Better with buy now, pay later?: A competitive analysis

Abstract

In this paper, we study the incentives of vertically differentiated firms to offer Buy Now, Pay Later (BNPL) in a competitive market. BNPL is a relatively new payment mechanism which, at the point of sale, allows consumers to pay for a product in interest-free installments spread out over a few weeks/months. For a monopolist, offering BNPL is essentially about expanding the market by offering financing to the consumers who cannot afford its product. Therefore, a monopolist is always better-off providing BNPL to its consumers. However, in a competitive environment, offering BNPL is a more complex strategic decision because retailers also need to consider strategic reactions from their competitors. We find that in a competitive situation either of the two retailers might refrain from offering BNPL. This is because when one retailer offers BNPL, the other firm not offering BNPL also benefits from competitive spillovers. Although a monopolist’s benefits from offering BNPL increases in its product quality, in a competitive environment, holding all else constant, a low-quality firm might have more to gain from offering BNPL. In addition to asymmetric equilibria, we also find that there is a symmetric equilibrium in which both retailers offer BNPL. In view of public concerns about possible negative impact of BNPL on consumers, we also study how BNPL consumers’ ignoring the cost of using BNPL can adversely affect them. We find that underestimation of these costs lowers consumers’ welfare, and this reduction in welfare stems from three different sources - (i) higher product prices, (ii) excessive purchase, and (ii) excessive upgrades to the higher quality product.

Appendix

Note: The equilibrium prices and profits in different cases are derived as usual. Details are available from the authors upon request.

1.1 Proof of Proposition 1

The YN case is an equilibrium if and only if the Condition YN below is satisfied.

Condition YN:

$$\begin{aligned} \frac{\rho _{1}(4q_{H}-q_{L})^{2}(\rho _{1}+\rho _{2}(1+\delta _{c})(1-\delta _{F}))}{(\rho _{1}(4q_{H}-q_{L})+4\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F}))^{2}}>1 \end{aligned}$$

(54)

We first show that $\pi _{H}^{YN*}>\pi _{H}^{NN*}$ and then show that $\pi _{L}^{YN*}>\pi _{L}^{YY*}$ when the YN condition is satisfied. After substituting $\delta _{H}=\delta _{F}$ and $\delta _{L}=\delta _{F}$, the profit expressions are as follows.

$$\begin{aligned} \pi _{H}^{YN*}= & {} \frac{4q_{H}(q_{H}\!-\!q_{L})(\rho _{1}\!+\!\rho _{2}(1\!-\!\delta _{F}))^{2}(\rho _{1}q_{H}\!+\!\rho _{2}(q_{H}\!-\!q_{L})(1\!+\!\delta _{c})(1\!-\!\delta _{F})q_{L})}{(\rho _{1}(4q_{H}\!-\!q_{L})\!+\!4\rho _{2}(q_{H}\!-\!q_{L})(1\!+\!\delta _{c})(1\!-\!\delta _{F}))^{2}}\\ \pi _{L}^{YN*}= & {} \frac{\rho _{1}q_{H}q_{L}(q_{H}-q_{L})(\rho _{1}+\rho _{2}(1-\delta _{F})){}^{2}}{(\rho _{1}(4q_{H}-q_{L})+4\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F}))^{2}}\\ \pi _{H}^{NN*}= & {} \frac{4q_{H}^{2}(q_{H}-q_{L})\rho _{1}}{(4q_{H}-q_{L})^{2}}\\ \pi _{L}^{YY*}= & {} \frac{q_{H}q_{L}(q_{H}-q_{L})(\rho _{1}+(1-\delta _{F})\rho _{2})^{2}}{(4q_{H}-q_{L}){}^{2}(\rho _{1}+(1+\delta _{c})(1-\delta _{F})\rho _{2})} \end{aligned}$$

