The Optimal Selling Strategy of Residential Real Estate

Abstract

This paper studies three selling strategies of residential real estate: delegation to a broker, cheap talk with a broker and For Sale By Owner (FSBO). We find that if the state of the market is very uncertain and information of the market is scarce, the seller should hire a broker and delegate the decision right to the broker, even if the broker has no informational advantage. In this case, delegation may induce the broker to over-invest in information acquisition, i.e., more than what the seller would do in FSBO. If the market is certain and information is easily available, FSBO is optimal. Cheap talk, on the other hand, is never optimal, as much information is lost in the cheap-talk communication, weakening the broker’s incentive to acquire information.

Appendix: Appendix: Proofs

Proof of Lemma 1

From Eqs. 11 and 12, we have:

$$\begin{array}{@{}rcl@{}} {\int}_{s_{i-1}}^{s_{i}} E[\tilde{\theta}|s, a]ds &=&{\int}_{s_{i-1}}^{s_{i}} \left[(\alpha a +\beta)(2s-1)+E[\tilde{\theta}]\right]ds \\ &=&\left[(\alpha a +\beta)(s_{i}+s_{i-1}-1)+E[\tilde{\theta}]\right](s_{i}-s_{i-1}) \\ &=&E\left[\tilde{\theta}| \frac{s_{i-1}+s_{i}}{2}, a\right]\left( s_{i}-s_{i-1}\right) \\ &=&y\left( [s_{i-1}, s_{i}], a\right)\left( s_{i}-s_{i-1}\right), \end{array} $$

(34)

where the last two equalities follows from Eq. 12. Then we have:

$$\begin{array}{@{}rcl@{}} &&{{\int}_{0}^{1}}{\int}_{\underline{\theta}}^{\overline{\theta}} (y(s)-\theta)^{2}f(\theta|s,a)d\theta ds \\ &=&\sum\limits_{i = 1}^{n}{\int}_{s_{i-1}}^{s_{i}}{\int}_{\underline{\theta}}^{\overline{\theta}} \left( y\left( [s_{i-1}, s_{i}], a\right)-\theta\right)^{2}f(\theta|s, a)d\theta ds \\ &=&\sum\limits_{i = 1}^{n}{\int}_{s_{i-1}}^{s_{i}} \left( y\left( [s_{i-1}, s_{i}], a\right)^{2} - 2y\left( [s_{i-1}, s_{i}], a\right)E[\tilde{\theta}|s, a] + E[\tilde{\theta}^{2}|s, a] \right)ds \\ &=&\sum\limits_{i = 1}^{n} y\left( [s_{i-1}, s_{i}], a\right)^{2} (s_{i}-s_{i-1})-2{\sum}_{i = 1}^{n} y\left( [s_{i-1}, s_{i}], a\right) {\int}_{s_{i-1}}^{s_{i}} E[\tilde{\theta}|s, a]ds \\ && + \sum\limits_{i = 1}^{n}{\int}_{s_{i-1}}^{s_{i}}E[\tilde{\theta}^{2}|s, a] ds \\ &=& -\sum\limits_{i = 1}^{n} y\left( [s_{i-1}, s_{i}], a\right)^{2} (s_{i}-s_{i-1}) + {{\int}_{0}^{1}} E[\tilde{\theta}^{2}|s, a] ds \\ &=& -\sum\limits_{i = 1}^{n} \left( (\alpha a + \beta)(s_{i-1}+s_{i}-1)+E[\tilde{\theta}]\right)^{2}(s_{i}-s_{i-1}) + E[\tilde{\theta}^{2}] \\ &=& -(\alpha a + \beta)^{2}\sum\limits_{i = 1}^{n}(s_{i}+s_{i-1}-1)^{2} (s_{i}-s_{i-1}) \\ && - 2E[\tilde{\theta}](\alpha a+\beta)\sum\limits_{i = 1}^{n}(s_{i}-s_{i-1})(s_{i}+s_{i-1}-1) - E^{2}[\tilde{\theta}] + E[\tilde{\theta}^{2}] , \end{array} $$

(35)

where the fourth equality follows from Eq. 34, and the fifth equality follows from Eq. 12. According to Eq. 15, we have:

$$\begin{array}{@{}rcl@{}} s_{i}-s_{i-1}&=&\frac{1}{n}+\frac{b}{\alpha a +\beta}(2i-1-n). \end{array} $$

