The value and direction of innovation

Abstract

This paper considers the allocation of innovators between two research lines that differ in their values of innovation and their probabilities of discovery. Innovators choose a research line to maximize their expected utility, and the high value research line may attract more or fewer innovators than the low value research line, depending on the difficulty of discovery. The equilibrium allocation is not efficient, as innovators ignore the effects of their choice of research lines on other innovators.

Appendix

Proof of (ii) of Lemma 1

To determine the sign of $\partial \pi (p, n)/\partial n$, rewrite (2) as

$$\begin{aligned} \pi \; n = 1 - (1 - p)^n\;\;\Longrightarrow \;\;1 - \pi \;n = (1 - p)^n, \end{aligned}$$

(25)

where the arguments of $\pi (p, n)$ are and will be suppressed when it does not create any confusion. Taking the natural log of both sides of (25),

$$\begin{aligned} ln\;(1 - \pi \;n) = n \;ln\;(1 - p). \end{aligned}$$

(26)

Using (25) and (26),

$$\begin{aligned} \pi _n(p, n) &= \frac{\partial \pi (p, n)}{\partial n} = \frac{1}{n^2}\;\{(1 - \pi \;n) [1 - ln\;(1 - \pi \;n)] - 1\}\nonumber \\ &= \frac{1}{n^2}\;\{ - \pi \;n -(1 - \pi \;n) \;ln\;(1 - \pi \;n)\}. \end{aligned}$$

(27)

The sign of $\pi _n(p, n)$ coincides with the sign of the numerator of (27), and the numerator can be rewritten as

$$\begin{aligned} \phi (x) \equiv - x -(1 - x) \;ln\;(1 - x) \end{aligned}$$

(28)

with $x \equiv \pi \;n.$ Given (25), it must be that $x \in (0, 1).$$\phi (x)$ has the following three properties. First, as x approaches zero,

$$\begin{aligned} lim_{x \rightarrow 0} \phi (x) = - 0 - \;ln\;(1) = 0. \end{aligned}$$

(29)

Second, differentiation of $\phi (x)$ gives

$$\begin{aligned} \phi '(x) = - 1 + ln\;(1 - x) -(1 - x)\;\frac{-1}{(1-x)} = ln\;(1 - x) < 0. \end{aligned}$$

(30)

Third, as x approaches one,

$$\begin{aligned} lim_{x \rightarrow 1} \phi (x) = - 1 - 0\;ln\;(0), \end{aligned}$$

which cannot be determined, as $ln\;(0) = - \infty .$ However, using L’Hospital’s rule,

$$\begin{aligned} lim_{x \rightarrow 1} \phi (x) &= - 1 - lim_{x \rightarrow 1} \frac{ln\;(1 - x)}{1/(1-x)} \nonumber \\ &= - 1 - lim_{x \rightarrow 1} \frac{-1/(1 - x)}{1/(1-x)^2} \nonumber \\&- 1 - lim_{x \rightarrow 1} [-(1 - x)] = - 1. \end{aligned}$$

(31)

These three properties of $\phi (x)$ imply that $\phi (x) < 0$ for all $x \in (0, 1)$ and hence $\pi _n(p, n) < 0$, establishing part (ii) of the lemma. $\square$

Proof of Proposition 1

For an interior case to occur, it must be that

$$\begin{aligned} U_1({\overline{n}}) = v_1 \pi (p_1, {\overline{n}}) < U_2(0) = v_2 \pi (p_2, 0), \end{aligned}$$

(32)

$$\begin{aligned} \;and\;\;U_1(0)= v_1 \pi (p_1, 0) > U_2({\overline{n}}) = v_2 \pi (p_2, {\overline{n}}). \end{aligned}$$

(33)

Using the definition of $\pi (p, n)$ in (2), $\pi (p_i, {\overline{n}}) = [1 - (1 -p_i)^{{\overline{n}}}]/{\overline{n}}$. However, $\pi (p_i, 0) = [1 - 1]/0 = 0/0$ is not defined, and using L’Hospital’s rule,

$$\begin{aligned} \pi (p_i, 0) = lim_{n_i \rightarrow 0} \pi (p_i, n_i) = lim_{n_i \rightarrow 0} \frac{- (1-p_i)^{n_i}\;ln\;(1 - p_i)}{1} = - ln\;(1 - p_i). \end{aligned}$$

