Policy convergence in a two-candidate probabilistic voting model

Abstract

We propose a generalization of the probabilistic voting model in two-candidate elections. We allow the candidates have general von Neumann–Morgenstern utility functions defined over the voting outcomes. We show that the candidates will choose identical policy positions only if the electoral competition game is constant-sum, such as when both candidates are probability-of-win maximizers or vote share maximizers, or for a small set of functions that for each voter define the probability of voting for each candidate, given candidate policy positions. At the same time, a pure-strategy local Nash equilibrium (in which the candidates do not necessarily choose identical positions) exists for a large set of such functions. Hence, if the candidate payoffs are unrestricted, the “mean voter theorem” for probabilistic voting models is shown to hold only for a small set of probability of vote functions.

Appendix

Proof of Theorem 1

Take arbitrary $z\in int(X)$. FOC for $z$ are:

$$\begin{aligned} D_1(U_1)=\sum _{l=0}^Nu^1_lD_1(P_l)=0, \quad D_2(U_2)=\sum _{l=0}^Nu^2_lD_2(P_{N-l})=0. \end{aligned}$$

Recall that $u_{1}^{0}=u_{2}^{0}=0$ and $u_{1}^{N}=u_{2}^{N}=1$ and suppose that for some $l=h$, we have $u^1_h+u^2_{N-h}\ne 1$. Now use the identity $\sum D(P_l)\equiv 0$ to exclude $P_h$ from FOC:

$$\begin{aligned} D_1(U_1)= \sum _{\begin{array}{c} l=0\\ l\ne h \end{array}}^N (u^1_l-u^1_h)D_1(P_l)=0, \quad D_2(U_2)=\sum _{\begin{array}{c} l=0\\ l\ne h \end{array}}^N (u^2_l-u^2_{N-h})D_2(P_{N-l})=0. \end{aligned}$$

Consider the following $(n+1)$-dimensional subspace of $\mathfrak {P}$ (spanned by $\alpha$ and $\{\delta _{i}\}$):

$$\begin{aligned} H_z(\alpha ,\delta )=\left( \begin{array}{l} P_{0}(x,y)=\alpha (y_1^{(1)}-y_2^{(1)})(y_1^{(1)}+y_2^{(1)}-2z^{(1)}+1-u^1_h)+\delta _{1}\\ P_{i}(x,y)=\delta _{i},\quad ( i\ne h)\\ P_{N}(x,y)=\alpha (y_1^{(1)}-y_2^{(1)})(y_1^{(1)}+y_2^{(1)}-2z^{(1)}+u^1_h)+\delta _{N}\\ P_h(x,y)=1-\sum _{k\ne h}P_k(y_1,y_2) \end{array}\right) , \end{aligned}$$

where $y_1^{(1)}$ is the first component of $y_1$.

Note that $\forall (\alpha ,\delta )\in \mathbb {R}^{N+1}$, we have $H_z(\alpha ,\delta )\subset \mathbf {P}(z)$, so $\lambda _{H}(\mathbf {P}(z)\cap H_z(\alpha ,\delta ))>0$. Indeed,

$$\begin{aligned} \begin{array}{lrl} d_{1}^{1}(P_{0}(y_1,y_2))=&{}\left. \frac{\partial P_{0}(y_1,y_2)}{\partial y_1^{(1)}}\right| _{y_1=y_2=z}&{}=\alpha (2y_1^{(1)}\!-\!2z^{(1)}\!+\!1\!-\!u^1_h)|_{y_1=z}=\alpha (1-u^1_h),\\ d_{1}^{1}(P_{i}(y_1,y_2))=&{}\left. \frac{\partial P_{i}(y_1,y_2)}{\partial y_1^{(1)}}\right| _{y_1=y_2=z}&{}=0\quad (i\ne h),\\ d_{1}^{1}(P_{N}(y_1,y_2))=&{}\left. \frac{\partial P_{N}(y_1,y_2)}{\partial y_1^{(1)}}\right| _{y_1=y_2=z}&{}=\alpha (2y_1^{(1)}-2z^{(1)}+u^1_h)|_{y_1=z}=\alpha u^1_h. \end{array} \end{aligned}$$