$\frac{\pi _{H}^{YN*}}{\pi _{H}^{YN*}}=\frac{(4q_{H}-q_{L})^{2}(\rho _{1}+\rho _{2}(1-\delta _{F}))^{2}(\rho _{1}q_{H}+\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F})q_{L})}{\rho _{1}q_{H}(\rho _{1}(4q_{H}-q_{L})+4\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F}))^{2}}$. This can be rewritten as

$$ \frac{\pi _{H}^{YN*}}{\pi _{H}^{YN*}}\!=\!\left( \!\frac{\rho _{1}q_{H}\!+\!\rho _{2}(q_{H}\!-\!q_{L})(1\!+\!\delta _{c})(1\!-\!\delta _{F})q_{L}}{\rho _{1}q_{H}}\!\right) \left( \!\frac{(4q_{H}\!-\!q_{L})^{2}(\rho _{1}\!+\!\rho _{2}(1\!-\!\delta _{F}))^{2}}{(\rho _{1}(4q_{H}\!-\!q_{L})\!+\!4\rho _{2}(q_{H}\!-\!q_{L})(1\!+\!\delta _{c})(1\!-\!\delta _{F}))^{2}}\!\right) $$

It is easy to see that $\frac{\rho _{1}q_{H}+\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F})q_{L}}{\rho _{1}q_{H}}>1$. For $\theta _{H}^{YN1}\in [0,1]$,

$$ \frac{(4q_{H}-q_{L})(\rho _{1}+\rho _{2}(1-\delta _{F}))}{(\rho _{1}(4q_{H}-q_{L})+4\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F}))}>1, $$

which also means that $\frac{(4q_{H}-q_{L})^{2}(\rho _{1}+\rho _{2}(1-\delta _{F}))^{2}}{(\rho _{1}(4q_{H}-q_{L})+4\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F}))^{2}}>1$. $\square$

After simplification, $\frac{\pi _{L}^{YN*}}{\pi _{L}^{YY*}}=\frac{(4q_{H}-q_{L}){}^{2}\rho _{1}(\rho _{1}+(1+\delta _{c})(1-\delta _{F})\rho _{2})}{(\rho _{1}(4q_{H}-q_{L})+4\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F}))^{2}}$. Therefore,

$$ \frac{(4q_{H}-q_{L}){}^{2}\rho _{1}(\rho _{1}+(1+\delta _{c})(1-\delta _{F})\rho _{2})}{(\rho _{1}(4q_{H}-q_{L})+4\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F}))^{2}}\gtrless 1\iff \frac{\pi _{L}^{YN*}}{\pi _{L}^{YY*}}\gtrless 1. $$

$\square$

1.1.1 Proof of Proposition 2

The NY case is an equilibrium iff the condition NY is satisfied.

Condition NY:

$$\begin{aligned} \frac{(4q_{H}\!-\!q_{L})^{2}\rho _{1}(\rho _{1}\!+\!\rho _{2}(1\!+\!\delta _{c})(1\!-\!\delta _{F}))(\rho _{2}(1\!-\!\delta _{F})(2q_{H}(1\!+\!\delta _{c})\!-\!q_{L}(1\!+\!2\delta _{c}))\!+\!2\rho _{1}q_{H})^{2}}{4q_{H}^{2}(\rho _{1}\!+\!\rho _{2}(1\!-\!\delta _{F}))^{2}(\rho _{1}(4q_{H}\!-\!q_{L})\!+\!4\rho _{2}(1\!+\!\delta _{c})(1\!-\!\delta _{F})(q_{H}\!-\!q_{L}))^{2}}\!>\!1 \end{aligned}$$

(55)

Here again, we first show that $\pi _{L}^{NY*}>\pi _{L}^{NN*}$ and then show that the NY condition is equivalent to $\pi _{H}^{NY*}>\pi _{H}^{YY*}$.