(36)

$$\begin{array}{@{}rcl@{}} s_{i}+s_{i-1}&=&\frac{2i-1}{n}+\frac{b}{\alpha a + \beta}\left[(2i-1)(i-n)-(i-1)\right]. \end{array} $$

(37)

From Eqs. 36 and 37, it is useful to compute the following:

$$\begin{array}{@{}rcl@{}} \sum\limits_{i = 1}^{n}(s_{i}-s_{i-1})&=&1, \end{array} $$

(38)

$$\begin{array}{@{}rcl@{}} \sum\limits_{i = 1}^{n}({s_{i}^{2}}-s_{i-1}^{2})&=&1, \end{array} $$

(39)

$$\begin{array}{@{}rcl@{}} \sum\limits_{i = 1}^{n}(s_{i}-s_{i-1})(s_{i}+s_{i-1})^{2}&=&\frac{1}{3}\left[\frac{b^{2}(1-n^{2})}{(\alpha a + \beta)^{2}}+ 4-\frac{1}{n^{2}}\right], \end{array} $$

(40)

$$\begin{array}{@{}rcl@{}} \sum\limits_{i = 1}^{n}(s_{i}-s_{i-1})(s_{i}+s_{i-1}-1)^{2} &=&\sum\limits_{i = 1}^{n}(s_{i}-s_{i-1})(s_{i}+s_{i-1})^{2} \\ &&\quad -2\sum\limits_{i = 1}^{n}({s_{i}^{2}}-s_{i-1}^{2})+\sum\limits_{i = 1}^{n}(s_{i}-s_{i-1})\\ &=&\frac{1}{3}\left[\frac{b^{2}(1-n^{2})}{(\alpha a + \beta)^{2}}+ 1-\frac{1}{n^{2}}\right], \end{array} $$

(41)

$$\begin{array}{@{}rcl@{}} \sum\limits_{i = 1}^{n}(s_{i}-s_{i-1})(s_{i}+s_{i-1}-1) &=&\sum\limits_{i = 1}^{n}({s_{i}^{2}}-s_{i-1}^{2})-\sum\limits_{i = 1}^{n}(s_{i}-s_{i-1})= 0. \end{array} $$

(42)

Substituting (41) and (42) into (35), one gets:

$$ {{\int}_{0}^{1}}{\int}_{\underline{\theta}}^{\overline{\theta}} (y(s)-\theta)^{2}f(\theta|s,a)d\theta ds=V(\tilde{\theta})-\frac{b^{2}(1-n^{2})}{3}-\frac{(\alpha a + \beta)^{2}}{3}\left( 1-\frac{1}{n^{2}}\right). $$

(43)

From Eq. 43, we have:

$$\begin{array}{@{}rcl@{}} U_{S}^{CT}(a)&=&{{\int}_{0}^{1}}{\int}_{\underline{\theta}}^{\overline{\theta}} u_{S}(y(s), \theta) f(\theta|s,a)d\theta ds \\ &=&K_{S}-M_{S}{{\int}_{0}^{1}}{\int}_{\underline{\theta}}^{\overline{\theta}} (y(s)-\theta)^{2}f(\theta|s,a)d\theta ds \\ &=&K_{S}-M_{S}\left[V(\tilde{\theta})-\frac{b^{2}(1-n^{2})}{3}-\frac{(\alpha a + \beta)^{2}}{3}\left( 1-\frac{1}{n^{2}}\right)\right]. \end{array} $$

(44)

In addition, we have:

$$ {{\int}_{0}^{1}}{\int}_{\underline{\theta}}^{\overline{\theta}} (y(s)-\theta)f(\theta|s,a)d\theta ds=E[y(\tilde{s})]-E[\tilde{\theta}] = 0. $$

(45)

Therefore, the broker’s welfare is:

$$\begin{array}{@{}rcl@{}} U_{B}^{CT}(a) &=&K_{B}-M_{B}\left[\iint (y(s)-\theta- b)^{2}f(\theta|s,a)d\theta ds\right] -c(a) \\ &=& K_{B}-M_{B}\left[\iint (y(s)-\theta)^{2}f(\theta|s,a)d\theta ds\right.\\&&\qquad\qquad\left.+b^{2}-2b\iint (y(s)-\theta)f(\theta|s,a)d\theta ds\right] -c(a) \\ &=&K_{B}-M_{B}\left[V(\tilde{\theta})-\frac{b^{2}(1-n^{2})}{3}-\frac{(\alpha a + \beta)^{2}}{3}\left( 1-\frac{1}{n^{2}}\right) +b^{2}\right] - a^{2}, \end{array} $$

where the third equality follows from Eqs. 43 and 44. This completes the proof. $\ \Box$

