Quality-Based Competition, Profitability, and Variable Costs Chester Chambers Cox School of Business Dallas, TX 75275
[email protected] 214-768-3151 Panos Kouvelis Olin School of Business Washington University
[email protected] 314-935-4604 John Semple Cox School of Business Dallas, TX 75275
[email protected]
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Appendix Equivalence of Utility Specifications. Consider (i) the utility function U (q, p : θ ) = q − θp where θ has the cumulative distribution function Fθ (•) corresponding to a uniform distribution on [ε ,1] versus (ii) the “classical” utility function U (q, p, α ) = αq − p where α has the (unknown) cumulative distribution Fα (t ) . The infinitesimal ε is introduced to avoid division by 0, although taking the limit ε → 0 poses no difficulties in our subsequent analysis. The two utility specifications are equivalent if and only if the proportion of the population preferring one product over another in our model equals the proportion of the population having identical preferences in the classical model. Straightforward algebra shows that these proportions are indeed equal if and only if Prα (α ≥ t ) = Prθ (θ ≤ 1 / t ) . However, 1 1 / t − ε 1 − Fα (t ) = Pr(α : α ≥ t ) = Pr(θ : θ ≤ 1 / t ) = Fθ (1 / t ) = 1 − ε 0
t 1/ ε
Differentiating each side with respect to t implies the equivalent density for the taste parameter 1 1 for 1 ≤ t ≤ 1 / ε and 0 elsewhere. In the special case in the classical model is f α (t ) = 1− ε t2 where ε → 0 , the equivalent density for the taste parameter in the classical model is f α (t ) = 1 / t 2 on [1, ∞) . Proposition 1. Condition 1. The terms qH , qL , and pL are considered fixed (given). Suppose p H < ( p L / q L )q H . Then me must have q L / p L < q H / p H < (q H − q L ) /( p H − p L ) (see Figure 1a) and H’s profit is π H = min(q H / p H ,1)( p H − c H ) . It is easy to show that H’s profit is strictly increasing on this interval. Thus p H ≥ ( p L / q L )q H , which is Condition 1. Condition 2. By Condition 1, we must have (q H − q L ) /( p H − p L ) ≤ q H / p H ≤ q L / p L . For p H such that (q H − q L ) /( p H − p L ) > 1 (see Figure 1c), H’s profit is π H = 1 ⋅ ( p H − c H ) , which is strictly increasing. Therefore, H will always respond with a price such that (q H − q L ) /( p H − p L ) ≤ 1 or equivalently pH ≥ pL + qH − qL . Best Response for H. Because p H must satisfy proposition 1, we have qH − qL ( p H − c H ) p p − L H
π H ( p H ) =
and
∂π H / ∂p H =
qH − qL cH − pL . p H − p L p H − p L
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If pL < cH , the function π H is strictly increasing and thus “ p H = ∞ ” (this price is simply large and finite if one takes θ ∈ [ε ,1] for ε > 0 and small). If p L = c H , then H’s profit is q H − q L , regardless of H’s price provided it is feasible (see Figure 2). Thus H selects any price satisfying p H ≥ max(c H + q H − q L , c H q H / q L ) . If p L > c H , then π H is strictly decreasing and H will set the lowest possible feasible price, which is max( p L + q H − q L , p L q H / q L ) . In summary, H’s best response is pH = ∞ R H ( p L ) = p H = max(c H + q H − q L , c H q H / q L ) p = max( p + q − q , p q / q ) L H L L H L H
pL < cH pL = cH pL > cH
Proposition 2. Given 0 < qL < qH let L chose a price cH , and H chose his best response. Thus, q − qL the total size of the served market is 1 and π L = 1 − H p H − pL q − qL . positive market share and profits as long as 1 > H p H − cH substitution.
