Dynamic Pricing with Stochastic Reference Price Effect

Abstract

We study a dynamic pricing problem of a firm facing stochastic reference price effect. Randomness is incorporated in the formation of reference prices to capture either consumers’ heterogeneity or exogenous factors that affect consumers’ memory processes. We apply the stochastic optimal control theory to the problem and derive an explicit expression for the optimal pricing strategy. The explicit expression allows us to obtain the distribution of the steady-state reference price. We compare the expected steady-state reference price to the steady-state reference price in a model with deterministic reference price effect, and we find that the former one is always higher. Our numerical study shows that the two steady-state reference prices can have opposite sensitivity to the problem parameters and the relative difference between the two can be very significant.

Appendix

1.1 Proof of Proposition 3.1

To solve the HJB equation (3.1), we start from solving a more general finite horizon problem. That is, let

$$\begin{aligned} V(r,t) = \max _{p(s)}\left[ \int _{t}^\mathrm{T} \mathrm{e}^{-\gamma t}F(r(s),p(s))\mathrm{d}s\right] \end{aligned}$$

be the value of optimal accumulated profit (profit-to-go function) from time t to the end of horizon T when the initial reference price is r. Notice that the value function in our problem (2.6) $V(r_0) = \lim _{T\rightarrow \infty }V(r_0,0)$.

From standard theory in stochastic optimal control, V(r, t) then satisfies the HJB equation

$$\begin{aligned} \gamma V(r,t)= \max _p \left[ F(r,p) + \frac{\partial V(r,t)}{\partial t} + \alpha (p-r)\frac{\partial V(r,t)}{\partial r} + \frac{\sigma ^2r}{2}\frac{\partial ^2 V(r,t)}{\partial r^2}\right] .\qquad \end{aligned}$$

(6.1)

Using first-order condition in (6.1) and with a slight abuse of notation, we can solve p as

$$\begin{aligned} p^*(r) = \frac{c}{2} + \frac{b + \eta r}{2(a + \eta )} + \frac{\alpha }{2(a+\eta )}\times \frac{\partial V(r,t)}{\partial r}. \end{aligned}$$

Substitute the above equation into (6.1), it follows

$$\begin{aligned}&\frac{\sigma ^2}{2}r\frac{\partial ^2 V}{\partial r^2} + \frac{\partial V}{\partial t} -\gamma V + \frac{\alpha ^2}{4(a+\eta )} \left( \frac{\partial V}{\partial r}\right) ^2 +\left[ \frac{\alpha c}{2} -\alpha r + \frac{\alpha (b+\eta r)}{2(a+\eta )}\right] \frac{\partial V}{\partial r}\\&\quad - \frac{c(b+\eta r)}{2}+ \frac{(b+\eta r)^2}{4(a+\eta )} + \frac{c^2(a+\eta )}{4} = 0. \end{aligned}$$

Introducing a few new notations, this can be written concisely as:

$$\begin{aligned} \frac{\partial V}{\partial t}-\gamma V+Ar\frac{\partial ^2 V}{\partial r^2} +B\left( \frac{\partial V}{\partial r}\right) ^2 + (p_{10} + p_{11}r)\frac{\partial V}{\partial r} + p_{20} + p_{21}r + p_{22}r^2 = 0, \end{aligned}$$

(6.2)

where

$$\begin{aligned}&\displaystyle A = \frac{\sigma ^2}{2},\\&\displaystyle B = \frac{\alpha ^2}{4(a+\eta )},\\&\displaystyle p_{10} = \frac{\alpha c}{2} + \frac{\alpha b}{2(a+\eta )},\\&\displaystyle p_{11} = -\alpha + \frac{\alpha \eta }{2(a+\eta )},\\&\displaystyle p_{20} = -\frac{bc}{2} + \frac{b^2}{4(a+\eta )} + \frac{c^2(a+\eta )}{4},\\&\displaystyle p_{21} = -\frac{c\eta }{2}+\frac{b\eta }{2(a+\eta )},\\&\displaystyle p_{22} = \frac{\eta ^2}{4(a+\eta )}. \end{aligned}$$

If we assume function V(r, t) has the following form:

$$\begin{aligned} V(r,t) = Q(t)r^2+R(t)r+M(t), \end{aligned}$$

(6.3)

then we get the following ordinary differential equations (ODEs):

$$\begin{aligned}&\displaystyle \frac{\mathrm{d}Q}{\mathrm{d}t}-\gamma Q + 4B Q^2 + 2p_{11}Q + p_{22} = 0, \end{aligned}$$

(6.4)

$$\begin{aligned}&\displaystyle \frac{\mathrm{d}R}{\mathrm{d}t}-\gamma R + 2AQ + 4BQR + 2p_{10}Q + p_{11}R + p_{21} = 0, \end{aligned}$$

(6.5)

$$\begin{aligned}&\displaystyle \frac{\mathrm{d}M}{\mathrm{d}t} -\gamma M + BR^2 +p_{10}R + p_{20} = 0, \end{aligned}$$

(6.6)

with terminal condition $Q(T) = R(T) = M(T) = 0$.