Proof of Lemma 2

From Eq. 17, it is obvious that $U_{B}^{CT}(a)$ is continuous in a on each of the intervals (a_i,a_i+ 1) for all i, because n(a) is constant on (a_i,a_i+ 1). Then we are left to prove that $U_{B}^{CT}(a)$ is continuous at a = a_i for all i. That is, we need to prove that $\lim _{a \rightarrow a_{i}^{-}}U_{B}^{CT}(a)=U_{B}^{CT}(a_{i})=\lim _{a\rightarrow a_{i}^{+}}U_{B}^{CT}(a)$, where $a\rightarrow a_{i}^{-}$ means a approaches a_i from below, and $a\rightarrow a_{i}^{+}$ means a approaches a_i from above. The first equality is obvious as $\lim _{a\rightarrow a_{i}^{-}}n(a)=n(a_{i})$. The second equality is less apparent because $\lim _{a\rightarrow a_{i}^{+}}n(a)=n(a_{i})+ 1=i + 1$. Nevertheless, according to Eq. 17 we have the following:

$$\begin{array}{@{}rcl@{}} \lim_{a \rightarrow a_{i}^{+}}U_{B}^{CT}(a)&=& K_{B}-M_{B}\left[V(\tilde{\theta})- \frac{b^{2}(1-(i + 1)^{2})}{3}-\frac{(\alpha a_{i} + \beta)^{2}}{3}\left( 1-\frac{1}{(i + 1)^{2}}\right) +b^{2} \right] - {a_{i}^{2}} \\ &=&K_{B}-M_{B}l[V(\tilde{\theta})-\frac{b^{2}(1-i^{2})}{3}-\frac{(\alpha a_{i} + \beta)^{2}}{3}\left( 1-\frac{1}{i^{2}}\right) +b^{2} \\ &&\quad + \frac{b^{2}(2i + 1)}{3} - \frac{(\alpha a_{i} + \beta)^{2}}{3}\left( \frac{1}{i^{2}}-\frac{1}{(i + 1)^{2}}\right) ] - {a_{i}^{2}} \\ &=&U_{B}^{CT}(a_{i})-M_{B}\frac{2i + 1}{3}\left[b^{2}-\frac{(\alpha a_{i} + \beta)^{2}}{i^{2}(1+i)^{2}}\right]. \end{array} $$

(46)

Note that

$$ i(1+i)=\left( \sqrt{\frac{1}{4}+\frac{\alpha a_{i} + \beta}{b}} -\frac{1}{2}\right)\left( \sqrt{\frac{1}{4}+\frac{\alpha a_{i} + \beta}{b}} +\frac{1}{2}\right)=\frac{\alpha a_{i} + \beta}{b}. $$

(47)

Substituting (47) into (46) gives $\lim _{a \rightarrow a_{i}^{+}}U_{B}^{CT}(a)=U_{B}^{CT}(a_{i})$. This completes the proof. $\ \Box$

Proof of Proposition 1

If β ≥ 2b, then according to Eq. 14 (taking strict equality), n(a) ≥ 2 for all a > 0. From Eq. 17, we have $\frac {\partial _{+} U_{B}^{CT}(a)}{\partial a}|_{a = 0}=M_{B}\frac {2\alpha \beta }{3}\left (1-\frac {1}{n(a)^{2}}\right )>0$. That is, a marginal increase of a at zero increases the broker’s expected utility, and therefore a_CT must be positive. Since n(a) ≥ 2, there are multiple equilibria.