( pL − cL ) . Since cH > cL L has
This is easily verified by direct
The following lemma is needed to simplify the analysis of L’s best response. It is based on straightforward algebra and therefore offered without proof. Lemma 1. If p L satisfies p L ≤ c L (a “market covering” price), then Condition 2 of proposition 1 implies Condition 1. If p L > c L , then Condition 1 implies Condition 2. Best response for L. Market is covered case. On the interval pL ≤ qL , L covers the market and q − qL the resulting profit function is π L = 1 − H ( pL − cL ) . Assuming the market is covered p H − pL implicitly requires cL ≤ qL ; if this is not the case, the market cannot be covered and one proceeds directly to the analysis of the uncovered market. For the covered market case, we have q − qL ∂π L / ∂pL = 1 − H ⋅ p − cL ) . The equation ∂π L / ∂pL = 0 has two roots. However, 2 ( H ( p H − pL )
only one of these roots is less than pH , and therefore it is the only relevant response. The root is pL# = pH − ( pH − cL )(qH − qL ) . Because of H’s price floor pH ≥ max(cH , cL + qH − qL , cL
qH ) one can show by direct qL
calculation that (i) p L# is an increasing function of p H , (ii) pL# ≥ cL , and (iii) Condition 2 of
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proposition 1 holds (observe that we must still show L’s price response forms a feasible pair so that the initial profit representation for L is valid.) If the root additionally satisfies pL# ≤ qL , then ( pH , pL# ) must be a feasible pair of prices by Lemma 1. In this case, we may make the stronger statement that pL# is a global maximum for q − qL all pL < pH . This follows from the fact that the function 1 − H ( pL − cL ) is concave on p p − H L the extended interval pL < pH and vanishes at most once on this interval. Thus for feasible prices satisfying pL > qL , we observe q L qH − q L qH − qL # q − qL p − cL ) ≥ 1 − H − 1 − ( p L − cL ) ≥ ( p L − cL ) # ( L pH − pL pH − pL pL pH − p L
The latter inequality ensures that pL# is a global maximum for π L if pL# ≤ qL . If pL# > qL , then the derivative of π L is positive on the interval pL ≤ qL and so π L is increasing on this interval. The maximum (and thus L’s best response) occurs at pL = qL . This response forms a feasible pair because (a) if Condition 2 of proposition 1 is satisfied for p L# then it is satisfied for p L = q L ≤ p L# and (b) Condition 2 and Condition 1 are identical if p L = q L . Market not covered case. If L has a positive market share and pL > qL , the market is not covered. Observe that pL > qL can only occur if pH > qH (otherwise the market is already covered by H see Figure 1b and Figure 2). Additionally, we may assume H has priced above q his price floor, pH ≥ max(cH , cL + qH − qL , cL H ) . For pL satisfying pL ≥ qL , L’s profit function qL q qH − q L q − qL cq is given by π L = L − H c − pH ) + L 2L . If ( pL − cL ) , and ∂π L / ∂pL = 2 ( L pL pL pH − pL ( p H − pL ) the derivative vanishes at an interior point pL > qL , then it is a global maximum on the interval pL ≥ qL since π L is strictly concave on this interval. Again, there are two possible roots for ∂π L / ∂pL = 0 . However, in this case only one root is positive:
pL## =
cL q L (qH − qL )( pH − cL ) cL q L 1+ (qH − qL )( pH − cL )
⋅ pH
It can be shown (we omit the tedious algebra) that ( pH , pL## ) satisfies Condition 1 of proposition q 1 provided pH ≥ cL H . Note that the latter condition is guaranteed by H’s price floor. qL
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Moreover, it can be shown the root is an increasing function of p H on the interval q pH ≥ max(cH , cL + qH − qL , cL H ) and further satisfies pL## ≥ cL . qL If pL## ≥ qL , then Lemma 1 implies Condition 2 must hold as well. Thus ( pH , pL## ) is a feasible pair of prices whenever pL## ≥ qL . This solution is also a global maximum for π L for reasons explained next. We observe first that the derivative for π L cannot vanish on both of the intervals pL ≤ qL and pL > qL . For suppose ∂π L / ∂pL = 0 for x * ≤ q L , then we have qH − qL * qL qH − qL q − qL x − cL ) ≥ 1 − H − 1 − ( p L − cL ) ≥ ( p L − cL ) * ( pH − x pH − pL pL p H − p L
