∂g_i^r

∂xi ≤0 fori= 1,2 when the SOC is satisfied i.e. at any Nash equilibrium.

From the above expressions we know that

∂g₁^r

∂x₂

∂g₂^r

∂x₁ ≤0 and at any Nash equilibrium we have

∂g₁^r

∂x₁

∂g₂^r

∂x₂ ≥0

2 Changing value of prize (v)

Let us see the impact of expected prize value on the equilibrium cost of efforts and probability of winning of the first player.

On derivating the FOC’s w.r.t v we get

∂g₁^r

∂v + ∂g₁^r

∂x₁

∂x^r∗₁

∂v + ∂g^r₁

∂x₂

∂x^r∗₂

∂v = 0

∂g₂^r

∂v + ∂g₂^r

∂x₁

∂x^r∗₁

∂v + ∂g^r₂

∂x₂

∂x^r∗₂

∂v = 0

where

∂g₁^r

∂v = m⁰p^r∗(1−p^r∗) x^r∗₁ >0

∂g₂^r

∂v = n⁰p^r∗(1−p^r∗) x^r∗₂ >0 The other partials are the same as given earlier.

On solving the system of linear equations, we get

∂x^r∗₁

∂v =

∂g^r₁

∂v

∂g^r₂

∂x2 −^∂g_∂v^r2∂g_∂x^r1

2

∂g₁^r

∂x2

∂g^r₂

∂x1 −_∂x^∂gr2

2

∂g₁^r

∂x1

(1)

∂x^r∗₂

∂v =

∂g^r₂

∂v

∂g^r₁

∂x1 −^∂g_∂v^r1∂g_∂x^r2

1

∂g₁^r

∂x2

∂g^r₂

∂x1 −_∂x^∂gr2

2

∂g₁^r

∂x1

(2) Let A = ^∂g_∂x¹

2

∂g2

∂x1 − ^∂g_∂x²

2

∂g1

∂x1. Using arguments given in the previous section, it is easy to see that at any NE, A < 0. Here we can use the partials which were calculated

earlier to obtain,

∂x^r∗₁

∂v =

−m⁰n⁰(p^r∗(1−p)^r∗)²v x^r∗₁ (x^r∗₂ )²

A >0

∂x^∗₂

∂v =

−m⁰n⁰(p^r∗(1−p)^r∗)²v (x^r∗₁ )²x^r∗₂

A >0 Also, since p^r∗ doesn’t depend onv directly, so we get

dp^r∗

dv = ∂p^r∗

∂x^r∗₁

∂x^r∗₁

∂v + ∂p^r∗

∂x^r∗₂

∂x^r∗₂

∂v dp^r∗

dv = m⁰p^r∗(1−p^r∗) x^r∗₁

∂x^r∗₁

∂v − n⁰p^r∗(1−p^r∗) x^r∗₂

∂x^r∗₂

∂v

Since in any equilibrium, we have ^x_m^r∗10 = ^x_n^r∗20 , thus we can simplify the above equation to get

dp^r∗

dv = m⁰

x^r∗₁ p^r∗(1−p^r∗) ∂x^r∗₁

∂v − ∂x^r∗₂

∂v

The dynamics of the above expression depend on ^∂x_∂v^r∗1 − ^∂x_∂v^∗2.

∂x^r∗₁

∂v −∂x^∗₂

∂v =

−m⁰n⁰(p^r∗(1−p)^r∗)²v

x^r∗₁ (x^r∗₂ )² − ^−m0n0_(x^(pr∗^{r∗(1−p)r∗)2v} 1 )²x^r∗₂

A

=

−m⁰n⁰(p^r∗(1−p)^r∗)²v x^r∗₁ x^r∗₂ (_x¹r∗

2 −_x¹r∗

1 ) A

Given that the denominator in the above expression is negative at any Nash equilib- rium, ^∂x_∂v^r∗1 − ^∂x_∂v^∗2 ≷0 as _x¹r∗

2 − _x¹r∗

1 ≷0

Note that in any equilibrium, ^x_m^r∗10 = ^x_n^r∗20 is true i.e. ^x_x^r∗1r∗

2 = ^m_n. So, m ≷ n =⇒ x^r∗₁ ≷ x^r∗₂ =⇒ _x¹r∗

2 ≷ _x^1r∗

1 =⇒ _x¹r∗

2 − _x¹r∗

1 ≷0.

Thus, ^∂x_∂v^r∗1 −^∂x_∂v^r∗2 ≷0 as m≷n =⇒ ^dp_dv^r∗ ≷0 as m≷n

2.1 Impact of change in natural advantage, (θ)

Here we have ^∂g_∂θ^1r = ^m0v(1−2p_x ^r∗)

1

∂p

∂θ

_xr∗

1 ,x^r∗₂ ; ^∂g_∂θ^2r = ^n0v(1−2p_x ^r∗)

2

∂p

∂θ

_xr∗

1 ,x^r∗₂ , where

∂p

∂θ

x^r∗₁ ,x^r∗₂ = x^m₁⁰xⁿ₂⁰ (θx^m₁⁰ +xⁿ₂⁰)²

x^r∗₁ ,x^r∗₂ = p^r∗(1−p^r∗) θ >0.

For this case, since ^x_m^r∗10 = ^x_n^r∗20 in an equilibrium, ^∂g_∂θ^r1 = ^∂g_∂θ^2r. The other partials are the same as in the previous case. The effect ofθ depends on the value of the equilibrium value of p, i.e., p^r∗.

∂x^r∗₁

∂θ =

∂g^r₁

∂θ

∂g^r₂

∂x2 − ^∂g_∂x^1r

2)

A =

∂g₁^r

∂θ

−n⁰vp^r∗(1−p^r∗) (x^r∗₂ )² )

A (3)

∂x^r∗₂

∂θ =

∂g^r₁

∂θ

∂g₁^r

∂x1 − ^∂g_∂x^r2

1)

A =

∂g^r₁

∂θ

−m⁰vp^r∗(1−p^r∗) (x^r∗₁ )² )

A (4)

It is easy to see that ^∂x_∂θ^r∗1 ≷ 0 as p^r∗ ≶ 1/2. Same is true for ^∂x_∂θ^r∗2 , i.e. ^∂x_∂θ^r∗2 ≷ 0 as p^r∗ ≶ 1/2. x^r∗₁ and x^r∗₁ are maximum when p^r∗ = 1/2 i.e. when the contest is symmetric.

For instance, when m⁰ = n⁰, an equilibrium say (x^r∗₁ , x^r∗₂ , p^r∗) = (_(θ+1)^mvθ2,_(θ+1)^mvθ2,_θ+1^θ ).

In this case, if θ <1 then p^r∗ <1/2 and hence ^∂x_∂θ^r∗1 >0 and ^∂x_∂θ^r∗1 > 0. The opposite holds true for θ >1. The closer θ is to 1 higher is the value.

Let’s look at how equilibrium probability of win, p^r∗, changes with θ, i.e., ^dp_dθ^r∗ > 0 holds.

dp^r∗

dθ = ∂p^r∗

∂θ + ∂p^r∗

∂x^r∗₁

∂x^r∗₁

∂θ + ∂p^r∗

∂x^r∗₂

∂x^r∗₂

∂θ

= ∂p^r∗

∂θ + m⁰p^r∗(1−p^r∗) x^r∗₁

∂x^r∗₁

∂θ − ∂x^r∗₂

∂θ

= ∂p^r∗

∂θ

1

1−(m⁰−n⁰)(1−2p^r∗)

In the literature, mostly the case m = n is considered in which case probability increases with natural advantage.