Substituting the equations above into the first candidate’s FOC we obtain:

$$\begin{aligned} \sum _{\begin{array}{c} l=0\\ l\ne h \end{array}}^N (u_{1}^{l}-u^1_h)d_{1}^{1}(P_{l}(z,z))&=(-u^1_h)d_{1}^{1}(P_{0}(z,z))+(1-u^1_h)d_{1}^{1}(P_{N}(z,z))\\&=\alpha ((-u^1_h)(1-u^1_h)+(1-u^1_h)u^1_h)=0. \end{aligned}$$

Now take arbitrary $P^{*}\in \mathfrak {P}$ and consider $\mathcal {P}(z)\cap (H_z(\alpha ,\delta )+P^{*})$. We state that this intersection is nonempty only for one $\alpha '\in \mathbb {R}$. Indeed, suppose the contrary: let $P',P''\in \mathcal {P}(z)$, $P'\in (H_z(\alpha ',\delta )+P^{*})$, and $P''\in (H_z(\alpha '',\delta )+P^{*})$. By taking the difference between the second player’s FOC for $P'$ and $P''$ at $y_1=y_2=z$ we obtain:

$$\begin{aligned}&\left. \sum _{\begin{array}{c} l=0\\ l\ne h \end{array}}^N (u^2_{N-l}-u^2_{N-h})d_{1}^{1}(P'_{l}(y_1,y_2))- \sum _{\begin{array}{c} l=0\\ l\ne h \end{array}}^N (u^2_{N-l}-u^2_{N-h})d_{1}^{1}(P''_{l}(y_1,y_2)) \right| _{y_1=y_2=z}\\&\quad =\left( (1-u^2_{N-h})d_{1}^{1}(P_{0}(z,z))+(-u^2_{N-h})d_{1}^{1}(P_{N}(z,z))\right) \\&\qquad -\left( (1-u^2_{N-h})d_{1}^{1}(P'_{0}(z,z))+(-u^2_{N-h})d_{1}^{1}(P'_{N}(z,z))\right) \\&\quad =\alpha '((1\!-\!u^2_{N-h})(1\!-\!u^1_h)\!+\!(-u^2_{N-h})u^1_h)\!-\!\alpha ''((1\!-\!u^2_{N-h})(1\!-\!u^1_h) + (-u^2_{N-h})u^1_h)\\&\quad =(\alpha '-\alpha '')(1-u^1_h-u^2_{N-h}). \end{aligned}$$

Since the constant-sum conditions are violated $(1-u^1_h-u^2_{N-h}\ne 0)$ and $d_{1}^{1}(P'_{i}(y_1,y_2))=d_{1}^{1}(P''_{i}(y_1,y_2))=0$, $i>0$, $i<N$, $i\ne h$ it implies that $\alpha '=\alpha ''$. Therefore, for all $P^{*}\in \mathfrak {P}$,

$$\begin{aligned} \lambda _{H}((\mathcal {P}(z)+P^{*})\cap H_z(\alpha ,\delta ))=0. \end{aligned}$$

Proof of Lemma 1

Take $p=(p_1,\ldots ,p_N)$. The idea is to construct an $n$-degree polynomial such that its roots are $\{p_{i}\}_{i\in \mathbf {N}}$. Indeed, consider

$$\begin{aligned} \prod _{i=1}^{n}(p-p_{i})=p^{n}+v_{1}p^{n-1}+\cdots +v_{n-1}p+v_{n} \end{aligned}$$

(16)