for all feasible prices pL ≥ qL . Observe that the two profit expressions agree at the crossover point pL = qL . However, the actual concave profit function that applies on the interval pL ≥ qL , qL q H − q L − ( pL − cL ) , is bounded above by the strictly decreasing concave function pL pH − pL qH − qL 1 − ( pL − cL ) . This prohibits ∂π L / ∂pL from vanishing on the interval pL > qL (if it p − p H L has already vanished for x * ≤ q L ). Thus if pL## > qL , π L must be increasing on [c L , q L ] ,
πL =
increasing on [q L , p L## ] , and then decreasing thereafter. This makes pL## a global maximum. If the solution satisfies pL## ≤ qL , then the profit function is decreasing on the interval pL ≥ qL and the maximum on this interval occurs at the endpoint pL = qL , which is feasible since pH > qH (see Figure 2). It can be shown that p L## < p L# provided H chooses pH above his price floor. Consequently, the best response for L is
pL = pL# RL ( pH ) = pL = pL## p =q L L
if pL# < qL if pL## > qL . otherwise
Theorem 1: Price Equilibrium. Existence of a price equilibrium. We observe that the best response curves always intersect. This can be shown in three steps, whose details are left to the reader. Step 1: The minimum point on H’s response curve has coordinates pL = cH , q pH = max(cH + qH − qL , cH H ) . This point occurs at the bottom of a vertical line segment qL positioned at pL = cH (see Figure 3). Step 2. If the minimum response on H’s curve is
pH = cH + qH − qL , then pL# (cH + qH − qL ) < cH , which implies RL (cH + qH − qL ) < cH . Step 3. If
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qH , then this implies qL < cH . One can then qL q q show pL## (cH H ) < qL < cH , which implies RL (cH H ) ≤ qL < cH . Since RL ( pH ) is an increasing qL qL the minimum response on H’s curve is pH = cH
function satisfying RL ( pH ) ↑ ∞ as pH ↑ ∞ , it must cross the infinite vertical segment of H’s q response curve (see Figure 3) at some point above pH = max(cH + qH − qL , cH H ) . Observe that qL a simple perturbation argument implies the existence of a similar intersection for the case where θ ∈ [ε ,1] , provided ε is sufficiently small. The price equilibrium. The only potential equilibrium solution occurs when pL = cH . In this q − qL case, the profit function becomes π H = H ( pH − cH ) = qH − qL . Consequently, H p H − cH appears indifferent to any feasible price pH since they all result in an identical profit of qH − qL . (this indifference disappears for θ ∈ [ε ,1] with ε > 0 and small). However, H must still choose his price carefully so that L accepts the price pL = cH and has no incentive to change. This is indeed the case if pH is set so that pL = cH is L’s optimal response. Analysis of the first term in L’s response curve shows that if cH < qL then the value of pH which drives pL to cH is
pH* = cH + 1/ 2(qH − qL ) + 1/ 4(qH − qL ) 2 + (cH − cL )(qH − qL )
One can readily check that pH* as defined above and pL* = cH form a feasible pair of prices (this ensures that our original profit representations are valid). These are the equilibrium prices when cH < qL . Analysis of the second term in L’s response curve shows that if cH > qL , the appropriate selection is qH − qL ) cH2 ( * 2 pH = cH + α / 2 + α / 4 + α ( cH − cL ) where α = cL q L One can check that pH* as defined above and pL* = cH form a feasible pair of prices. These are the equilibrium prices when cH > qL . Finally, if cH = qL , then there are an infinity of price equilibrium solutions. One can calculate a value for pH* using either of the preceding formulas, although any price selected between these two values will also suffice. This occurs because the finite vertical segment in L’s best response curve perfectly coincides with the infinite vertical segment in H’s response curve (see Figure 3). We note that this situation does not occur when the price sensitivity parameter satisfies θ ∈ [ε ,1] ε > 0 . The solution with the lowest price for H is the limiting price as ε → 0 . Theorem 2: Theorem 1.