On the other hand, if β < 2b, then according to Eq. 14, n(a) = 1 and $\frac {\partial U_{B}^{CT}(a)}{\partial a}=-2a<0$ for all $a\leq \frac {2b-\beta }{\alpha }$. Then to prove that a_CT = 0, it suffices to prove that $\frac {\partial U_{B}^{CT}(a)}{\partial a}\leq 0$, $\forall a\geq \frac {2b-\beta }{\alpha }$. We have:

$$\begin{array}{@{}rcl@{}} \frac{\partial U_{B}^{CT}(a)}{\partial a}&=&M_{B}\frac{2\alpha (\alpha a + \beta)}{3}\left( 1-\frac{1}{n(a)^{2}}\right)-2 a < M_{B}\frac{2\alpha (\alpha a+ \beta)}{3}-2 a \\ &=&\left( \frac{2\alpha^{2}M_{B}}{3}-2\right)a + \frac{2\alpha \beta M_{B}}{3}, \end{array} $$

(48)

where the inequality follows from the fact that n(a) ≥ 2 if $a\geq \frac {2b-\beta }{\alpha }$. Therefore it suffices to prove that

$$ \left( \frac{2\alpha^{2}M_{B}}{3}-2\right)a + \frac{2\alpha \beta M_{B}}{3}\leq0, \quad \forall a\geq\frac{2b-\beta}{\alpha}. $$

(49)

One can readily check that (49) holds if and only if $b\geq \frac {3\beta }{2(3-M_{B}\alpha ^{2})}$ and 3 − M_Bα² > 0.

Since a_CT = 0 and $b >\frac {\beta }{2}$, then n(a_CT) = 1 according to Eq. 14. Therefore, the only equilibrium is the “babbling” equilibrium where the seller completely ignores the broker’s advice. $\ \Box$

Proof of Proposition 2

If b ≤ β, then Case 2 is relevant for all values of a. Taking derivative of Eq. 21 in a at a = 0, we have:

$$\begin{array}{@{}rcl@{}} \frac{\partial {U_{B}^{D}}(a)}{\partial a}&=&M_{B}\left( \frac{2\alpha(\alpha a+\beta)}{3} +\frac{4\alpha b^{3}}{3(\alpha a+\beta)^{2}}\right)-2 a \\ &=&M_{B}\left( \frac{2\alpha \beta}{3}+\frac{4\alpha b^{3}}{3\beta^{2}}\right)>0,\quad \text{at a = 0}. \end{array} $$

(50)

That is, the broker benefits from a marginal increase of a from zero, and therefore a_D > 0.

On the other hand, if $b\geq \frac {\beta }{1-M_{B}\alpha ^{2}}>\beta$, then the model switches from Case 1 to Case 2 as a increases from 0 and exceeds $\frac {b-\beta }{\alpha }$. In Case 1, the broker’s expected utility ${U_{B}^{D}}(a)$ is characterized by Eq. 19, while in Case 2, ${U_{B}^{D}}(a)$ is characterized by Eq. 21. Figure 8 shows ${U_{B}^{D}}(a)$ as a function of a. When a is small, i.e., $a<\frac {b-\beta }{\alpha }$, ${U_{B}^{D}}(a)$ strictly decreases with a. Therefore to prove that ${U_{B}^{D}}(a)$ is maximized at a = 0, it suffices to prove that ${U_{B}^{D}}(a)$ in Case 2–as characterized in Eq. 21–is strictly concave in a and that $\frac {\partial {U_{B}^{D}}(a)}{\partial a}\leq 0$ at $a=\frac {b-\beta }{\alpha }$. Indeed, from Eq. 21 we have:

$$\begin{array}{@{}rcl@{}} \frac{\partial^{2} {U_{B}^{D}}(a)}{\partial a^{2}}& = &\frac{2\alpha^{2}M_{B}}{3}\left( 1-\frac{4b^{3}}{(\alpha a+\beta)^{3}}\right)-2 \leq 2 \left( \frac{\alpha^{2}M_{B}}{3}-1\right)\!\leq\!0. \end{array} $$

(51)

$$\begin{array}{@{}rcl@{}} \frac{\partial {U_{B}^{D}}(a)}{\partial a}|_{a=\frac{b-\beta}{\alpha}}&=&M_{B}\left( \frac{2\alpha(\alpha a +\beta)}{3} + \frac{4\alpha b^{3}}{3(\alpha a + \beta)^{2}}\right)-2a|_{a=\frac{b-\beta}{\alpha}} \\ &=&\frac{2(M_{B}\alpha^{2} b-b+\beta)}{\alpha}\leq0. \end{array} $$

(52)