Using the Viete’s formulas we obtain:

$$\begin{aligned} \begin{array}{l} v_{1}=-(p_{1}+\cdots +p_{n})\\ v_{2}=(p_{1}x_{2}+\cdots p_{1}p_{n})+(p_{2}p_{3}+p_{2}p_{n})+\cdots +p_{n_{1}}p_{n}\\ \cdots \\ v_{n-1}=(-1)^{n-1}(p_{2}p_{3}\cdots +p_{n}+p_{1}p_{3}\ldots p_{n}+\cdots +p_{1}\ldots p_{n-1})\\ v_{n}=(-1)^{n}p_{1}\ldots p_{n}. \end{array} \end{aligned}$$

We have $v_{n}=(-1)^{n}P_{n}$. Note that

$$\begin{aligned} p_{2}p_{3}\ldots p_{n}\!+\!p_{1}p_{3}\ldots p_{n}\!+\cdots +\!p_{1}\ldots p_{n-1}\!=\!\sum _{l=1}^{n}\!\left( \!\!\left( \prod _{i\ne l}p_{i}\!\right) (1\!-\!p_{l})\!\right) \!+\!np_{1}\ldots p_{n}, \end{aligned}$$

so $v_{n-1}=(-1)^{n-1}(P_{n-1}-nP_{n})$. For an arbitrary $l$ we have:

$$\begin{aligned}&\sum _{S\subseteq \mathbf {N},\;|S|=l}\left( \,\prod _{i\in S}p_{i}\prod _{i\notin S}(1-p_{i})\right) \\&\quad = {\mathop {\mathop {\sum }\limits _{S\subseteq \mathbf {N},}}\limits _{|S|=l}} \left( \prod _{i\in S}p_{i}\right) +\sum _{j=l+1}^{n}\left( (-1)^{j-l}\left( \begin{array}{l} l\\ j \end{array}\right) {\mathop {\mathop {\sum }\limits _{S\subseteq \mathbf {N},}}\limits _{|S|=j}} \left( \prod _{i\in S}p_{i}\right) \right) \end{aligned}$$

Here $\left( \begin{array}{l} l\\ j \end{array}\right)$ denotes the number of $l$ combinations from $j$. Indeed, take a size $j>l$ product of $\{p_{i}\}$: $p_{i_{1}}\ldots p_{i_{i_{j}}}$ and count how many times it occurs in $\sum _{S\subseteq \mathbf {N},\;|S|=l}\left( \prod _{i\in S}p_{i}\prod _{i\notin S}(1-p_{i})\right)$. By fixing the first $l$ multiples in $p_{i_{1}}\ldots p_{i_{i_{j}}}$ we get a unique case, and there’re a $\left( \begin{array}{l} l\\ j \end{array}\right)$ ways of selecting $l$ multiples.

Now it’s easy to write a recurrent formula for $v_{l}$:

$$\begin{aligned} v_{l}=(-1)^{l}\left( P_{l}-\sum _{j=l+1}^{n}\left( (-1)^{j}\left( \begin{array}{l} l\\ j \end{array}\right) v_{j}\right) \right) ,\qquad v_{n}=(-1)^{n}P_{n}. \end{aligned}$$

Since an $n$-degree polynomial can’t have more than $n$ roots we obtain that roots of this polynomial (they exist because the $(P_{0},\ldots ,P_{n})$ vector is proper) form a unique generating set $\{p_{i}\}_{i\in \mathbf {N}}$.

Proof of Lemma 2

First, if an $N$-degree polynomial has $N$ distinct roots then it has a nonzero derivative at each root, therefore the implicit function theorem can be applied for $v_{0},\ldots , v_{N}$ (see proof of Lemma 1). Second, note that the coefficients of polynomial (16) are a non-degenerate linear transformation of $P_{0}^{0},\ldots ,P_{N}^{0}$ (again, see proof of Lemma 1).

Proof of Lemma 3

Note that if $p^a(y_1,y_2)$ and $p^b(y_1,y_2)$ are marginally neutral at convergent positions, then $p^a+p^b$, $p_ap_b$ and $\frac{1}{p_a}$ ($p^a\ne 0$) are also marginally neutral at convergent positions. Now the first statement follows immediately, while the second one also requires the implicit differentiation theorem (applied the same way as in the proof of Lemma 2).

