A theory of unanimous jury voting with an ambiguous likelihood

Abstract

We examine collective decision-making in a jury voting game under the unanimity rule when voters have ambiguous beliefs. Unlike in existing studies (Ellis in Theoretical Economics 11:865–895, 2016; Fabrizi et al., in: AUT Economics Working Paper, 2021; Ryan in Theory and Decision 90:543–577, 2021), the locus of ambiguity is the likelihood function (signal precision) rather than the prior. This significantly alters the properties of symmetric equilibria. While prior ambiguity may induce multiple equilibria (Fabrizi et al., in: AUT Economics Working Paper, 2021; Ryan in Theory and Decision 90:543–577, 2021) we show that, under likelihood ambiguity, there exists a unique non-trivial symmetric responsive equilibrium that takes the same form as in the absence of ambiguity (Feddersen and Pesendorfer in The American Political Science Review 92:23–35, 1998). Moreover, likelihood ambiguity partially offsets the pernicious effects of pivotality on decision quality: the frequency of Type I error (convicting the innocent) is typically lower than in the absence of ambiguity.

Appendices

Appendix

Proofs

Proof of Proposition 1

Let $\sigma _{a}^{*}=\sigma _{b}^{*}=x$. If $x=0$ we already observed that $\sigma ^{*}$ is an equilibrium. Suppose $x>0$. Then $\rho ^{*} _{a}=\rho ^{*} _{b}=x^{N}>0$ and $V_{t}$ is bilinear for each $t\in T$. By the minimax theorem, it follows that if $\sigma ^{*}$ is an equilibrium then we must have $x\in \Sigma _{a}\left( r^{*}\right)$ for some $r^{*}\in [{\underline{r}},{\overline{r}}]$. Since $x>0$ and $V_{a}\left( \sigma _{a}^{i},r^{*}\right)$ is linear in $\sigma _{a}^{i}$, this means that the coefficient on $\sigma _{a}^{i}$ must be non-negative. That is:

$$\begin{aligned}&-r^{*}q\rho ^{*} _{a}+(1-r^{*})(1-q)\rho ^{*} _{b}\ \ge \ 0\nonumber \\&\Leftrightarrow \ \ \ (1-r^{*})(1-q)\ \ge \ r^{*}q \end{aligned}$$

(7)

where we have used the fact that $\rho ^{*} _{a}=\rho ^{*} _{b}>0$. But inequality (7) cannot hold, since $r^{*}>\frac{1}{2}$ and $q\ge \frac{1}{2}$. $\square$

Proof of Proposition 2

The fact that $h(x)=0$ iff (4) follows by straightforward calculation.

Suppose $\sigma ^{*}=\left( \sigma _{a}^{*},\sigma _b^*\right)$ is a responsive equilibrium. Since each $V_t$ is linear in $\sigma ^{i}_{t}$ and

$$\begin{aligned} \frac{\partial V_{b}}{\partial \sigma ^{i}_{b}}= & {} (1-q)x\rho ^{*}_{b}-q(1-x)\rho ^{*}_{a}\\\ge & {} (1-q)(1-x)\rho ^{*}_{b}-qx\rho ^{*}_{a} \ \ =\ \ \frac{\partial V_{a}}{\partial \sigma ^{i}_{a}} \end{aligned}$$

(recalling that $r>\frac{1}{2}$) we have $\sigma _{a}^{*}\le \sigma _{b}^{*}$. Responsiveness therefore implies $\sigma ^{*}_{a}<1$. Similarly, responsiveness excludes $\sigma ^{*}=(0,0)$ so $0<\rho _{a}^{*}<\rho _{b}^{*}$ and therefore

$$\begin{aligned} \frac{\partial {V_a}}{\partial {\sigma ^{i}_{a}}}\ <\ \frac{\partial {V_b}}{\partial {\sigma ^{i}_{b}}}. \end{aligned}$$

It follows that if $\sigma ^{*}_{b}<1$ then $\sigma ^{*}_{a}=0$. However, Assumption 1 ensures

$$\begin{aligned} \frac{\partial {V_b}}{\partial {\sigma ^{i}_{b}}}\ >\ 0 \end{aligned}$$

when $\sigma ^{*}_{a}=0$, so $\sigma ^{*}_{b}=1$ in any responsive equilibrium. In other words, any responsive equilibrium is an FP profile.