The second inequality in Eq. 51 follows from the condition that α² < 3/M_B. The inequality in Eq. 52 follows from the condition that $b\geq \frac {\beta }{1-M_{B}\alpha ^{2}}$. Therefore, we have proven that a_D = 0 in the equilibrium. And as shown in Fig. 8, the equilibrium is Case 1. $\ \Box$

Proof of Proposition 4

Proposition 4 can be proved by a sequence of lemmas as follows. $\ \Box$

Fig. 8

Graph of ${U_{B}^{D}}(a)$ When $b\geq \frac {\beta }{1-M_{B}\alpha ^{2}}$

Lemma 4

Ifa_CT > 0,a_FSBO > a_CT.

Lemma 5

Ifa_CT > 0,a_D > a_CT.

Lemma 6

Ifa_FSBO > 0 anda_D > 0,thena_D < a_FSBOifand only if$\frac {3\beta }{3-2\alpha ^{2}}> b\left (\frac {2(1+r)}{1-r}\right )^{1/3}$.

Proof of Lemma 4

If a_CT > 0, then

$$\begin{array}{@{}rcl@{}} \frac{\partial U_{S}^{FSBO}(a)}{\partial a}|_{a=a_{CT}}&=&\frac{2\alpha M_{S}(\alpha a_{CT}+\beta)}{3}-2 a_{CT} \\ &&>\frac{2\alpha M_{S}(\alpha a_{CT}+\beta)}{3}\left( 1-\frac{1}{n(a_{CT})^{2}}\right)-2 a_{CT} \\ &&=\frac{\partial U_{B}^{CT}(a)}{\partial a}|_{a=a_{CT}}= 0, \end{array} $$

(53)

where the first equality follows from Eq. 27, the inequality follows from the fact that n(a_CT) ≥ 1, and the second equality follows from Eq. 18. Since $U_{S}^{FSBO}(a)$ is concave in a, Eq. 53 implies that a_FSBO > a_CT. $\ \Box$

Proof of Lemma 5

The proof takes two steps. In Step 1, we prove that if a_CT > 0 then a_D > 0. In Step 2, we prove that if a_CT > 0 then a_D > a_CT.

• Step 1. We approve that if a_D = 0 then a_CT = 0 by contradiction. Assume a_D = 0, then according to Proposition 2, b > β = αa_D + β. Then according to Eq. 19, ${U_{B}^{D}}(0)=K_{B}-M_{B}(V(\tilde {\theta })+b^{2})$. Assume for contradiction that a_CT > 0, then n(a_CT) ≥ 2, or equivalently αa_CT + β ≥ 2b according to Eq 14 (taking strict equality). Then we have:

  $$\begin{array}{@{}rcl@{}} U_{B}^{CT}(a_{CT})&=&K_{B}-M_{B}\left[V(\tilde{\theta})-\frac{(\alpha a_{CT} + \beta)^{2}}{3}\left( 1-\frac{1}{n(a_{CT})^{2}}\right)\right.\\&&\quad\qquad\qquad\left.- \frac{b^{2}(1-n(a_{CT})^{2})}{3}+b^{2}\right] - a_{CT}^{2} \\ &&<K_{B}-M_{B}\left[V(\tilde{\theta})-\frac{(\alpha a_{CT} + \beta)^{2}}{3} + 2b^{2}\right]- a_{CT}^{2} \\ &&<K_{B}-M_{B}\left[V(\tilde{\theta})-\frac{(\alpha a_{CT} + \beta)^{2}}{3} + \frac{4b^{3}}{3(\alpha a_{CT} +\beta)}\right]- a_{CT}^{2} \\ &&={U_{B}^{D}}(a_{CT}) \leq {U_{B}^{D}}(a_{D}) = {U_{B}^{D}}(0) =U_{B}^{CT}(0), \end{array} $$

  where the first equality is Eq. 17, the first inequality follows from the fact n(a_CT) ≥ 2 if a_CT > 0, the second inequality follows from the fact that αa_CT + β ≥ 2b if a_CT > 0, the second equality is Eq. 21 since αa_CT + β ≥ 2b > b, the third inequality follows from the fact that ${U_{B}^{D}}$ is maximized at a = a_D, the third equality follows from the condition that a_D = 0, and the last equality follows from the fact that ${U_{B}^{D}}(0)=U_{B}^{CT}(0)=K_{B}-M_{B}\left (V(\tilde {\theta }) + b^{2}\right )$. We thus arrive at a contradiction with the assumption that a_CT > 0. Therefore, it must be true that a_CT = 0.

