Proof of Lemma 1

proof. Part (i). Recall that the logconcavity of f implies that vˆi−1−F(vˆi)f(vˆi) is strictly increasing in vˆi. By the definition of v˜ and μ˜ , we have vˆi−1−F(vˆi)f(vˆi)≤0 if μ∈[μ˜,1], with the strict inequality if μ≠μ˜. Since xivˆi[1−F(vˆi)]≥0 for any μ, we reach the conclusion that ∂Πi∂μi<0 if μ∈(μ˜,1].

Part (ii). A necessary condition for an equilibrium μi∗ is ∂Πi∂μi(μi∗)=0. Following part (i) we must have μi∗∈[0,μ˜], if equilibrium exists.

Part (iii). We first show that ∂2Πi∂μi2<0 for μi∈[0,μ˜). By eq. (5),

Now inspect each term in the above expression. Recall that, for μi∈[0,μ˜),vˆi−1−F(vˆi)f(vˆi)>0 and is strictly decreasing in μi. The term mi(μi,μj) is also positive and strictly decreasing in μi. By previous analysis, vˆi[1−F(vˆi)]=R(μi) is strictly increasing in μi for μi∈[0,μ˜). Therefore, ∂2Πi∂μi2<0 for μi∈[0,μ˜).

Next we show that ∂2Πi∂μi∂μj>0. Again by eq. (5),

which is strictly greater than 0 for μi∈[0,μ˜).

Proof of Lemma 2

proof. Recall that, by part (iii) of Lemma 1, μi and μj are strategic complements. And the domain of [0,μ˜] is compact. By Theorem 4.2(ii) of Vives (1990), the set of Nash equilibria is non-empty. To show the uniqueness of Nash equilibrium, we add eqs (6) and (7), and get

Using the definition of Z(v), the above equation can be written compactly as xAZ(vˆA∗)+xBZ(vˆB∗)=1. Now suppose there are two Nash equilibria: (μA1∗,μB1∗) and (μA2∗,μB2∗). By Theorem 4.2(ii) Vives (1990), the two Nash equilibria can be ordered. Without loss of generality, suppose (μA1∗,μB1∗)\lt(μA2∗,μB2∗). But by the monotonicity of Z(v), according to Claim 1, we have

which contradicts eq. (12). Therefore, there is a unique equilibrium.

Proof of Proposition 1

proof. Part (i). Suppose, to the contrary, μB∗≤μA∗. Since xA>xB, we have xAμA∗>xBμB∗. Taking the difference between eqs (6) and (7), we have

The RHS of the above equation, (13), is strictly greater than 0, which follows from the presumption that μB∗≤μA∗, and the facts that Z(v)>0 and is increasing in μ (by Claim 1) for μ∈[0,μ˜), and xA>xB. This contradicts the result that the LHS of eq. (13) is strictly less than 0. Thus we must have μB∗>μA∗. It immediately follows that vˆA∗>vˆB∗.

To show xAμA∗>xBμB∗, suppose, to the contrary, xAμA∗≤xBμB∗. This means that the LHS of eq. (13) is greater than or equal to 0. Now consider the RHS of eq. (13). By the fact that μ˜>μB∗>μA∗ and by Claim 1, we have W(vˆB∗)\gtW(vˆA∗)>0. Combined with the presumption that xAμA∗≤xBμB∗, we draw the conclusion that the RHS of eq. (13) is strictly less than 0, a contradiction. Therefore, we must have xAμA∗>xBμB∗. It follows immediately that mA(μA∗,μB∗)<mB(μA∗,μB∗).

The difference in equilibrium profits can be written as

Since αBmB∗>αAmA∗,vˆB∗[1−F(vˆB∗)]−vˆA∗[1−F(vˆA∗)]≥0 implies that ΠB∗−ΠA∗>0. Recall that v[1−F(v)] is decreasing in v for v∈[v˜,v‾]. Combining with the result in part (i) that vˆA∗>vˆB∗, we have vˆB∗[1−F(vˆB∗)]−vˆA∗[1−F(vˆA∗)]≥0. This proves that ΠA∗<ΠB∗.