We first explicitly solve ODE (6.4) by rewriting it as:

$$\begin{aligned} \frac{\mathrm{d}Q}{\mathrm{d}t} = -4B(Q-Q_1)(Q-Q_2), \end{aligned}$$

where $Q_1<Q_2$ are the two distinct roots of the equation:

$$\begin{aligned} 4BQ^2-(\gamma -2p_{11})Q+p_{22}=0. \end{aligned}$$

Namely:

$$\begin{aligned} Q_1= & {} \frac{\gamma -2p_{11}-\sqrt{(\gamma -2p_{11})^2-16 Bp_{22}}}{8B},\\ Q_2= & {} \frac{\gamma -2p_{11}+\sqrt{(\gamma -2p_{11})^2-16 Bp_{22}}}{8B}. \end{aligned}$$

Therefore,

$$\begin{aligned}&\frac{\mathrm{d}Q}{(Q-Q_1)(Q-Q_2)}=-4B\mathrm{d}t \nonumber \\&\quad \Rightarrow \quad \frac{dQ}{Q_1-Q_2}\left[ \frac{1}{Q-Q_1} -\frac{1}{Q-Q_2}\right] = -4B \mathrm{d}t\nonumber \\&\quad \Rightarrow \quad \ln \frac{Q-Q_1}{Q-Q_2}=-4B(Q_1-Q_2)t + C\nonumber \\&\quad \Rightarrow \quad \frac{Q-Q_1}{Q-Q_2} = D\cdot \mathrm{e}^{-4B(Q_1-Q_2)t}, \end{aligned}$$

(6.7)

where C and $D = e^C$ are constants to be determined. By $Q(T)=0$, we can solve

$$\begin{aligned} D=\frac{Q_1}{Q_2}\mathrm{e}^{4B(Q_1-Q_2)T}. \end{aligned}$$

Substitute D back into (6.7), it follows

$$\begin{aligned} Q(t) = \frac{Q_1\mathrm{e}^{4B(Q_1-Q_2)T}-Q_1\mathrm{e}^{4B(Q_1-Q_2)t}}{Q_1/Q_2 \mathrm{e}^{4B(Q_1-Q_2)T}-\mathrm{e}^{4B(Q_1-Q_2)t}}. \end{aligned}$$

(6.8)

With the expressions for Q(t), expressions for R(t) and M(t) can then be obtained by solving () and (6.6). Consequently, $p^*(r)$ can be determined as well. One can easily verify using Theorem 4.1 in chapter VI of [30] that $p^*(r)$ solved in this way is indeed optimal and V(r, t) is given by (6.3).

The solution to (3.1) is then obtained by letting $T\rightarrow \infty$. Since $Q_1<Q_2$, we have $Q:= Q_1 = \lim _{T\rightarrow \infty }Q(t)$, where by substituting the expressions for $B, p_{11}$ and $p_{22}$

$$\begin{aligned} Q=\frac{\gamma }{2\alpha ^2}(a+\eta )+\frac{2a+\eta }{2\alpha } -\frac{a+\eta }{2\alpha ^2}\Delta \end{aligned}$$

and $\Delta$ is given by:

$$\begin{aligned} \Delta =\sqrt{\gamma ^2+2\alpha \frac{2a(\gamma +\alpha )+\gamma \eta }{\eta +a}}. \end{aligned}$$

Correspondingly, one can also obtain $R:=\lim _{T\rightarrow \infty }R(t)$ as

$$\begin{aligned} R= & {} \frac{2p_{10}Q + p_{21} + 2AQ}{\gamma -4BQ-p_{11}}\\= & {} \left[ \frac{b}{\alpha }+\frac{\sigma ^2(a+\eta )}{\alpha ^2} + \frac{c(a+\eta )}{\alpha }\right] \frac{\gamma -\Delta }{\gamma +\Delta } +\left[ b + ca +\frac{\sigma ^2(2a+\eta )}{2\alpha } + \right] \frac{2}{\gamma +\Delta }. \end{aligned}$$

Similarly, $M:=\lim _{T\rightarrow \infty }M(t)$ can be computed.

Now, we have explicitly solved (3.1), where $V(r) = Qr^2 + Rr + M$ and the optimal pricing strategy can then be obtained as (3.2).

Finally, we remark that both $Q_1$ and $Q_2$ are positive solutions to the Algebraic Riccati Equation (ARE): $4BQ^2-(\Gamma -2p_{11})Q+p_{22}=0$, which has no negative solution. This deviates significantly from standard linear quadratic control theory and is the primal reason we need to solve from the finite horizon problem instead of solving ARE directly.

1.2 Proof of Proposition 3.2

Note that $r^*(t)$ follows (3.3), which is a square-root diffusion process. It is not difficult to show that $\lambda , \mu >0$. For such square-root diffusion process, it is known that $r^*(t)$ converges in distribution to a steady state which follows a Gamma distribution with shape parameter $\frac{2\lambda \mu }{\sigma ^2}$ and rate parameter $\frac{2\lambda }{\sigma ^2}$ (see, for instance, [15]).

1.3 Proof of Corollary 3.3

By Proposition 3.2, $R_s^*$ follows a Gamma distribution and its mean and variance can then be computed as

$$\begin{aligned}&\displaystyle {{E}}[R_s^*] = \frac{2\lambda \mu }{\sigma ^2}\frac{\sigma ^2}{2\lambda } = \mu ,\\&\displaystyle \text {var}(R_s^*) = \frac{2\lambda \mu }{\sigma ^2}\left( \frac{\sigma ^2}{2\lambda }\right) ^2 = \frac{\mu }{2\lambda }\sigma ^2. \end{aligned}$$

Substituting the expressions for Q, R and $\lambda$ into $\mu$, with cumbersome algebraic manipulations, one can further obtain

$$\begin{aligned} \mu = \frac{(\gamma +\bar{\alpha })b}{2a(\gamma +\bar{\alpha })+\gamma \eta } + \frac{\sigma ^2}{2a(\gamma +\bar{\alpha })+\gamma \eta } \left[ \frac{a+\eta }{\bar{\alpha }}\left( \frac{\gamma }{2} -\frac{\Delta }{2}\right) +\frac{2a+\eta }{2}\right] . \end{aligned}$$