Proof of Lemma 4

Let’s prove that there exists $A\subset \mathbf {\mathtt {P}}(z,z)$ such that $A$ is relatively open in $\mathbf {P}(z,z)$. Take some arbitrary distinct nonzero probabilities $(\hat{p}_{1},\ldots ,\hat{p}_{N})$, for example, $\hat{p}_{1}=\frac{1}{N+1}$,..., $\hat{p}_{N}=\frac{N}{N+1}$ and construct $\hat{P}=(\hat{P}_{0},\ldots ,\hat{P}_{N})$. Clearly, $\hat{P}\in \mathbf {\mathtt {P}}(z,z)$. Take $B_{\varepsilon }(\hat{P})$—$\varepsilon$-neighborhood of $\hat{P}$ in $C^{2}$ metric. Consider $A=\mathbf {P}(z,z)\cap B_{\varepsilon }(\hat{P})$, it is relatively open in $\mathbf {P}(z,z)$. We state that there exists $\varepsilon >0$ such that $A\subset \mathbf {\mathtt {P}}(z,z)$. Indeed, take sufficiently small $\varepsilon$ and apply Lemmas 1, 2 and 3: all of the polynomial (16) roots remain distinct, therefore $p_i(y_1,y_2)$ can be reconstructed for all $y_1,y_2\in X$, $\max |\hat{p}_i-p_i(y_1,y_2)|<\varepsilon$; also $p_i(y_1,y_2)$ are smooth and marginally neutral at convergent positions.

Now lets define $\bar{P}(y_1,y_2)$ as:

$$\begin{aligned} \begin{array}{l} \bar{P}_0(y_1,y_2)=\hat{P}_0+\alpha (||y_1-z||^2-||y_2-z||^2),\\ \bar{P}_i(y_1,y_2)=\hat{P}_i, \quad i=1,\ldots ,N-1,\\ \bar{P}_N(y_1,y_2)=\hat{P}_N+\alpha (-||y_1-z||^2+||y_2-z||^2). \end{array} \end{aligned}$$

We are assuming that $\alpha >0$. Note that we defined $\bar{P}_j$ in such a way that they are marginally neutral at convergent positions and $u^1(y_1,y_2)$ is strictly concave in $y_1$—that’s why $y_1=y_2=z$ the second order conditions for candidate 1 hold. Moreover, if $\alpha$ is small enough $\bar{P}(y_1,y_2)$ is proper (see argument above). Take $B_{\varepsilon }(\bar{P})$—$\varepsilon$-neighborhood of $\bar{P}$ in $C^{2}$ metric. Consider $A=\mathbf {P}(z,z)\cap B_{\varepsilon }(\bar{P})$, which is relatively open in $\mathbf {P}(z,z)$. We state that there exists $\varepsilon >0$ such that $A\subset \overline{\mathcal {P}}(z,z)$. The first order conditions are satisfied by construction, therefore we only need to check that there are no problems with the second order conditions—but it is so because no sufficiently small $C^2$-variation of $\bar{P}$ can undermine strict concavity of $u^1(y_1,y_2)$ in $y_1$.

Proof of Theorem 2

The key part of the argument is borrowed from the proof of Lemma 4. Take $\bar{P}(y_1,y_2)$ as above. Recall that $u^j(y_1,y_2)$ are strictly concave in $y_j$—that’s why $y_1=y_2=z$ is a local Nash equilibrium. Take $B_{\varepsilon }(\bar{P})$—$\varepsilon$-neighborhood of $\bar{P}$ in $C^{2}$ metric. Consider $A=\mathbf {P}(z,z)\cap B_{\varepsilon }(\bar{P})$, which is relatively open in $\mathbf {P}(z,z)$. We state that there exists $\varepsilon >0$ such that $A\subset \overline{\mathcal {P}}(z,z)$. The first order conditions are satisfied by construction, therefore we only need to check that there are no problems with the second order conditions—but it is so because no sufficiently small $C^2$-variation of $\bar{P}$ can undermine strict concavity of $u^j(y_1,y_2)$ in $y_j$. Note that we are not limited to this specific example—the argument stands for any $P$ such that an interior local convergent Nash equilibrium exists.