Part (ii). The sign of pB∗−pA∗ can be expressed as

By the above expression, pB∗−pA∗>0 is equivalent to

Recall that μvˆ is strictly decreasing in vˆ for vˆ∈[v˜,v‾), which means that μB∗vˆB∗>μA∗vˆA∗ , or μA∗μB∗<vˆB∗vˆA∗. Thus, to prove pA∗<pB∗, it is enough to show that 1+1−αBαB1−qq1+1−αAαA1−qq≤μA∗μB∗. The fact that xAμA∗>xBμB∗ implies that μA∗μB∗>xBxA. Thus it is sufficient that 1+1−αBαB1−qq1+1−αAαA1−qq≤yByA.

Part (iii). When both xA and xB goes to 0, we have mi∗→12 and μi∗→μ˜, for i=A,B. Therefore,

where the last inequality follows the fact q\lt1/2.

Proof of Proposition 2

proof. Part (i). A consumer’s total net nuisance cost incurred on platform i is yiμi∗. The fact that yAμA∗>yBμB∗ follows directly from the result that xAμA∗>xBμB∗ (part (i) of Proposition 1).

Part (ii). Consider a type v firm advertising on both platforms: v≥vˆA∗>vˆB∗. Its profit per consumer and total profit on platform i are αiq(v−vˆi∗) and αiq(v−vˆi∗)mi∗, respectively. Since αA<αB,vˆA∗>vˆB∗, and mA∗<mB∗, we conclude that αAq(v−vˆA∗)\ltαBq(v−vˆB∗) and αAq(v−vˆA∗)mA∗<αBq(v−vˆB∗)mB∗.

Proof of Proposition 3

proof. Part (i). Since an increase in αB implies a decrease in xB, it is sufficient to show that ∂μA∗∂xB>0 and ∂μB∗∂xB<0. Define Wi≡W(vˆi∗). Differentiating eqs (6) and (7) with respect to xB, we get

From the above two equations, we can solve for ∂μB∗∂xB as follows:

where the inequality follows dWAdμA∗>0 and dWBdμB∗>0. From eqs (14) and (16), the sign of ∂μA∗∂xB can be expressed as

Part (ii). It is sufficient to show that ∂(xAμA∗−xBμB∗)∂xB<0. More explicitly,

where the inequality follows eq. (14), which implies that μB∗+xB∂μB∗∂xB>xA∂μA∗∂xB.

Part (iii). Recall that vˆμ>0 and it is increasing in μ for μ∈[0,μ˜]. Thus by part (i), we have vˆA∗′μA∗′<vˆA∗μA∗ and vˆB∗′μB∗′>vˆB∗μB∗. By eq. (4), Πi∗∝αimi∗vˆi∗μi∗. Since αA remains the same, mA∗′<mA∗ by part (ii), and vˆA∗′μA∗′<vˆA∗μA∗, it must be the case that ΠA∗′<ΠA∗. Similarly, since αB′>αB,mB∗′>mB∗ by part (ii), and vˆB∗′μB∗′>vˆB∗μB∗, we must have ΠB∗′>ΠB∗.

Part (iv). To show pA∗′>pA∗, it is sufficient to show that ∂pA∗∂xB<0. By eq. (1), ∂pA∗∂xB∝dvˆA∗dμA∗∂μA∗∂xB<0, where the inequality follows ∂μA∗∂xB>0 in part (i).

The sign of pB∗′−pB∗ can be expressed as

By the above expression, pB∗′−pB∗>0 is equivalent to

Since vˆμ>0 and it is increasing in μ for μ∈[0,μ˜], we have vˆB∗′μB∗′−vˆB∗μB∗>0, or μB∗μB∗′<vˆB∗′vˆB∗. Thus to show that pB∗−pB∗′<0, it is enough to show that 1+1−αB′αB′1−qq1+1−αBαB1−qq≤μB∗μB∗′. Note that xB′μB∗′<xBμB∗. To see this, by eqs (16) and (17), we have

This implies that ∂(xBμB∗)∂αB<0 and xB′μB∗′, since μB∗′>μB∗ by part (i) of Proposition 3. The fact that xB′μB∗′<xBμB∗ implies yB′yB<μB∗μB∗′. Therefore, to show pB∗−pB∗′<0 it is sufficient that 1+1−αB′αB′1−qq1+1−αBαB1−qq≤yB′yB.