(34)

Substitution of the expression of $\pi (p_i, {\overline{n}})$ and that of $\pi (p_i, 0)$ into (32) and (33) leads to the condition in the proposition. $\square$

Proof of Lemma 2

As $v_2 > v_1$, the equilibrium condition (4) implies $\pi (p_2, n_2^*) < \pi (p_1, n_1^*).$ Since $\pi _p(p, n) > 0$ from part (i) of Lemma 1, and since $p_2 \ge p_1$ by hypothesis, $\pi (p_2, n_2^*) \ge \pi (p_1, n_2^*).$ The two inequalities imply

$$\begin{aligned} \pi (p_1, n_2^*) \le \pi (p_2, n_2^*) < \pi (p_1, n_1^*). \end{aligned}$$

(35)

The first term in (35) and the last one have the same probability of discovery $p_1$, but the first one is smaller. As $\pi (p, n)$ decreases in n by part (ii) of Lemma 1, it must be that $n_2^* > n_1^*$, establishing the lemma. $\square$

Proof of Proposition 2

Since $n_1^*(p_2, v_2) < n_2^*(p_2, v_2)$ when $p_2 = p_1$ by Lemma 2, and since $\partial n_1^*(p_2, v_2)/\partial p_2 < 0$ by (7), there exists $p_2^* \equiv p_2^*(p_1, v_1, v_2)$ such that $n_1^*(p_2^*, v_2) = n_2^*(p_2^*, v_2)$ with $p_2^* < p_1$. In addition, due to the inequality in (7), $n_1^*(p_2, v_2) < (>)\;n_2^*(p_2, v_2)$ for $p_2 > (<)\; p_2^*,$ establishing the proposition. $\square$

Proof of Lemma 3

Evaluation of the planner’s FOC (8) at the equilibrium allocation $(n_1^*, n_2^*) = (n_1^*, {\overline{n}} - n_1^*)$ gives

$$\begin{aligned} \frac{\partial W}{\partial n_1} \mid _{(n_1 = n_1^*)}= & {} v_1 \lambda _n(p_1, n_1^*) - v_2\lambda _n(p_2, n_2^*) \nonumber \\= & {} v_1 \pi (p_1, n_1^*)\;\left[ \frac{\lambda _n(p_1, n_1^*)}{\pi (p_1, n_1^*)} - \frac{\lambda _n(p_2, n_2^*)}{\pi (p_2, n_2^*)}\right] \nonumber \\= & {} v_1 \pi (p_1, n_1^*)\;[\eta (p_1, n_1^*) - \eta (p_2, n_2^*)] > 0, \end{aligned}$$

(36)

where the first equality uses the equilibrium condition (4) and $\eta (p, n)$ is defined as

$$\begin{aligned} \eta (p, n) \equiv \lambda _n(p, n)/\pi (p, n). \end{aligned}$$

(37)

Since $p_2 \ge p_1$ and hence $n_2^* > n_1^*$ by Lemma 2, and since $\eta (p, n)$ is decreasing in p and n, as shown below, (36) is positive. This implies $n_1^* < {\hat{n}}_1$ and $n_2^* > {\hat{n}}_2,$ establishing the lemma. $\square$

Proof of $\eta _p(p, n) < 0$ and $\eta _n(p, n) < 0$

Ignoring $1/[\pi (p, n)]^2,$ the sign of $\eta _p(p, n)$ is identical to that of

$$\begin{aligned}&\eta _p(p, n) \cong \pi \lambda _{np} - \lambda _n \pi _p \nonumber \\&\quad = [1 - (1-p)^n] n (1-p)^{(n-1)} [1 + n ln\;(1-p)] \nonumber \\&\qquad -\, n [ - (1-p)^n ln\;(1-p)] n (1-p)^{(n-1)}\nonumber \\&\quad = n (1-p)^{(n-1)}\;Q(p), \end{aligned}$$

(38)

where

$$\begin{aligned} Q(p) \equiv n\; ln\;(1-p) + 1 - (1-p)^n. \end{aligned}$$

(39)

Q(p) has three properties:

$$\begin{aligned}&lim_{p \rightarrow 0} Q(p) = Q(0) = 0, \end{aligned}$$

(40)

$$\begin{aligned}&Q'(p) = - \frac{n}{1-p} + n (1-p)^{(n-1)} = \frac{n}{1-p} [(1-p)^n - 1] < 0, \end{aligned}$$

(41)

$$\begin{aligned}&lim_{p \rightarrow 1} Q(p) = Q(1) = - \infty . \end{aligned}$$

(42)

Thus, $Q(p) < 0$ for all $p > 0$, and hence $\eta _p(p, n) < 0$ in (38).