Since $\sigma ^{*}_{a}<\sigma ^{*}_{b}=1$ we must have:

$$\begin{aligned} \frac{\rho ^{*}_{a}}{\rho ^{*}_{b}}\ \ge \ \frac{(1-q)(1-x)}{xq} \quad\Leftrightarrow & {} \quad \frac{x\sigma _{a}^{*}+(1-x)}{(1-x)\sigma _{a}^{*}+x}\ \ge \ A(x) \\&\\\Leftrightarrow & {} \ \ \ \left[ x(1+A(x))-A(x)\right] \sigma _{a}^{*}\ \ge \ x(1+A(x))-1. \end{aligned}$$

Now observe that

$$\begin{aligned} x(1+A(x))-A(x)>0\ \ \ \Leftrightarrow \ \ \ A(x)<\frac{x}{1-x} \end{aligned}$$

which is evidently true, and verifies (3). Therefore:

$$\begin{aligned} \sigma _{a}^{*}\ \ge \ \frac{x(1+A(x))-1}{x(1+A(x))-A(x)}. \end{aligned}$$

The result now follows directly.

$\square$

Proof of Lemma 1

Suppose that $\sigma ^{*}$ is an equilibrium and $\sigma _{a}^{*}> \sigma _{b}^{*}$. Note that $V_a$ is linear in $\sigma ^{i}_{a}$ with slope equal to:

$$\begin{aligned} (1-q)\left( 1-r\right) \rho _{b}^{*}-qr\rho _{a}^{*}. \end{aligned}$$

Since $\sigma _{a}^{*}> \sigma _{b}^{*}$ it follows that $\rho _{a}^{*}> \rho _{b}^{*}$ for any $r>\frac{1}{2}$, so $V_a$ is strictly decreasing in $\sigma ^{i}_a$ for any $r>\frac{1}{2}$. Hence

$$\begin{aligned} \min _{r\in \left[ {\underline{r}},{\overline{r}}\right] }\ V_{a}\left( \sigma _{a}^{i},r\right) \end{aligned}$$

is strictly decreasing in $\sigma ^{i}_a$ which means that

$$\begin{aligned} \left\{ 0\right\} \ =\ \arg \max _{\sigma _{a}^{i}\in [0,1]}\ \left[ \min _{r\in [{\underline{r}},{\overline{r}} ]}V_{a}\left( \sigma _{a}^{i},r\right) \right] . \end{aligned}$$

This contradicts the assumption $\sigma ^{*}$ is an equilibrium, since $\sigma ^{*}_{a}>0$. $\square$

Proof of Lemma 2

Given Assumption 1, if $\sigma ^{*}=\left( 0,\sigma _{b}^{*}\right)$ with $\sigma _{b}^{*}>0$ then it is straightforward to verify (and intuitive) that $V_{b}\left( \sigma ^{i}_{b},r\right)$ is strictly increasing in $\sigma ^{i}_{b}$ for any $r\in [{\underline{r}},{\overline{r}}]$. Hence, the lower envelope

$$\begin{aligned} \min _{r\in [{\underline{r}},{\overline{r}}]}\ V_{b}\left( \sigma _{b}^{i},r\right) \end{aligned}$$

is also strictly increasing in $\sigma ^{i}_{b}$. $\square$

Proof of Lemma 3

If $\sigma ^{*}$ is non-responsive the claim follows by Proposition 1.

Let us therefore assume that $\sigma ^{*}$ is responsive. Hence $\sigma ^{*}_{a}<\sigma ^{*}_{b}$ by Lemma 1. Note that $V_{b}$ is a linear function of $\sigma ^{i}_{b}$ whose slope is strictly increasing in r (as a direct calculation will easily verify).