Proof of Theorem 3

We will take some $\bar{P} \in \mathfrak {P}$ and show that it’s $\varepsilon$-neighborhood is a subset of ${\mathfrak {P}}(u_1,u_2)$.¹² Let’s construct the probability of getting $l$-votes the same way as in the proof of Theorem 2:

$$\begin{aligned} \begin{array}{l} \bar{P}_0(y_1,y_2)=\hat{P}_0+\alpha (||y_1-z||^2-||y_2-z||^2),\\ \bar{P}_i(y_1,y_2)=\hat{P}_i, \quad i=1,\ldots ,N-1,\\ \bar{P}_N(y_1,y_2)=\hat{P}_N+\alpha (-||y_1-z||^2+||y_2-z||^2). \end{array} \end{aligned}$$

As it has been shown above, $(z,z)$ here is a local Nash equilibrium. Let $\mathcal {B}_\varepsilon (z)$ be a small enough closed $\varepsilon$-ball round $z$: since $(z,z)$ is a local equilibrium and $P\in C^2(X)$, $D_j^2(u^j(y_1,y_2))<0$ ¹³ for all $y_j$ in $\mathcal {B}_\varepsilon (z)$.

Now take some $\tilde{P}\in \mathfrak {P}$ such that $\rho _{C^2}(\tilde{P},\bar{P})<\delta$ and choose $\delta$ small enough for $D^2_{j}(u_j(\tilde{P}(y_1,y_2)))<0$ for all $y_j$ in $\mathcal {B}_\varepsilon (z)$.

Restrict both players’ strategy sets to $\mathcal {B}_\varepsilon (z)$. As $u^j$ is concave in $y_j$ for $j=1,2$, there exists a Nash equilibrium $(\tilde{y} _1, \tilde{y} _2)$ in this game. In order to show that this is a local Nash equilibrium for the game with unrestricted strategy sets, we must show that $(\tilde{y} _1, \tilde{y} _2)$ is interior to $\mathcal {B}_\varepsilon (z)$.

Consider both players’ reaction functions: $\tilde{y}_1(y_2)$ and $\tilde{y}_2(y_1)$. Since $u^j$ is strictly concave in $y_j$ the best-reply strategy is unique. Second, it can be shown that if best-reply is unique then $\tilde{y}_1(y_2)$ and $\tilde{y}_2(y_1)$ are continuous. And third, $\forall \varepsilon '>0$ $\exists \delta '>0$ such that if $\rho _{C^2}(\tilde{P},\bar{P})<\delta '$ then $\rho _{C}(\tilde{y}(\cdot ),\bar{y}(\cdot ))<\varepsilon '$, where $\bar{y}_1(\cdot )$ and $\bar{y}_2(\cdot )$ are constant functions. All of these three statements can be easily proved by contradiction. Now take $\varepsilon ' = \varepsilon /4$ and minimum of $\delta$ and $\delta '$—the third statement implies that $(\tilde{y}_1, \tilde{y}_2)$ cannot lie on the boundary of $\mathcal {B}_\varepsilon (z)\times \mathcal {B}_\varepsilon (z)$. Hence, $(\tilde{y} _1, \tilde{y} _2)$ is a local Nash equilibrium for some open neighborhood of ${\mathfrak {P}}(u_1,u_2)$