In a manner analogous to (38), it is straightforward to verify that

$$\begin{aligned}&\eta _n(p, n) \cong \pi \lambda _{nn} - \lambda _n \pi _n\nonumber \\&\quad = -\, (1-p)^{n}\;ln\;(1-p)\;\;Q(p) < 0. \end{aligned}$$

(43)

Proof of Proposition 3

Consider (36) in the proof of Lemma 3. Since the sign of (36) coincides with that of the expression inside the pair of square brackets, and since the relationship between $n_1^*$ and $n_2^*$ depends on the magnitude of $p_2$ by Lemma 2, define

$$\begin{aligned} \Omega (p_2) \equiv \eta (p_1, n_1^*) - \eta (p_2, n_2^*). \end{aligned}$$

(44)

At $p_2 = p_2^*$ and hence $n_1^* = n_2^*$ by Lemma 2, (44) becomes

$$\begin{aligned} \Omega (p_2^*) = \eta (p_1, n_1^*) - \eta (p_2^*, n_1^*) < 0, \end{aligned}$$

(45)

as $p_2^* < p_1$ by Lemma 2 and $\eta (p, n)$ is decreasing in p. Next, at $p_2 = p_1$,

$$\begin{aligned} \Omega (p_1) = \eta (p_1, n_1^*) - \eta (p_1, n_2^*) > 0 \end{aligned}$$

(46)

because Lemma 3 holds when $p_2 \ge p_1.$ Thus, there is ${\hat{p}}_2$ such that $\Omega (p_2) > (=, <)\;0$ for $p_2 > (=, <)\;{\hat{p}}_2,$ establishing the proposition. $\square$

Derivation of (21)

$$\begin{aligned} d W= & {} [(n_1 + \triangle ) U_1(n_1 + \triangle ) - n_1 U_1(n_1)] \nonumber \\&-\, [n_2 U_2(n_2) - (n_2 - \triangle ) U_2(n_2 - \triangle )] \nonumber \\= & {} v_1 [\lambda (e_1(n_1 + \triangle ) p, n_1 + \triangle ) - \lambda (e_1(n_1)p, n_1)] \nonumber \\&-\, [(n_1+\triangle )e_1(n_1+\triangle ) - n_1 e_1(n_1)]\nonumber \\&-\, v_2 [\lambda (e_2(n_2)p, n_2) - \lambda (e_2(n_2 -\triangle )p, n_2 - \triangle )] \nonumber \\&+\, [n_2 e_2(n_2) - (n_2 - \triangle )e_2(n_2 - \triangle )], \end{aligned}$$

(47)

where $(n_1, n_2) = (n_1^*, n_2^*)$ and $e_i(n_i) = e_i^*(v_i, p, n_i)$ with the arguments $v_i$ and p dropped. For a small $\triangle ,$dW reduces to $W'$ in (21). $W'$ includes $[v_1 \lambda _e(e_1(n_1 + \triangle ) p, n_1 + \triangle ) - (n_1+\triangle )]\; (\partial e(n_1+\triangle )/\partial (n_1+\triangle )) - [v_2 \lambda _e(e_2(n_2 - \triangle ) p, n_2 - \triangle ) - (n_2 -\triangle )]\; (\partial e(n_2-\triangle )/\partial (n_2-\triangle ))$. The expression in each pair of square brackets becomes FOC (20) and hence zero for a small $\triangle$ once divided by $n_1 + \triangle$ and $n_2 - \triangle$, respectively.