If $V_{b}$ is a non-decreasing function of $\sigma ^{i}_{b}$ when $r={\underline{r}}$ it follows that $V_{b}$ is strictly increasing in $\sigma ^{i}_{b}$ for any $r>{\underline{r}}$ and therefore the lower envelope

$$\begin{aligned} \min _{r\in [{\underline{r}},{\overline{r}}]}\ V_{b}\left( \sigma _{b}^{i},r\right) \end{aligned}$$

is a strictly increasing function of $\sigma ^{i}_{b}$. In this case

$$\begin{aligned} \{1\}\ =\ \arg \max _{\sigma _{b}^{i}\in [0,1]}\ \left[ \min _{r\in [{\underline{r}},{\overline{r}}]}V_{b}\left( \sigma _{b}^{i},r\right) \right] . \end{aligned}$$

Since $\sigma ^{*}$ is an equilibrium, we must therefore have $\sigma ^{*}_{b}=1$.

It remains to consider the case in which $V_{b}$ is strictly decreasing in $\sigma ^{i}_{b}$ when $r={\underline{r}}$. It follows that $V_{a}$ (which, recall, is linear in $\sigma ^{i}_{a}$) is also strictly decreasing in $\sigma ^{i}_{a}$ when $r={\underline{r}}$ since¹⁵

$$\begin{aligned} \frac{\partial {V_a}}{\partial {\sigma ^{i}_{a}}}\ \le \ \frac{\partial {V_b}}{\partial {\sigma ^{i}_{b}}} \end{aligned}$$

for any $r\in \left( \frac{1}{2},1\right)$. We will show that this implies $\sigma ^{*}_{a}=0$. First, note that $\partial V_{a}/ \partial \sigma ^{i}_{a}$ is a linear function of $\sigma ^{i}_{a}$ whose vertical intercept is strictly increasing in r. If it is downward sloping at ${\underline{r}}$, then the lower envelope

$$\begin{aligned} \min _{r\in [{\underline{r}},{\overline{r}}]}\ V_{a}\left( \sigma _{a}^{i},r\right) \end{aligned}$$

is a strictly decreasing function of $\sigma ^{i}_{a}$. Since $\sigma ^{*}$ is an equilibrium, we must have $\sigma ^{*}_{a}=0$ as claimed. $\square$

Proof of Lemma 4

Observe that

$$\begin{aligned} \frac{\partial V_a\left( \sigma _{a}^{i},r\right) }{\partial r} \ =\ -q\rho ^{*}_{a}{\sigma _{a}^{i }}+(1-q)(1-\rho ^{*}_{b}{\sigma _{a}^{i})-r}q{\sigma _{a}^{i}}\frac{\partial \rho ^{*}_{a}}{\partial r}+(1-q)(1-r){\sigma _{a}^{i}} \frac{\partial \rho ^{*}_{b}}{\partial r} \end{aligned}$$

(8)

Since $\sigma _{a}^{*}<1=\sigma ^{*}_{b}$ we have

$$\begin{aligned}\frac{\partial \rho ^{*}_{b}}{\partial r}>0>\frac{\partial \rho ^{*}_{a}}{\partial r}.\end{aligned}$$

Therefore, given $\sigma _{a}^{i}\ge 0$, $q\in \left[ \frac{1}{2}, 1\right)$ and $r\in \left( \frac{1}{2}, 1\right)$, it follows that (8) is strictly positive if

$$\begin{aligned} -q\left[ \rho ^{*}_{a}{\sigma _{a}^{i}}+{r}{\sigma _{a}^{i}}\frac{\partial \rho ^{*}_{a}}{\partial r}\right] \ \ge \ 0\text {.} \end{aligned}$$

This obviously holds when $\sigma ^{i}_{a}=0$, and when $\sigma ^{i}_{a}>0$ it is equivalent to

$$\begin{aligned} \rho ^{*}_{a}+{r}\frac{\partial \rho ^{*}_{a}}{\partial r}\ \le \ 0. \end{aligned}$$