After substituting $\delta _{H}=\delta _{F}$ and $\delta _{L}=\delta _{F}$, the profit expressions are as follows.

$$\begin{aligned} \pi {}_{H}^{NY*}= & {} \frac{\rho _{1}(q_{H}-q_{L})(2\rho _{1}q_{H}+\rho _{2}(1-\delta _{F})(2q_{H}(1+\delta _{c})-q_{L}(1+2\delta _{c})))^{2}}{(\rho _{1}(4q_{H}-q_{L})+4\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F})){}^{2}}\\ \pi _{H}^{YY*}= & {} \frac{4q_{H}^{2}(q_{H}-q_{L})(\rho _{1}+(1-\delta _{F})\rho _{2})^{2}}{(4q_{H}-q_{L})^{2}(\rho _{1}+(1+\delta _{c})(1-\delta _{F})\rho _{2})^{2}}\\ \pi {}_{L}^{NY*}= & {} \frac{q_{L}(q_{H}-q_{L})(\rho _{1}+2\rho _{2}(1-\delta _{F}))^{2}(\rho _{1}q_{H}+(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F})\rho _{2})}{(\rho _{1}(4q_{H}-q_{L})+4\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F})){}^{2}}\\ \pi _{L}^{NN*}= & {} \frac{\rho _{1}q_{H}q_{L}(q_{H}-q_{L})}{(4q_{H}-q_{L}){}^{2}} \end{aligned}$$

$\dfrac{\pi _{L}^{NY*}}{\pi {}_{L}^{NN*}}\!=\!\dfrac{(4q_{H}-q_{L}){}^{2}(\rho _{1}\!+\!2\rho _{2}(1-\delta _{F}))^{2}(\rho _{1}q_{H}+(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F})\rho _{2})}{\rho _{1}q_{H}(\rho _{1}(4q_{H}-q_{L})+4\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F})){}^{2}}$, which can be rewritten as

$$ \frac{\pi _{L}^{NY*}}{\pi _{L}^{NN*}}\!=\!\!\left( \!\frac{\rho _{1}q_{H}\!+\!\rho _{2}(q_{H}\!-\!q_{L})(1\!+\!\delta _{c})(1\!-\!\delta _{F})q_{L}}{\rho _{1}q_{H}}\!\right) \!\left( \!\frac{(4q_{H}\!-\!q_{L})^{2}(\rho _{1}\!+\!2\rho _{2}(1\!-\!\delta _{F}))^{2}}{(\rho _{1}(4q_{H}\!-\!q_{L})\!+\!4\rho _{2}(q_{H}\!-\!q_{L})(1\!+\!\delta _{c})(1\!-\!\delta _{F}))^{2}}\!\right) \!. $$

It is easy to see that $\dfrac{\rho _{1}q_{H}+\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F})q_{L}}{\rho _{1}q_{H}}>1.$ After simplification and rearranging terms,

$$ \frac{(4q_{H}\!-\!q_{L})^{2}(\rho _{1}\!+\!2\rho _{2}(1\!-\!\delta _{F}))^{2}}{(\rho _{1}(4q_{H}\!-\!q_{L})\!+\!4\rho _{2}(q_{H}\!-\!q_{L})(1\!+\!\delta _{c})(1\!-\!\delta _{F}))^{2}}\!-\!1\!=\!\frac{4(2q_{H}(1\!-\!\delta _{c})\!+\!q_{L}(1\!+\!2\delta _{c}))\rho _{2}(1\!-\!\delta _{F})\eta }{(\rho _{1}(4q_{H}\!-\!q_{L})\!+\!4\rho _{2}(q_{H}\!-\!q_{L})(1\!+\!\delta _{c})(1\!-\!\delta _{F}))^{2}} $$

where $\eta =2q_{H}(2\rho _{1}+\rho _{2}(3+\delta _{c})(1-\delta _{F}))-q_{L}(\rho _{1}+\rho _{2}(3+2\delta _{c})(1-\delta _{F}))>0$. Therefore,

$$ \frac{(4q_{H}-q_{L})^{2}(\rho _{1}+2\rho _{2}(1-\delta _{F}))^{2}}{(\rho _{1}(4q_{H}-q_{L})+4\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F}))^{2}}>1 $$