• Step 2. If a_CT > 0, then from Step 1 above we know that a_D > 0. Therefore, $U_{B}^{CT}$ and ${U_{B}^{D}}$ are characterized by Eqs. 17 and 21 respectively. Taking derivative of Eqs. 17 and 21 in a respectively, we have:

  $$\begin{array}{@{}rcl@{}} \frac{\partial U_{B}^{CT}(a)}{\partial a} & =&M_{B}\left[\frac{2\alpha(\alpha a +\beta)}{3}\left( 1-\frac{1}{n^{2}}\right)\right]-2 a \\ &&< M_{B}\left[\frac{2\alpha(\alpha a +\beta)}{3} + \frac{4\alpha b^{3}}{3(\alpha a + \beta)^{2}}\right]\\&&-2 a = \frac{\partial {U_{B}^{D}}(a)}{\partial a}, \quad \forall a \text{ and } n. \end{array} $$

  (54)

  Since both $U_{B}^{CT}(a)$ and ${U_{B}^{D}}(a)$ are concave in a, it must be true that a_CT < a_D.

$\ \Box$

Proof of Lemma 6

Since ${U_{B}^{D}}(a)$ is concave in a, a_D < a_FSBO if and only if:

$$ \frac{\partial {U_{B}^{D}}(a)}{\partial a}=M_{B}\left( \frac{2\alpha(\alpha a +\beta)}{3}+\frac{4\alpha b^{3}}{3(\alpha a +\beta)^{2}}\right) -2a <0, \quad \text{at \(a=a_{FSBO}\)}, $$

(55)

where the equality follows from Lemma 3 and (21), and a_FSBO is characterized by Eq. 28. Substituting (28) into (55) and rearranging, one gets that a_D < a_FSBO if and only if $\frac {3\beta }{3-2\alpha ^{2}}> b\left (\frac {2(1+r)}{1-r}\right )^{1/3}$. This completes the proof. $\ \Box$

Proof of Proposition 5

We prove ${U_{S}^{D}}\geq U_{S}^{CT}$ for two cases: (i) a_CT = 0 and (ii) a_CT > 0.

• (i) If a_CT = 0, then according to Proposition (1), $b>\frac {\beta }{2}$. Then Eq. 14 implies that n(a_CT) = 1 and Eq. 16 implies that $U_{S}^{CT}(a_{CT})=K_{S}-M_{S} V(\tilde {\theta })$. On one hand, if b ≥ αa_D + β, then Eq. 20 implies that ${U_{S}^{D}}(a_{D})=K_{S}-M_{S} V(\tilde {\theta })=U_{S}^{CT}(a_{CT})$. If, on the other hand, b < αa_D + β, i.e., $a_{D}>\frac {b-\beta }{\alpha }$, then ${U_{S}^{D}}(a_{D})$ follows (22). Equation 22 implies that:

  $$ \frac{\partial {U_{S}^{D}}(a)}{\partial a}=M_{S}\left( \frac{2\alpha(\alpha a+\beta)}{3} - \frac{2\alpha b^{3}}{3(\alpha a+\beta)^{2}}\right)>0, \quad \forall \alpha a+\beta> b. $$

  That is, ${U_{S}^{D}}(a)$ increases in a for all $a > \frac {b-\beta }{\alpha }$. Then we have:

  $$ {U_{S}^{D}}(a_{D})> {U_{S}^{D}}\left( \frac{b-\beta}{\alpha}\right)=K_{S}-M_{S} V(\tilde{\theta})=U_{S}^{CT}(a_{CT}), $$

  (56)

  where the first inequality follows from the condition that $a_{D}> \frac {b-\beta }{\alpha }$, the first equality follows from substituting $a=\frac {b-\beta }{\alpha }$ into Eq. 22.