Proof of Theorem 4

The first claim is a direct implication of the implicit function theorem; let’s turn to the second one. The Jacobian is degenerate if and only if:

$$\begin{aligned} |J(u)|= \left| \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} \frac{\partial ^2 u^1(\hat{y}_1,\hat{y}_2) }{\partial y_1^1 \partial y_1^1}&{}...&{}\frac{\partial ^2 u^1(\hat{y}_1,\hat{y}_2) }{\partial y_1^1 \partial y_1^M}&{}\frac{\partial ^2 u^1(\hat{y}_1,\hat{y}_2) }{\partial y_1^1 \partial y_2^1}&{}...&{}\frac{\partial ^2 u^1(\hat{y}_1,\hat{y}_2) }{\partial y_1^1 \partial y_2^M}\\ \vdots &{} &{}\vdots &{}\vdots &{} &{}\vdots \\ \frac{\partial ^2 u^1(\hat{y}_1,\hat{y}_2) }{\partial y_1^M \partial y_1^1}&{}...&{}\frac{\partial ^2 u^1(\hat{y}_1,\hat{y}_2) }{\partial y_1^M \partial y_1^M}&{}\frac{\partial ^2 u^1(\hat{y}_1,\hat{y}_2) }{\partial y_1^M \partial y_2^1}&{}...&{}\frac{\partial ^2 u^1(\hat{y}_1,\hat{y}_2) }{\partial y_1^M \partial y_2^M}\\ \frac{\partial ^2 u^2(\hat{y}_1,\hat{y}_2) }{\partial y_2^1 \partial y_1^1}&{}...&{}\frac{\partial ^2 u^2(\hat{y}_1,\hat{y}_2) }{\partial y_2^1 \partial y_1^M}&{}\frac{\partial ^2 u^2(\hat{y}_1,\hat{y}_2) }{\partial y_2^1 \partial y_2^1}&{}...&{}\frac{\partial ^2 u^2(\hat{y}_1,\hat{y}_2) }{\partial y_2^1 \partial y_2^M}\\ \vdots &{} &{}\vdots &{}\vdots &{} &{}\vdots \\ \frac{\partial ^2 u^2(\hat{y}_1,\hat{y}_2) }{\partial y_2^M \partial y_1^1}&{}...&{}\frac{\partial ^2 u^2(\hat{y}_1,\hat{y}_2) }{\partial y_2^M \partial y_1^M}&{}\frac{\partial ^2 u^2(\hat{y}_1,\hat{y}_2) }{\partial y_2^M \partial y_2^1}&{}...&{}\frac{\partial ^2 u^2(\hat{y}_1,\hat{y}_2) }{\partial y_2^M \partial y_2^M} \end{array} \right| =0 \end{aligned}$$

Now note that both $\tilde{\mathcal{P}}(\hat{y}_1,\hat{y}_2)$ and $\mathcal{P}(\hat{y}_1,\hat{y}_2)$ are closed (in $C^2$ metric) and therefore Borel; also note that $\mathcal{P}(\hat{y}_1,\hat{y}_2)$ is convex.

Assume that there exist $\bar{P}\in \mathcal { P}(\hat{y}_1,\hat{y}_2)$ and $\bar{\bar{P}}\in \mathcal { P}(\hat{y}_1,\hat{y}_2)$ such that $\bar{P}\ne \bar{\bar{P}}$ and $|J(u(\bar{\bar{P}})-u(\bar{P}))|\ne 0$. Recall that $u$ depend linearly on $P$. Let’s construct (affine) subspace $H$ the following way: $H(\alpha )=\bar{P}+\alpha (\bar{\bar{P}}-\bar{P})$. Clearly, if $\alpha \in [0,1]$ then $H(\alpha )\in \mathcal{P}(y_1,y_2)$. Now let’s take arbitrary $x\in \mathfrak {P}$ and show that $|\{\alpha : x+\alpha (\bar{\bar{P}}-\bar{P})\in \tilde{\mathcal{P}}(\hat{y}_1,\hat{y}_2)\}|$ is finite.