(9)

This says that the probability of Type I error, when i votes to convict following signal $t_{i}=a$, is weakly decreasing in signal precision. Note that

$$\begin{aligned}\frac{\partial \rho ^{*}_{a}}{\partial r}<0\end{aligned}$$

when $\sigma ^{*}_{a}<1$. Using $\sigma ^{*}_{b}=1$, condition (9) may be expressed as follows:

$$\begin{aligned} \sigma _{a}^{*}\ \le \ \frac{(N+1)r-1}{(N+1)r}. \end{aligned}$$

Since $r>\frac{1}{2}$ we have

$$\begin{aligned} \frac{(N+1)r-1}{(N+1)r}>\frac{N-1}{N+1}\ge 0 \end{aligned}$$

(10)

so it suffices to show that

$$\begin{aligned} \frac{x(1+A(x))-1}{x(1+A(x))-A(x)}\ \le \ \frac{N-1}{N+1}. \end{aligned}$$

(11)

Recall that A(x) depends on q. It is easily verified that he left-hand side of ( 11) is weakly decreasing in q so if the inequality holds for $q= \frac{1}{2}$, then it holds for any $q\ge \frac{1}{2}$. Evaluating the left-hand side at $q=\frac{1}{2}$, we get

$$\begin{aligned} \frac{x\left( \frac{1-x}{x}\right) ^{1/N}-\left( 1-x\right) }{x-(1-x)\left( \frac{1-x}{x}\right) ^{1/N}}\ \le \ \frac{N-1}{N+1}. \end{aligned}$$

(12)

It is straightforward to verify that for $N>1$ the left-hand side of (12) is strictly decreasing in $x\in \left( \frac{1}{2},1\right)$. Using l’Hôpital’s rule,

$$\begin{aligned}\lim _{x\downarrow \frac{1}{2}}\ \ \frac{x\left( \frac{1-x}{x}\right) ^{1/N}-\left( 1-x\right) }{x-(1-x)\left( \frac{1-x}{x}\right) ^{1/N}}\ =\ \frac{N-1}{N+1} \end{aligned}$$

so (12) holds for any $x\in \left( \frac{1}{2},1\right)$. $\square$

Proof of Theorem 1

We first show that $\sigma ^{*}=\left( h\left( {\underline{r}}\right) ,1\right)$ is an equilibrium. Using Proposition 2 and Lemma 4 we have

$$\begin{aligned} \sigma _{a}^{*} \ \in \ \arg \max _{\sigma _{a}^{i}\in [0,1]}\ V_{a}\left( \sigma _{a}^{i},{\underline{r}}\right) \ =\ \arg \max _{\sigma _{a}^{i}\in [0,1]}\ \left[ \min _{r\in [{\underline{r}},{\overline{r}}]}V_{a}\left( \sigma _{a}^{i},r\right) \right] . \end{aligned}$$

It remains to show that

$$\begin{aligned} 1 \ \in \ \arg \max _{\sigma _{b}^{i}\in [0,1]}\ \left[ \min _{r\in [{\underline{r}},{\overline{r}}]}V_{b}\left( \sigma _{b}^{i},r\right) \right] . \end{aligned}$$

(13)

If $\sigma ^{*}_{a}=0$ (i.e., ${\underline{r}}\le {\hat{r}}$) this follows from Lemma 2.