and $\frac{\pi _{L}^{NY*}}{\pi _{L}^{NN*}}>1.$ $\square$

$\frac{\pi {}_{H}^{NY*}}{\pi {}_{H}^{YY*}}=\frac{(4q_{H}-q_{L})^{2}\rho _{1}(\rho _{1}+\rho _{2}(1+\delta _{c})(1-\delta _{F}))(2\rho _{1}q_{H}+\rho _{2}(1-\delta _{F})(2q_{H}(1+\delta _{c})-q_{L}(1+2\delta _{c})))^{2}}{4q_{H}^{2}(\rho _{1}+\rho _{2}(1-\delta _{F}))^{2}(\rho _{1}(4q_{H}-q_{L})+4\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F})){}^{2}}$.

Therefore,

$$\begin{aligned} \frac{(4q_{H}-q_{L})^{2}\rho _{1}(\rho _{1}+\rho _{2}(1+\delta _{c})(1-\delta _{F}))(2\rho _{1}q_{H}+\rho _{2}(1-\delta _{F})(2q_{H}(1+\delta _{c})-q_{L}(1+2\delta _{c})))^{2}}{4q_{H}^{2}(\rho _{1}+\rho _{2}(1-\delta _{F}))^{2}(\rho _{1}(4q_{H}-q_{L})+4\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F})){}^{2}}&\gtrless 1\\ \iff \frac{\pi {}_{H}^{NY*}}{\pi {}_{H}^{YY*}}&\gtrless 1. \end{aligned}$$

1.1.2 Proof of Proposition 3

A YY equilibrium can occur iff either Condition YN or Condition NY is violated.

As proofs of Propositions 1 and 2 show, for each firm, offering BNPL is the dominant strategy when the other firm does not offer BNPL. When Condition YN is violated, $\pi _{L}^{YN*}<\pi _{L}^{YY*}$ and Firm L will deviate to offering BNPL when Firm H offers BNPL. Along similar lines, when Condition NY is violated, $\pi _{H}^{NY*}<\pi _{H}^{YY*}$and Firm H will deviate to offering BNPL when Firm L offers BNPL. Thus, when either condition is violated, the YY case is an equilibrium.$\square$

1.2 Proof of Result 1

If $\delta _{c}>\frac{q_{L}}{2q_{H}-q_{L}}$, then only YN equilibrium cannot exist. If $\delta _{c}<\frac{q_{L}}{2q_{H}-q_{L}}$, then only NY equilibrium cannot exist.

We compare Conditions YN and NY to identify the conditions under which one of them is more stringent than the other. We define $\Lambda$ as

$$ \Lambda =\frac{2\rho _{1}q_{H}+\rho _{2}(1-\delta _{F})(2q_{H}(1+\delta _{c})-q_{L}(1+2\delta _{c}))}{2q_{H}(\rho _{1}+\rho _{2}(1-\delta _{F}))}. $$

$$ \Lambda -1=\frac{\rho _{2}(1-\delta _{F})(2q_{H}\delta _{c}-q_{L}(1+2\delta _{c}))}{2q_{H}(\rho _{1}+\rho _{2}(1-\delta _{F}))}. $$

When $\delta _{c}>(<)\frac{q_{L}}{2(q_{H}-q_{L})}$, the numerator of $\Lambda -1$ is positive (negative), and $\Lambda >(<)1$.

Taking the ratio of the LHS of Condition NY to the LHS of Condition YN, we get,

$$ \frac{(2\rho _{1}q_{H}+\rho _{2}(1-\delta _{F})(2q_{H}(1+\delta _{c})-q_{L}(1+2\delta _{c})))^{2}}{4q_{H}^{2}(\rho _{1}+\rho _{2}(1-\delta _{F}))^{2}}=\Lambda ^{2}. $$

Therefore, when $\delta _{c}>(<)\frac{q_{L}}{2q_{H}-q_{L}}$, Condition NY is less (more) stringent than Condition YN and it is possible that Condition NY (YN) is satisfied but Condition YN (NY) is not.$\square$