Relationship between $e_2^*$ and $e_1^*$

Solving (17) for e,

$$\begin{aligned} e^* = \frac{1}{p}\;[1 - (vp)^{1/(1-n)}]. \end{aligned}$$

(48)

To rule out negative effort, it is assumed that

$$\begin{aligned} vp > 1 \end{aligned}$$

(49)

(that is, $v_1p_1 > 1$ and $v_2p_2 > 1$). Total differentiation of equilibrium condition (19), along with (48), gives

$$\begin{aligned} \frac{\partial n_2}{\partial p_2}= & {} - \frac{1}{D} v_2^2 \frac{1}{n_2 - 1} (p_2v_2)^{(2n_2 - 1)/(1-n_2)}> 0,\nonumber \\ \frac{\partial n_2}{\partial v_2}= & {} - \frac{1}{D} v_2^2 \frac{1}{n_2 (n_2 - 1)} [n_1 - 1 + (p_2v_2)^{(n_2)/(1-n_2)}] > 0, \end{aligned}$$

(50)

where

$$\begin{aligned} D\equiv & {} - \frac{v_2}{n_2^2} \left[ \frac{n_2}{(1- n_2)^2} (p_2v_2)^{n_2/(1-n_2)} ln (p_2v_2) + 1 - (p_2v_2)^{n_2/(1-n_2)}\right] \\&- \frac{v_1}{n_1^2} \left[ \frac{n_1}{(1- n_1)^2} (p_1v_1)^{n_1/(1-n_1)} ln (p_1v_1) + 1 - (p_1v_1)^{n_1/(1-n_1)}\right] < 0. \end{aligned}$$

Since $U_1(n_1) = v_1 \pi (e_1(n_1)p_1, n_1) - e_1(n_1) = U_2(n_2) = v_2 \pi (e_2(n_2)p_2, n_2) - e_2(n_2)$ in equilibrium,

$$\begin{aligned} e_2(n_2) - e_1(n_1)= & {} v_2 \pi (e_2(n_2)p_2, n_2) - v_1 \pi (e_1(n_1)p_1, n_1)\nonumber \\= & {} v_2 \frac{1}{n_2} [1 - (p_2v_2)^{n_2/(1-n_2)}] - v_1 \frac{1}{n_1} [1 - (p_1v_1)^{n_1/(1-n_1)}] \nonumber \\\equiv & {} f(v_2, p_2: v_1, p_1). \end{aligned}$$

(51)

Differentiation of f(.), along with (48), yields

$$\begin{aligned} \frac{\partial f(v_2, p_2: v_1, p_1)}{\partial p_2}= & {} v_2^2 \frac{1}{n_2 - 1} (p_2v_2)^{(2n_2-1)/(1-n_2)} \nonumber \\&-\, \frac{\partial n_2}{\partial p_2}\left[ v_2 \frac{1}{n_2 (1-n_2)^2} (p_2v_2)^{n_2/(1-n_2)} ln\; (p_2v_2) \right. \nonumber \\&\left. +\, v_1 \frac{1}{n_1 (1-n_1)^2} (p_1v_1)^{n_1/(1-n_1)} ln\; (p_1v_1)\right] \nonumber \\= & {} -\, \frac{1}{D} v_2^2 \frac{1}{n_2 - 1} (p_2v_2)^{(2n_2-1)/(1-n_2)} \left\{ \frac{v_2}{n_2^2} [1 - (p_2v_2)^{n_2/(1-n_2)}] \right. \nonumber \\&\left. +\, \frac{v_1}{n_1^2} [1 - (p_1v_1)^{n_1/(1-n_1)}]\right\}> 0, \nonumber \\ \frac{\partial f(v_2, p_2: v_1, p_1)}{\partial v_2}= & {} \frac{1}{n_2}[1 - (p_2v_2)^{n_2/(1-n_2)}] + \frac{1}{n_2 -1}(p_2v_2)^{(n_2)/(1-n_2)} \nonumber \\&-\, \frac{\partial n_2}{\partial v_2}\left[ v_2 \frac{1}{n_2 (1-n_2)^2} (p_2v_2)^{n_2/(1-n_2)} ln\; (p_2v_2) \right. \nonumber \\&\left. +\, v_1 \frac{1}{n_1 (1-n_1)^2} (p_1v_1)^{n_1/(1-n_1)} ln\; (p_1v_1)\right] \nonumber \\= & {} -\, \frac{1}{D} \left\{ \frac{1}{n_2}[1 - (p_2v_2)^{n_2/(1-n_2)}] + \frac{1}{n_2 -1}(p_2v_2)^{(n_2)/(1-n_2)}\right\} \; \nonumber \\&\left\{ \frac{v_2}{n_2^2} [1 - (p_2v_2)^{n_2/(1-n_2)}] + \frac{v_1}{n_1^2} [1 - (p_1v_1)^{n_1/(1-n_1)}]\right\} > 0. \end{aligned}$$