Part (v). Since αA remains the same, mA∗′<mA∗, and μA∗′<μA∗, the total number of ads on platform A decreases. Regarding platform B, when t→∞, we have mB∗→12,mB∗′→12,μB∗→μ˜, and μB∗′→μ˜. Therefore,

since q\lt1/2. This means that for t big enough, the total number of ads on platform B decreases.

Now we show that the total combined relevant ads increase if the distribution of v is either uniform or exponential. Define ui≡vˆi−1−F(vˆi)f(vˆi) and ni≡vˆiμi. Note that ∂ni∂μi=ui>0, and ∂ui∂μi<0. Now the FOCs of eqs (6) and (7) can be written as:

Taking derivatives w.r.t. αB, we get

where k=12tλq+γ1−2q>0. The total combined relevant ads are given by αAqμA∗mA∗+αBqμB∗mB∗. Taking derivatives w.r.t. αB and using the above equations, we get

Since ∂μA∗∂αB<0, it is enough to show that the term in the braces is negative, which is equivalent to

Using the facts that, αA<αB,xA>xB,mA∗<mB∗, and ∂ui∂μi<0, the following condition is sufficient

Suppose v is uniformly distributed on [0,1], then ui∗=1−2μi∗, and ∂ui∗∂μi=−2. The inequality eq. (18) becomes

which obviously holds.

Now suppose v has an exponential distribution with density f(v)=ξe−ξx,ξ>0. In this case, ui∗=1ξ(−lnμi∗−1) and ∂ui∗∂μi=−1ξμi∗. The inequality eq. (18) becomes

The above inequality holds because −(μB∗lnμB∗)\gt−(μA∗lnμA∗) by the facts that −μi∗lnμi∗=ξni∗,∂ni∗∂μi>0, and μB∗>μA∗.

Proof of Proposition 4

proof. Part (i). Following Proposition 3, we have μA∗′<μA∗. Since αA, and hence yA, does not change, it implies that yAμA∗′<yAμA∗. The fact that mA∗′<mA∗ means that yB′μB∗′−yAμA∗′<yBμB∗−yAμA∗. The above two inequalities imply that yB′μB∗′−yBμB∗<yAμA∗′−yAμA∗<0.

Part (ii). Firms with v∈(vˆB∗′,vˆB∗) are clearly better off since they were not participating on any platform initially. Firms with v∈[vˆB∗,vˆA∗) are also better off. To see this, note that before and after the change they only participate on platform B. A type v firm’s profit on platform B equals to αBqmB∗(v−vˆB∗). Since αB′>αB,mB∗′>mB∗,vˆB∗′<vˆB∗, we have αB′qmB∗′(v−vˆB∗′)\gtαBqmB∗(v−vˆB∗).

For firms participating on both platforms after the change (v≥vˆA∗′), a type v firm’s total profit on two platforms is given by αAqmA∗(v−vˆA∗)+αBqmB∗(v−vˆB∗). Taking derivatives w.r.t. αB, and using similar logic as in the proof of part (iii), in the case of uniform distribution we get

which holds, since v≥vˆA∗′≥1/2.

In the case of exponential distribution,

The term αAmA∗xBf(vˆA∗)−αBmB∗xAf(vˆB∗)∝αAnA∗f(vˆA∗)uA∗−αBnB∗f(vˆB∗)uB∗=αA1−F(vˆA∗)f(vˆA∗)W(vˆA∗)−αB1−F(vˆB∗)f(vˆB∗)W(vˆB∗)<0, since the hazard rate is constant and W(⋅) is decreasing. Now the original inequality boils down to