Suppose ${\underline{r}}>{\hat{r}}$ so that $\sigma _{a}^{*}>0$. The fact that $\sigma _{a}^{*}$ maximizes $V_{a}\left( \sigma _{a}^{i},{\underline{r}}\right)$ therefore implies

$$\begin{aligned}\{ 1\} \ =\ \arg \max _{\sigma _{b}^{i}\in [0,1]}\ V_{b}\left( \sigma _{b}^{i},{\underline{r}}\right) .\end{aligned}$$

By straightforward calculation

$$\begin{aligned} \frac{\partial ^{2}V_{b}\left( \sigma ^{i}_{b},r\right) }{\partial \sigma ^{i}_{b}\partial r}\ \ge \ 0\ \ \text {for any}\, r\in \left( \frac{1}{2},1\right) \, \text {and any}\, \sigma ^{i}_{b}\in (0,1) \end{aligned}$$

from which we deduce

$$\begin{aligned}1 \ \in \ \arg \max _{\sigma _{b}^{i}\in [0,1]}\ V_{b}\left( \sigma _{b}^{i},r\right) \ \ \text {for any }\, r\in [{\underline{r}},{\overline{r}}]\end{aligned}$$

and (13) follows.

This establishes that $\left( h\left( {\underline{r}}\right) ,1\right)$ is an equilibrium. We next rule out the existence of a responsive equilibrium that is not an FP profile. By Proposition 1 and Lemma 1, a non-trivial equilibrium $\sigma ^{*}$ must be responsive with $\sigma ^{*}_{b}>\sigma ^{*}_{a}$. By Lemma 3, it must also have $\sigma ^{*}_{b}=1$ or $\sigma ^{*}_{a}=0$. In the former case $\sigma ^{*}$ is obviously an FP profile, and in the latter it is an FP profile by Corollary 1.

Finally, we must show that $\left( h\left( {\underline{r}}\right) ,1\right)$ is the only FP profile that is an equilibrium.

Let $\sigma ^{*}$ be an FP profile that is an equilibrium. We will prove that $\sigma ^{*}_{a}= h\left( {\underline{r}}\right)$.

First, some preliminaries:

• For each $r\in \left( \frac{1}{2},1\right)$, define $g_{r}:\left[ 0,1\right] \rightarrow {\mathbb {R}}$ by

  $$\begin{aligned} g_{r}\left( \sigma _{a}^{i}\right) =V_{a}\left( \sigma _{a}^{i},r\right) \end{aligned}$$

  so that

  $$\begin{aligned} \min _{r\in \left[ {\underline{r}},{\overline{r}}\right] }\ V_{a}\left( \sigma _{a}^{i},r\right) \ =\ \min _{r\in \left[ {\underline{r}},\overline{r}\right] }\ g_{r}\left( \sigma _{a}^{i}\right) \qquad \qquad \qquad \qquad \qquad (*) \end{aligned}$$

  is the lower envelope of the $g_{r}$ functions. Recall that each $g_{r}$ is linear, $g_{r}\left( 0\right)$ is strictly increasing in r, and the mapping $r\mapsto g_{r}^{\prime }$ is continuous.

• From the definition of h in (the proof of) Proposition 2 we observe the following: if $0<\sigma _{a}^{*}<1$ then $\sigma _{a}^{*}=h\left( r\right)$ iff $V_{a}\left( \sigma _{a}^{i},r\right)$ is constant in $\sigma _{a}^{i}$ (i.e., $g_{r}^{\prime }=0$); and if $\sigma _{a}^{*}=0$ then $\sigma _{a}^{*}=h\left( r\right)$ iff $V_{a}\left( \sigma _{a}^{i},r\right)$ is non-increasing in $\sigma _{a}^{i}$ (i.e., $g_{r}^{\prime }\le 0$).

We start by proving that $\sigma ^{*}=\left( h\left( r\right) ,1\right)$ for some $r\in \left( \frac{1}{2},1\right)$ and then use Lemma 4 to show that this implies $\sigma ^{*}=\left( h\left( {\underline{r}}\right) ,1\right)$. To prove that $\sigma ^{*}=\left( h\left( r\right) ,1\right)$ for some $r\in \left( \frac{1}{2},1\right)$, we argue by contradiction.

Suppose $\sigma ^{*}\ne \left( h\left( r\right) ,1\right)$ for all $r\in \left( \frac{1}{2},1\right)$.