Indeed

$$\begin{aligned} |J(u(x+\alpha (\bar{\bar{P}}-\bar{P})))|&= |J(u(x))+\alpha J(u(\bar{\bar{P}})-u(\bar{P}))|\\&= |J(u(\bar{\bar{P}})\!-\!u(\bar{P}))|\cdot |J^{-1}(u(\bar{\bar{P}})\!-\!u(\bar{P}))J(u(x))\!+\!\alpha E|=0 \end{aligned}$$

But $|J(u(\bar{\bar{P}})-u(\bar{P}))|\ne 0$ and matrix $J^{-1}(u(\bar{\bar{P}})-u(\bar{P}))J(u(x))$ can have only finite number of eigenvalues.¹⁴ Thus $x+\alpha (\bar{\bar{P}}-\bar{P})\in \tilde{\mathcal{P}}(\hat{y}_1,\hat{y}_2)$ only for a finite number of $\alpha$.

We will now show, why our assumption is correct and $\bar{P}, \bar{\bar{P}}$ we need indeed exist. First assume that $\hat{y}_1=\hat{y}_2=z$. Then take:

$$\begin{aligned} \begin{array}{l} \bar{P}_0(y_1,y_2)=\hat{P}_0+\alpha (||y_1-z||^2-||y_2-z||^2),\\ \bar{P}_i(y_1,y_2)=\hat{P}_i, \quad i=1,\ldots ,(N-1),\\ \bar{P}_N(y_1,y_2)=\hat{P}_N+\alpha (-||y_1-z||^2+||y_2-z||^2). \end{array}\\ \begin{array}{l} \bar{\bar{P}}_0(y_1,y_2)=\hat{P}_0+2\alpha (||y_1-z||^2-||y_2-z||^2),\\ \bar{\bar{P}}_i(y_1,y_2)=\hat{P}_i, \quad i=1,\ldots ,(N-1),\\ \bar{\bar{P}}_N(y_1,y_2)=\hat{P}_N+2\alpha (-||y_1-z||^2+||y_2-z||^2). \end{array} \end{aligned}$$

It can now be easily verified that

$$\begin{aligned} |J(u(\bar{\bar{P}}(z,z))-u(\bar{P}(z,z)))|= \left| \begin{array}{c|c} -\alpha E &{} 0\\ \hline 0&{}-\alpha E \end{array} \right| \ne 0. \end{aligned}$$

Now let $\hat{y}_1\ne \hat{y}_2$. Take such small $\varepsilon >0$ that no convergent positions lie inside $\varepsilon$-neighborhood of $\hat{y}_1, \hat{y}_2$. Define $\bar{P},\bar{\bar{P}}$ in $\varepsilon$-neighborhood of $\hat{y}_1, \hat{y}_2$ the following way:

$$\begin{aligned} \begin{array}{l} \bar{P}_0(y_1,y_2)=\hat{P}_0+\alpha (||y_1-\hat{y}_1||^2-||y_2- \hat{y}_2||^2),\\ \bar{P}_i(y_1,y_2)=\hat{P}_i, \quad i=1,\ldots ,(N-1),\\ \bar{P}_N(y_1,y_2)=\hat{P}_N+\alpha (-||y_1-\hat{y}_1||^2+||y_2- \hat{y}_2||^2). \end{array}\\ \begin{array}{l} \bar{\bar{P}}_0(y_1,y_2)=\hat{P}_0+2\alpha (||y_1-\hat{y}_1||^2-||y_2- \hat{y}_2||^2),\\ \bar{\bar{P}}_i(y_1,y_2)=\hat{P}_i, \quad i=1,\ldots ,(N-1),\\ \bar{\bar{P}}_N(y_1,y_2)=\hat{P}_N+2\alpha (-||y_1-\hat{y}_1||^2+||y_2- \hat{y}_2||^2). \end{array} \end{aligned}$$

Outside of $\varepsilon$-neighborhood of $\hat{y}_1, \hat{y}_2$ $\bar{P}$ $\bar{\bar{P}}$ can be continued (in a $C^2$ class) in an arbitrary way, only the marginal neutrality at convergent positions has to be satisfied.

Now we can see that $|J(u(\bar{\bar{P}}(\hat{y}_1, \hat{y}_2))-u(\bar{P}(\hat{y}_1, \hat{y}_2)))|\ne 0$ for the same reason as above.