The result follows from direct substitution into equation (2) using the results from
Theorem 3: Since q L < q H by definition, the market is necessarily uncovered unless * p L = c H ≤ q H , which is therefore the only situation to assume. For notational convenience,
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recall c H = c(q H ) . The proof proceeds by showing that L’s market share for the uncovered market as q L ↑ c H (see (10)) exceeds L’s market share for the entire covered market q L ≥ c H (see (9)). Since profit margins are also larger on the uncovered market side, it follows that the profit for L as q L ↑ c H (see 10) will dominate all profits for (9) with q L ≥ c H . According to (10), the market share for L as q L ↑ c H is q L 2c(q L ) / c(q H ) 2c(c H ) / c(q H ) 1 − = 1 − Lim q L ↑ c(q H ) 4c(q L )q L [c(q H ) − c(q L )] 4c(c H )[c(q H ) − c(c H )] c(q H ) 1+ 1+ 1+ 1+ 2 c(q H )[q H − c H ] [ ] [ ] − c ( q ) q q H H L 2 . Since c(q H ) > 1 , = 1 − c (c ) 2 H c(q H ) c(q H ) 4c(q H )[c(q H ) − c(c H )] + + c (c H ) c(c H )[q H − c H ] c (c H ) 2 2 ≥ 1 − 1 − . 2 [ ] − 4 c ( q ) c ( q ) c ( c ) H H H c( q H ) c(q H ) 4c(q H )[c(q H ) − c(c H )] 1+ 1+ + + − c ( c ) q c [ ] H H H [ ] c ( c ) c ( c ) c ( c ) q c − H H H H H Because c(q) / q is log-concave, so is c(q) . Thus for any y ≥ x we must have
c ′( x) c ′( y ) ≥ . c( x) c( y )
c( y ) − c( x) ≥ c ′( x) , where the difference y−x quotient at y = x is interpreted as c ′( x) . Thus for y = q H , x = c H ≤ q H and z ≤ q H we have Because c(q) is convex, for any y ≥ x we must have
c(q H ) − c( z ) c ( q H ) c ( q H ) − c (c H ) , ≥ c ′(q H ) ≥ ⋅ qH − z c (c H ) qH − cH
where the last inequality follows from the convexity of c(q ) . It follows that for any z ≤ q H ,
2 2 . ≥ 1− 1 − c(q H ) − c( z ) 1 + 1 + 4c(q H )[c(q H ) − c(c H )] 1 + 1 + 4 c(c H )[q H − c H ] qH − z The last term is precisely L’s market share in (9) for z in the interval q H ≥ z ≥ c H . This demonstrates that the market share for L as q L ↑ c H exceeds that for all q L ≥ c H where (9)
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applies. This implies the maximum profit occurs over the region q L ≤ c H , i.e., the uncovered market region whose profit is determined by (10). We introduce three lemmas that will help with the proof of Theorem 4. Lemma 2. Suppose c(q) / q is convex and log-concave. Then the ratio function c(q) / q is non-decreasing. r (q) = c(q max ) − c(q ) q max − q Proof. Observe that r (q) is continuous on [0, q max ] with r (0) = q max c ′(0) / c(q max ) and r (q max ) = c(q max ) /[c ′(q max )q max ] . Moreover, r (q) is differentiable on (0, q max ) and therefore it is
non-decreasing on [0, q max ] if and only if its derivative is nonnegative on (0, q max ) . The latter is equivalent, after suitable algebraic manipulations, to the inequality condition c(q max ) c ′(q ) c(q max ) − c(q ) . q≥ q max c(q ) q max − q
We will now show the latter inequality is true. Observe that we may write c(q) = qh(q) with h(q) convex and log-concave. Log-concavity of h(q) implies h′(q) / h(q) is a decreasing function of q. Therefore, beginning with the left hand side of the previous inequality condition h ′(q max ) c(q max ) c ′(q ) h ′(q ) q = h(q max ) 1 + q ≥ h(q max ) 1 + q = {h(q max ) + qh ′(q max )} h(q max ) q max c(q ) h( q )
But h(q ) is convex, so the final term in curly brackets satisfies
{h(q max ) + qh′(q max )} ≥ h(q max ) + q
h(q max ) − h(q ) q max h(q max ) − qh(q ) c(q max ) − c(q max ) = . = q max − q q max − q q max − q
Lemma 3. Suppose f ( x) is nonnegative, differentiable, and strictly concave on [a, b] . Suppose g ( x) is differentiable, non-increasing, and positive on [a, b] . Let the maximum of f ( x) on [a, b] occur at the point x *f , and let a global maximum of f ( x) g ( x) on [a, b] occur at
the point x *fg . Then x *fg ≤ x *f . Proof.
For x ∈ ( x *f , b] , f ′( x) < 0 .