• (ii) If a_CT > 0, we must have n(a_CT) ≥ 2 because otherwise the “babbling” equilibrium is the only equilibrium, and a_CT > 0 can not be a solution. In addition, according to Proposition 4, a_D > a_CT > 0. Then we have:

  $$\begin{array}{@{}rcl@{}} U_{S}^{CT}(a_{CT})&=&K_{S}-M_{S}\left( V(\tilde{\theta})-\frac{(\alpha a_{CT} + \beta)^{2}}{3}\left( 1-\frac{1}{n(a_{CT})^{2}}\right)- \frac{b^{2}(1-n(a_{CT})^{2})}{3}\right) \\ &&< K_{S}-M_{S}\left( V(\tilde{\theta})-\frac{(\alpha a_{CT} +\beta)^{2}}{3} + b^{2}\right) \\ &&< K_{S}-M_{S}\left( V(\tilde{\theta})-\frac{(\alpha a_{CT} +\beta)^{2}}{3} + b^{2} - \frac{2b^{3}}{3(\alpha a_{D}+\beta)} \right) \\ &&< K_{S}-M_{S}\left( V(\tilde{\theta})- \frac{(\alpha a_{D} +\beta)^{2}}{3} + b^{2} - \frac{2b^{3}}{3(\alpha a_{D}+\beta)} \right) \\ &&={U_{S}^{D}}(a_{D}), \end{array} $$

  (57)

  where the first equality is Eq. 16, the first inequality follows from the fact that $2\leq n_{CT}<\infty$, the second inequality follows from the fact that $\frac {2b^{3}}{3(\alpha a_{D}+\beta )}>0$, the third inequality follows from the result that a_D > a_CT, and the last equality is Eq. 22.

$\ \Box$

Proof of Proposition 6

From Eq. 22, we have:

$$\begin{array}{@{}rcl@{}} \frac{\partial {U_{S}^{D}}(a)}{\partial a}&=&\frac{2\alpha M_{S}}{3}\left( (\alpha a+ \beta)-\frac{b^{3}}{(\alpha a+ \beta)^{2}}\right). \end{array} $$

(58)

$$\begin{array}{@{}rcl@{}} \frac{\partial^{2} {U_{S}^{D}}(a)}{\partial a^{2}}&=&\frac{2\alpha^{2} M_{S}}{3}\left( \alpha + \frac{2b^{3}}{(\alpha a+ \beta)^{3}}\right)>0. \end{array} $$

(59)

Formula (59) suggests that ${U_{S}^{D}}(a)$ is convex in a. And Eq. 58 implies that $\frac {\partial {U_{S}^{D}}(a)}{\partial a}= 0$ at a_b, where a_b is defined as such that αa_b + β = b. That is, ${U_{S}^{D}}(a)$ reaches its minimum at a_b. From Lemma 3, we know that a_D > a_b. Figure 9 demonstrates ${U_{S}^{D}}(a)$ as a function of a. The rest of the proof is composed of two parts.

Fig. 9

Graph of ${U_{S}^{D}}(a)$

• Part 1. Prove that $U_{S}^{FSBO}>{U_{S}^{D}}$if Eqs. 31 and 32hold.

If Eq. 32 holds true, then we have:

$$ r\left( V(\tilde{\theta})-\frac{b^{2}}{3}\right) \leq \frac{2-r}{3}b^{2}. $$

(60)

From Eq. 28, we also have:

$$ \alpha a_{FSBO}+\beta=\frac{3\beta}{3-2\alpha^{2}}>\frac{3\beta}{3-2\alpha^{2}}\left( \frac{1-r}{2(1+r)}\right)^{1/3}>b, $$

(61)

where the first equality follows from Eq. 28, the first inequality follows from the fact that $0<\left (\frac {1-r}{2(1+r)}\right )^{1/3}<1$ and the last inequality follows from Eq. 31.

Condition (61) implies that:

$$\begin{array}{@{}rcl@{}} r\left( V(\tilde{\theta})-\frac{b^{2}}{3}\right) & \geq& r\left( V(\tilde{\theta})-\frac{(\alpha a_{FSBO}+\beta)^{2}}{3}\right), \quad \text{and} \end{array} $$

(62)

$$\begin{array}{@{}rcl@{}} \frac{2-r}{3}b^{2} &\leq& (2-r)b^{2}\left( 1-\frac{2b}{3(\alpha a_{FSBO}+\beta)}\right). \end{array} $$

(63)

From Eqs. 60, 62 and 63, we have:

$$ r\left( V(\tilde{\theta})-\frac{(\alpha a_{FSBO}+\beta)^{2}}{3}\right) \leq (2-r)b^{2}\left( 1-\frac{2b}{3(\alpha a_{FSBO}+\beta)}\right). $$