If $\sigma ^{*}=\left( 0,1\right)$ then $g_{{\underline{r}}}^{\prime }\le 0$, since otherwise ($*$) would be upward sloping in a neighbourhood of 0 (recall the three properties of $g_{r}$ noted above) so $\sigma _{a}^{i}=0$ could not maximise ($*$). Hence $\sigma _{a}^{*}=h\left( {\underline{r}} \right)$, which is the desired contradiction.

If $\sigma _{a}^{*}>0$, then $g_{r}^{\prime }\ne 0$ for all $r\in \left( \frac{1}{2},1\right)$, since $\sigma _{a}^{*}\ne h\left( r\right)$ for all $r\in \left( \frac{1}{2},1\right)$. The mapping $r\mapsto g_{r}^{\prime }$ is continuous, so the sign of $g_{r}^{\prime }$ must therefore be the same for all r. If $g_{r}^{\prime }<0$ for all r, we must have $\sigma _{a} ^{i}=0$ to maximise ($*$). This contradicts the facts that $\sigma _{a}^{*}>0$ and $\sigma ^{*}$ is an equilibrium. If $g_{r}^{\prime }>0$ for all r, then $\sigma _{a}^{i}=1$ is the unique maximiser of ($*$), which means that $\sigma ^{*}=\left( 1,1\right)$. This contradicts the fact that $\sigma ^{*}$ is an FP profile.

We have therefore shown that $\sigma ^{*}=\left( h\left( r\right) ,1\right)$ for some $r\in \left( \frac{1}{2},1\right)$. The proof of Lemma 4 implies that $\partial V_{a}/\partial r<0$ everywhere on the domain of $V_{a}$ so

$$\begin{aligned} \min _{r\in \left[ {\underline{r}},{\overline{r}}\right] }\ V_{a}\left( \sigma _{a}^{i},r\right) \ \equiv \ V_{a}\left( \sigma _{a}^{i},{\underline{r}} \right) \text {.} \end{aligned}$$

It follows that $\sigma _{a}^{*}$ maximises $V_{a}\left( \sigma _{a} ^{i},{\underline{r}}\right)$. Since $\sigma _{a}^{*}<1$ this means $g_{{\underline{r}}}^{\prime }\le 0$, with equality if $\sigma _{a}^{*}>0$, so $\sigma _{a}^{*}=h\left( {\underline{r}}\right)$.

This completes the proof. $\square$

Proof of Lemma 5

When $h(x)>0$ we have

$$\begin{aligned} h'(x)=\frac{1-A(x)^2+A'(x)(2x-1)}{[x(1+A(x))-A(x)]^2}. \end{aligned}$$

For Lemma 5 to hold, it suffices to show that the numerator of this expression is strictly decreasing. This implies that if $h^{\prime }(x) = 0$ at some x with $h(x)>0$, then $h^{\prime }(x^{*}))> 0$ for any $x^{*}<x$ with $h(x^{*})>0$. By direct calculation we have:

$$\begin{aligned} \frac{d(1-A(x)^2+A'(x)(2x-1))}{dx}=-2 A(x) A'(x)+2 A'(x)+(2 x-1) A''(x) \end{aligned}$$

where

$$\begin{aligned} A'(x)=-\frac{1}{N}\frac{A(x)}{x(1-x)} \end{aligned}$$

and

$$\begin{aligned} A''(x)=-\frac{1}{N}\frac{A'(x)x(1-x)-A(x)(1-2x)}{(x(1-x))^2}=\frac{1}{N}\frac{\frac{1}{N}A(x)+A(x)(1-2x)}{(x(1-x))^2}. \end{aligned}$$

Substituting $A'(x)=-\frac{1}{N}\frac{A(x)}{x(1-x)}$ and $A''(x)=\frac{1}{N}\frac{\frac{1}{N}A(x)+A(x)(1-2x)}{(x(1-x))^2}$ and simplifying gives

$$\begin{aligned} \gamma _1(x)\equiv & {} \frac{d(1-A(x)^2+A'(x)(2x-1))}{dx}\\= & {} -\frac{A(x) (2 N (x-1) x A(x)+2 N (x-1) x+N-2 x+1)}{N^2 (x-1)^2 x^2}. \end{aligned}$$

Then $\gamma _1(x)<0$ if and only if $-(2 N (x-1) x A(x)+2 N (x-1) x+N-2 x+1)<0$ or, equivalently, if and only if $\gamma _2(x)\equiv 2N(1-x)x\left( A(x)+1\right) -N+2x-1<0$.