After applying the product rule and the various sign ′ conditions stated in the theorem, it follows that ( fg ) ( x) < 0 on this interval as well. This proves the result. Lemma 4. Suppose n(x) and d (x) are nonnegative, continuous functions on [a, b] that are
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n( x ) is continuous d ( x) n( x ) on [a, b] , non-decreasing on [a, b] , and differentiable on (a, b) , then is continuous α + d ( x) and non-decreasing on [a, b] for any α > 0 . n( x ) n( x ) is non-decreasing and on [a, b] is clear. Since Proof. Continuity of α + d ( x) d ( x) differentiable on ( a, b) , n ′( x)d ( x) − n( x)d ′( x) ≥ 0 on ( a , b) , and so n( x ) is non-decreasing on [a, b] . n′( x)[α + d ( x)] − n( x)d ′( x) ≥ 0 on (a, b) , which implies α + d ( x)
also differentiable on (a, b) . Assume n(x) is non-decreasing. If the ratio
Theorem 4. Part (a). By lemma 2, r (q L ) is non-decreasing on [0, q max ] . Thus r (q)c(q) / 4c(q) is non-decreasing (with a removable singularity at 0). The following chain of non-decreasing (non-d for short) functions is implied:
r (q)c(q) / 4c(q) =
(c(q L ) )2
c − c( q L ) 4c(q L )q L max q max − q L
c( q L )
⇒ 2 c max
(c(q L ) )2
Lemma 4
non-d ⇒
c
2 max
c − c(q L ) + 4c(q L )q L max q max − q L c(q L )
Lemma 4
c − c(q L ) + 4c(q L )q L max q max − q L ⇒
non-d ⇒
2 cmax + cmax
c − c(q L ) + 4c(qL )q L max qmax − q L
2c(q L ) / c max 4c(q L )q L c max − c(q L ) 1+ 1+ 2 q max − q L c max
non-d
non-d
non-d.
Consequently, 1-
2c(q L ) / c max 4c(q L )q L c max − c(q L ) 1+ 1+ 2 q max − q L c max
= g (q L )
is seen to be positive and non-increasing. If we define f (q L ) = (c max − c(q L ))(q L / c max ) , then part (a) of Theorem 4 follows immediately from Lemma 3. For part (b) of the theorem, observe that an upper bound on L’s profit in (12) is π L* = π L (q L* ) ≤ c H . If H leapfrogs L, the same inequality applies where q L represents the high quality position. Since q L* < q K* , c(q L* ) < c(q K* ) , H’s best profit from leapfrogging L is bounded
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above by c(q K* ) . Part (b) follows by comparing the lower bound on H’s profit ensured by part (a) and the upper bound on profit c(q K* ) for leapfrogging L. For part (c), observe that the profit for L is bounded above by the expression q max K [ 0,qmax ] (qc(q )) . If H leapfrogs L, then the same bound applies for the optimal position that H takes below L: q L* K [ 0,q* ] (qc(q)) , where q L* is L’s current position. It is immediately clear that L
q K [ 0,q* ] (qc(q)) ≤ q K* K [ 0,q* ] (qc(q)) . * L
L
L
We
will
be
done
if
we
can
show
that
K [ 0,q* ] (qc(q)) ≤ K [ 0,qmax ] (qc(q)) . The condition stated in part (c) plays an essential role. L
Construct the associated function Consider an arbitrary cost function c(q) ∈ Χ . c (λ q ) v (λ , q ) = λ − defined for 0 ≤ λ ≤ 1 and 0 < q . Then c(q )
∂v(λ , q ) −λ c′(λ q )c(q ) + c(λ q )c′(q ) = = 2 ∂q [ c(q )]
c(λ q )c(q ) c′(q ) c′(λ q ) λq q− q c(λ q ) c(q)
[ c(q ) ]
2
≥ 0.
The last inequality follows because qc ′(q) /(c(q) is assumed to be non-decreasing, and so the ∂v(λ , q) bracketed term must be nonnegative, too. Because ≥ 0 , it follows that ∂q c(λq max ) c(λq L* ) = K [ 0,q* ] (c(q )) . ≥ Max λ − K [ 0,qmax ] (c(q )) = Max λ − L λ∈[ 0 ,1] c(q max ) λ∈[ 0,1] c(q L* )
Since q ∈ Χ and c(q) ∈ Χ , we must have qc(q) ∈ Χ , and consequently the same result applies for qc(q ) . The maximum profit H can obtain by leapfrogging L is therefore bounded above by q K* K [ 0,qmax ] (qc(q )) . Part (c) now follows by insisting that the upper bound on H’s profit for leapfrogging is no better than the upper bound on H’s profit (as previously established in part (a)) for staying at q max .
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