(52)

Since $f(v_1, p_1: v_1, p_1) = 0, f(v_2, p_1: v_1, p_1) > 0,$ given $v_2 > v_1.$ As $f(v_2, p_2: v_1, p_1)$ is increasing in $p_2, f(v_2, p_2: v_1, p_1) > 0$ for all $p_2 \ge p_1.$ If $p_2 < p_1$, there is a critical value of $p_2, {\tilde{p}}_2$, such that $f(v_2, p_2: v_1, p_1) > (=, <)\;0$ for $p_2 > (=,<)\;{\tilde{p}}_2 < p_1$. In summary,

$$\begin{aligned}&e_2(n_2) - e_1(n_1)> 0\;\;when\;\;p_2 \ge p_1,\nonumber \\&\quad e_2(n_2) - e_1(n_1)> (=,<)\;0\quad for \quad p_2 > (=,<)\;{\tilde{p}}_2\quad when \quad p_2 < p_1. \end{aligned}$$

(53)

Proof of Lemma 4

Evaluation of the planner’s FOC (20) at the equilibrium allocation $(n_1^*, n_2^*)$ gives

$$\begin{aligned} \frac{\partial W}{\partial n_1} \mid _{(n_1 = n_1^*)}= & {} v_1 \lambda _n(e_1^*p_1, n_1^*) - v_2\lambda _n(e_2^*p_2, n_2^*) + (e_2^* - e_1^*) \nonumber \\= & {} v_1 \pi (e_1^*p_1, n_1^*) \left[ \frac{\lambda _n(e_1^* p_1, n_1^*)}{\pi (e_1^*p_1, n_1^*)} - \frac{\lambda _n(e_2^* p_2, n_2^*)}{\pi (e_2^*p_2, n_2^*)}\right] \nonumber \\&+\, [e_2^* - e_1^*]\;\left[ 1 - \frac{\lambda _n(e_2^* p_2, n_2^*)}{\pi (e_2^*p_2, n_2^*)}\right] > 0. \end{aligned}$$

(54)

Letting $z \equiv ep$, $\eta (z, n) \equiv \lambda _n(z, n)/\pi (z, n)$ is decreasing in z and n, as in (38) and (43). Since $p_2 \ge p_1$ by hypothesis in Lemma 4, $e_2^* > e_1^*$ by (53) and hence $e_2^* p_2 > e_1^*p_1$. In addition, as $p_2 \ge p_1,$$n_2^* > n_1^*$ by Lemma 2. As a result, $\eta (e_1^* p_1, n_1^*) > \eta (e_2^* p_2, n_2^*)$ in the second line of (54). Also, $1 - \frac{\lambda _n(e_2^* p_2, n_2^*)}{\pi (e_2^*p_2, n_2^*)} > 0$ in the last line of (54), as explained in footnote 13. Thus, the inequality of (54) holds, establishing the lemma.

For the subsequent analysis, it is also necessary to compare $e_2^*p_2$ and $e_1^*p_1$. $\square$

Relationship between $e_2^*p_2$ and $e_1^*p_1$

Since $f(.) = e_2 - e_1$ and hence $e_2 = e_1 + f(.)$,

$$\begin{aligned} e_2(n_2)p_2 - e_1(n_1)p_1 = p_2 f(.) + (p_2 - p_1) e_1(n_1) \equiv g(v_2, p_2: v_1, p_1). \end{aligned}$$

(55)

Since $f(.) > 0$ for $p_2 \ge p_1$ by (53), $g(.) > 0$ for $p_2 \ge p_1.$ In addition, $f(v_2, {\tilde{p}}_2: v_1, p_1) = 0$ by (53), and ${\tilde{p}}_2 < p_1$, so $g(v_2, {\tilde{p}}_2: v_1, p_1) < 0.$ Therefore, there is a critical value, denoted $p_2' \in ({\tilde{p}}_2, p_1)$, such that $g(v_2, p_2: v_1, p_1) \ge 0$ for $p_2 \ge p_2'.$ Unlike $f(v_2, p_2: v_1, p_1), g(v_2, p_2: v_1, p_1)$ is not necessarily a monotonic function of $p_2$, so there may be multiple values of $p_2$ that makes $g(v_2, p_2': v_1, p_1) = 0$, and it is sufficient to choose the maximum of such values of $p_2$ for $g(v_2, p_2: v_1, p_1) \ge 0$ when $p_2 \ge p_2'$. In summary,

$$\begin{aligned} e_2(n_2)p_2 - e_1(n_1)p_1 \ge 0\quad for\quad p_2 \ge p_2' \in ({\tilde{p}}_2, p_1). \end{aligned}$$