We will show that $d\gamma _2(x)/dN<0$ and $d\gamma _2(x)/dq<0$ for $N\ge 1$, $q\in (\frac{1}{2},1)$ and $x\in (\frac{1}{2},1)$, and that $\lim _{q\downarrow \frac{1}{2}}\gamma _2(x)=0$ for $N=1$. These results imply that $\gamma _2(x)<0$ for all $N\ge 1$, $q\in (\frac{1}{2},1)$ and $x\in (\frac{1}{2},1)$. Note that

$$\begin{aligned} dA(x)/dN=-\frac{A(x) \log \left( \frac{(1-q) (1-x)}{q x}\right) }{N^2} \end{aligned}$$

so that

$$\begin{aligned} \frac{d\gamma _2(x)}{dN}&=2(1-x)x(A(x)+1)-1+2N(1-x)x\frac{dA(x)}{dN}\\&=2(1-x)x\left( A(x)+1+N\frac{dA(x)}{dN}\right) -1\\&=2(1-x)x\left( 1+A(x)\left[ 1-\frac{\log \left( \frac{(1-q) (1-x)}{q x}\right) }{N}\right] \right) -1 \end{aligned}$$

Thus, $\frac{d\gamma _2(x)}{dN}<0$ if and only if

$$\begin{aligned} f(N)\equiv \left( \frac{(1-q) (1-x)}{q x}\right) ^\frac{1}{N}\left[ 1-\frac{\log \left( \frac{(1-q) (1-x)}{q x}\right) }{N}\right] <\frac{1-2(1-x)x}{2(1-x)x}. \end{aligned}$$

Note that

$$\begin{aligned} f'(N)=\frac{\left( \frac{(1-q) (1-x)}{q x}\right) ^\frac{1}{N}\left[ \log \left( \frac{(1-q) (1-x)}{q x}\right) \right] ^2}{N^3}>0 \end{aligned}$$

and

$$\begin{aligned} \lim _{N\rightarrow +\infty }\left( \frac{(1-q) (1-x)}{q x}\right) ^\frac{1}{N}\left[ 1-\frac{\log \left( \frac{(1-q) (1-x)}{q x}\right) }{N}\right] =1 \end{aligned}$$

Note also that $2x(1-x)\in (0,\frac{1}{2})$ for all $x\in (\frac{1}{2},1)$ so that

$$\begin{aligned} \frac{1-2(1-x)x}{2(1-x)x}>1 \end{aligned}$$

for any $x\in (\frac{1}{2},1)$. It follows that $\frac{d\gamma _2(x)}{dN}<0$.

Next, for $N\ge 1$, $q\in (\frac{1}{2},1)$, and $x\in (\frac{1}{2},1)$,

$$\begin{aligned} \frac{d\gamma _2(x)}{dq}=-\frac{\left( \frac{(1-q) (1-x)}{q x}\right) ^{1/N}}{N (1-q) q}<0. \end{aligned}$$

so that $\gamma _2(x)$ is strictly decreasing in q for $N\ge 1$, $q\in (\frac{1}{2},1)$, and $x\in (\frac{1}{2},1)$.

Finally, because $\gamma _2(x)$ decreases strictly in N and q, we only need to check whether $\gamma _2(x)= 0$ for $q=\frac{1}{2}$ and $N=1$. Indeed, evaluating $\gamma _2(x)$ at $q=\frac{1}{2}$ and $N=1$ gives $\gamma _2(x)=0$. Hence, for all $N\ge 1$ and all $q\in (\frac{1}{2},1)$, the numerator of $h'(x)$ is decreasing. $\square$