(64)

Adding $(2-r)\left (V(\tilde {\theta })-\frac {(\alpha a_{FSBO}+\beta )^{2}}{3}\right )$ to both sides of Eq. 64, one gets:

$$ 2\left( V(\tilde{\theta})-\frac{(\alpha a_{FSBO}+\beta)^{2}}{3}\right) \leq (2-r)b^{2}\left( V(\tilde{\theta})-\frac{(\alpha a_{FSBO}+\beta)^{2}}{3}-\frac{2b^{3}}{3(\alpha a_{FSBO}+\beta)}+b^{2}\right). $$

(65)

Then we have:

$$\begin{array}{@{}rcl@{}} {U_{S}^{D}}(a_{D})&=&K_{S}-M_{S}\left( V(\tilde{\theta}) - \frac{(\alpha a_{D}+ \beta)^{2}}{3} - \frac{2b^{3}}{3(\alpha a_{D}+\beta)} + b^{2}\right) \\ & \leq &{U_{S}^{D}}(a_{FSBO}) \\ &= &K_{S}-M_{S}\left( V(\tilde{\theta}) - \frac{(\alpha a_{FSBO}+ \beta)^{2}}{3} - \frac{2b^{3}}{3(\alpha a_{FSBO}+\beta)} + b^{2}\right) \\ &<&\left( \bar{P}-\frac{\hat{b}^{2}}{2}\right)-(2-r)\left( V(\tilde{\theta}) - \frac{(\alpha a_{FSBO}+ \beta)^{2}}{3} - \frac{2b^{3}}{3(\alpha a_{FSBO}+\beta)} + b^{2}\right) \\ & \leq& \left( \bar{P}-\frac{\hat{b}^{2}}{2}\right)-2\left( V(\tilde{\theta})-\frac{(\alpha a_{FSBO}+\beta)^{2}}{3}\right) \\ &=&U_{S}^{FSBO}, \end{array} $$

(66)

where the first equality follows from Eq. 22, the first inequality follows from the result that a_FSBO > a_D (due to Eq. 31) and ${U_{S}^{D}}(a)$ is convex in a (Fig. 9), the second equality follows from Eq. 22, the second inequality follows from the fact that $K_{S}=(1-r)\left (\bar {P}-\frac {\hat {b}^{2}}{2-r}\right )<\bar {P}-\frac {\hat {b}^{2}}{2}$ and M_S = 2 − r, the third inequality follows from Eq. 65.

• Part 2. Prove that ${U_{S}^{D}}>U_{S}^{FSBO}$if Eq. 33holds.

$$\begin{array}{@{}rcl@{}} {U_{S}^{D}}(a_{D})&=&K_{S}-M_{S}\left( V(\tilde{\theta}) - \frac{(\alpha a_{D}+ \beta)^{2}}{3} - \frac{2b^{3}}{3(\alpha a_{D}+\beta)} + b^{2}\right) \\ &>&{U_{S}^{D}}(a_{b}) \\ &=& K_{S}-M_{S}\left( V(\tilde{\theta}) - \frac{(\alpha a_{b}+ \beta)^{2}}{3} - \frac{2b^{3}}{3(\alpha a_{b}+\beta)} + b^{2}\right) \\ &=&K_{S}-M_{S} V(\tilde{\theta}) \\ &=&(1-r)\left( \bar{P}-\frac{\hat{b}^{2}}{2-r}\right) - (2-r)V(\tilde{\theta}) \\ &\geq& \left( \bar{P}-\frac{\hat{b}^{2}}{2}\right)- 2 \left( V(\tilde{\theta}) -\frac{\beta^{2}}{3-2\alpha^{2}}\right) \\ &=&U_{S}^{FSBO}, \end{array} $$

(67)

where the first equality follows from Eq. 22, the first inequality follows from the fact that a_D > a_b and ${U_{S}^{D}}(a)$ is minimized at a = a_b (Fig. 9), the second equality follows from Eq. 22, the third equality follows from the definition of a_b, i.e., αa_b + β = b, the fourth equality follows from substituting in Eqs. 4 and 5, the last second equality follows from Eq. 33, and the last equality follows from Eq. 29. $\ \Box$