1.3 Proof of Proposition 4

Firm H prefers the YN equilibrium to the NY equilibrium: $\pi _{H}^{YN*}>\pi _{H}^{NY*}$, whereas Firm L prefers the NY equilibrium to the YN equilibrium: $\pi _{L}^{YN*}<\pi _{L}^{NY*}$.

$$ \pi _{L}^{NY*}-\pi _{L}^{YN*}\!=\!\!\frac{q_{L}(q_{H}\!-\!q_{L})((\rho _{1}\!+\!2\rho _{2}(1\!-\!\delta _{F}))^{2}(\rho _{1}q_{H}\!+\!(q_{H}\!-\!q_{L})(1\!+\!\delta _{c})(1\!-\!\delta _{F})\rho _{2})\!-\!\rho _{1}q_{H}(\rho _{1}\!+\!(1\!-\!\delta _{F})\rho _{2})^{2})}{(\rho _{1}(4q_{H}\!-\!q_{L})\!+\!4\rho _{2}(q_{H}\!-\!q_{L})(1\!+\!\delta _{c})(1\!-\!\delta _{F})){}^{2}\!} $$

The denominator of the above expression is positive. Therefore we need to show that $((\rho _{1}+2\rho _{2}(1-\delta _{F}))^{2}(\rho _{1}q_{H}+(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F})\rho _{2})-\rho _{1}q_{H}(\rho _{1}+(1-\delta _{F})\rho _{2})^{2})>0$.

$$ \begin{array}{l} ((\rho _{1}\!+\!2\rho _{2}(1\!-\!\delta _{F}))^{2}(\rho _{1}q_{H}\!+\!(q_{H}\!-\!q_{L})(1\!+\!\delta _{c})(1\!-\!\delta _{F})\rho _{2})\!-\!\rho _{1}q_{H}(\rho _{1}\!+\!(1\!-\!\delta _{F})\rho _{2})^{2})\\ =\!((\rho _{1}\!+\!2\rho _{2}(1\!-\!\delta _{F}))^{2}(\rho _{1}q_{H})\!-\!\rho _{1}q_{H}(\rho _{1}\!+\!(1\!-\!\delta _{F})\rho _{2})^{2})\!+\!((\rho _{1}\!+\!2\rho _{2}(1\!-\!\delta _{F}))^{2}(q_{H}\!-\!q_{L})(1\!+\!\delta _{c})(1\!-\!\delta _{F})\rho _{2})\\ =\!\rho _{1}\rho _{2}q_{H}(1\!-\!\delta _{F})(2\rho _{1}\!+\!3(1\!-\!\delta _{F})\rho _{2})\!+\!((\rho _{1}\!+\!2\rho _{2}(1\!-\!\delta _{F}))^{2}(q_{H}\!-\!q_{L})(1\!+\!\delta _{c})(1\!-\!\delta _{F})\rho _{2})>0 \end{array} $$

Thus, $\pi _{L}^{YN*}-\pi _{L}^{NY*}>0$. $\square$

$$ \pi _{H}^{YN*}-\pi _{L}^{NY*}=\frac{\rho _{2}(q_{H}-q_{L})(1-\delta _{F})(4\rho _{1}^{2}q_{H}^{2}+k_{1}+k_{2}+k_{3})}{(\rho _{1}(4q_{H}-q_{L})+4\rho _{2}(q_{H}-q_{L})(1+\delta _{c})(1-\delta _{F})){}^{2}},\textrm{where} $$

$$\begin{aligned} k_{1}= & {} \rho _{2}(1-\delta _{F})(8q_{H}^{2}\rho _{1}-4q_{H}q_{L}\rho _{1}-q_{L}^{2}\rho _{1}+4\rho _{2}q_{H}(q_{H}-q_{L})(1-\delta _{F})),\\ k_{2}= & {} 4(q_{H}-q_{L})\delta _{c}(q_{H}((1-\delta _{F})^{2}-\rho _{1}^{2})+\rho _{1}\rho _{2}q_{L}(1-\delta _{F})),\\ k_{3}= & {} -4\rho _{1}\rho _{2}(q_{H}-q_{L})^{2}\delta _{c}^{2}(1-\delta _{F}). \end{aligned}$$