(56)

As $f(.) = e_2(n_2) - e_1(n_1) \le 0$ for $p_2 \le {\tilde{p}}_2$ and ${\tilde{p}}_2 < p_1$ in (53),

$$\begin{aligned} e_2(n_2)p_2 - e_1(n_1)p_1 \le 0\quad for\quad p_2 \le {\tilde{p}}_2 < p_1. \end{aligned}$$

(57)

Proof of Proposition 5

Part (i):

A sufficient condition for (54) to continue to hold is $e_2^*p_2 \ge e_1^*p_1, n_2^* \ge n_1^*$ and $e_2^* \ge e_1^*.$ Since each inequality holds for $p_2 \ge p_2', p_2 \ge p_2^*$ and $p_2 \ge {\tilde{p}}_2$, (54) holds and $W' \ge 0$ for $p_2 \ge p_2^o \equiv$ max $\{p_2', p_2^*, {\tilde{p}}_2\} =$ max $\{p_2', p_2^*\},$ because $p_2' > {\tilde{p}}_2.$

Part (ii):

A sufficient condition for the inequality of (54) to be reversed is $e_2^*p_2 \le e_1^*p_1, n_2^* \le n_1^*$ and $e_2^* \le e_1^*.$ Since the first and the third inequalities hold for $p_2 \le {\tilde{p}}_2$ in (57), all three hold for $p_2 \le {\tilde{p}}_2$ and $p_2 \le p_2^*$. Thus, $W' \le 0$ for $p_2 \le p_2^{oo} \equiv$ min $\{p_2^*, {\tilde{p}}_2\}.$

The sign of the third line of (24)

$$\begin{aligned} \lambda _e(e p, n) = \frac{d}{d e} v [(1 - (1-ep)^n] = vpn (1-ep)^{(n-1)} = n^2 e. \end{aligned}$$

(58)

The last equality uses FOC (23). Differentiation of the FOC gives

$$\begin{aligned} \frac{\partial e^*}{\partial n} = \frac{(1-ep)e}{(1 + nep - 2 ep) n}\;[n\;ln\;(1 - ep) - 1] < 0. \end{aligned}$$

(59)

Substituting (58) and (59) into the first term of the third line of (24) with research line subscript i omitted,

$$\begin{aligned}&(\lambda _e(e p, n) - ne)\;\frac{\partial e^*}{\partial n} \nonumber \\&\quad = (n^2 e - n e) \frac{(1-ep)e}{(1 + nep - 2 ep) n}\;[n\;ln\;(1 - ep) - 1]\nonumber \\&\quad = \frac{(n-1)(1-ep)e^2}{(1 + nep - 2 ep)}\;[n\;ln\;(1 - ep) - 1] \equiv g(n, e). \end{aligned}$$

(60)

Differentiation of (60) can show

$$\begin{aligned} \frac{\partial g(n, e)}{\partial n}= & {} \frac{1}{(1 + nep - 2 ep)^2}\;\{(1- ep)[n\;ln\;(1 - ep) - 1] \\&+ (1 + nep - 2 ep) (n-1) \;ln\;(1 - ep)\}< 0, \\ \frac{\partial g(n, e)}{\partial e}= & {} \frac{1}{(1 + nep - 2 ep)^2}\;\{[p(n-2)e^2(1-2ep) + e(2 - 3ep)]\\&[n\;ln\;(1 - ep) - 1] - (1 + nep - 2 ep) e^2 n p\} < 0, \end{aligned}$$

if $n \ge 2$ and $(1 - 2ep) \ge 0.$ Since $n_2 > n_1$ in equilibrium and $e_2 > e_1$ by hypothesis, $g(n_1, e_1) > g(n_2, e_2)$ and hence the third line of (24) is positive. $\square$