Note that $k_{1}>0$ and $k_{3}<0$. Because $\delta _{c}\in (0,1)$, $k_{1}+k_{3}>k_{1}+\frac{k_{3}}{\delta _{c}^{2}}$. It can be shown that $k_{1}+\frac{k_{3}}{\delta _{c}^{2}}=\rho _{1}\rho _{2}(1-\delta _{F})(4q_{H}^{2}+4q_{H}q_{L}-5q_{L}^{2})+4\rho _{2}^{2}(1-\delta _{F})^{2}q_{H}(q_{H}-q_{L})>0$. Therefore, $k_{1}+k_{3}>0$. $k_{2}+4\rho _{1}^{2}q_{H}^{2}=4\rho _{1}^{2}q_{H}^{2}(1-\delta _{c})+4\delta _{c}(q_{H}q_{L}\rho _{1}^{2}+(q_{H}-q_{L})q_{L})(1-\delta _{F})\rho _{1}\rho _{2}+q_{H}(q_{H}-q_{L})(1-\delta _{F})^{2}\rho _{2}^{2})>0.$Therefore, $\pi _{H}^{YN*}-\pi _{L}^{NY*}>0$.$\square$

1.4 Proof of Proposition 5

For the parameter values for which both asymmetric equilibria can coexist with pure strategies, $\psi _{H}^{*}=\dfrac{(\pi _{L}^{NY*}-\pi _{L}^{NN*})}{(\pi _{L}^{NY*}-\pi _{L}^{NN*})+(\pi _{L}^{YN*}-\pi _{L}^{YY*})}$ and $\psi _{L}^{*}=$ $\dfrac{(\pi _{H}^{YN*}-\pi _{H}^{NN*})}{(\pi _{H}^{YN*}-\pi _{H}^{NN*})+(\pi _{H}^{NY*}-\pi _{H}^{YY*})}.$

When Firm H plays BNPL with probability $\psi _{H},$ Firm L’s profit from offering BNPL is

$$\begin{aligned} \psi _{H}\pi _{L}^{YY*}+(1-\psi _{H})\pi _{L}^{NY*} \end{aligned}$$

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Firm L’s profit from not offering BNPL is

$$\begin{aligned} \psi _{H}\pi _{L}^{YN*}+(1-\psi _{H})\pi _{L}^{NN*} \end{aligned}$$

(57)

Equating the above two profits and solving for $\psi _{H}$, we get the equilibrium value as,

$$ \psi _{H}^{*}\!=\!\frac{(\pi _{L}^{NY*}\!-\!\pi _{L}^{NN*})}{(\pi _{L}^{NY*}\!-\!\pi _{L}^{NN*})\!+\!(\pi _{L}^{YN*}\!-\!\pi _{L}^{YY*})};\;\psi _{L}^{*}\!=\!\frac{(\pi _{H}^{YN*}\!-\!\pi _{H}^{NN*})}{(\pi _{H}^{YN*}\!-\!\pi _{H}^{NN*})\!+\!(\pi _{H}^{NY*}\!-\!\pi _{H}^{YY*})} $$

We can write $\psi _{H}^{*}=\frac{1}{1+z_{L}}$ where $z_{L}=\frac{\pi _{L}^{YN*}-\pi _{L}^{YY*}}{\pi _{L}^{NY*}-\pi _{L}^{NN*}}$ and $\psi _{L}^{*}=\frac{1}{1+z_{H}}$ where $z_{H}=\frac{\pi _{H}^{NY*}-\pi _{H}^{YY*}}{\pi _{H}^{YN*}-\pi _{H}^{NN*}}$. $\square$