Robust Trust

## A. 6 Proof of Proposition 1

By Lemma 1 and Corollary 2, the choice of an optimal strategy for the agent reduces to optimization over the extreme points $\underline{\mu}, \bar{\mu}$ of the trust interval with the corresponding payoff:

$$
\begin{aligned}
& V(\underline{\mu}, \bar{\mu}) \triangleq \\
& \alpha\left(\int_{0}^{\underline{\mu}}\left(U(\underline{\mu})+U^{\prime}(\underline{\mu})(\mu-\underline{\mu})\right) \tau(\mu) d \mu+\int_{\underline{\mu}}^{\bar{\mu}} U(\mu) \tau(\mu) d \mu+\int_{\bar{\mu}}^{1}\left(U(\bar{\mu})+U^{\prime}(\bar{\mu})(\mu-\bar{\mu})\right) \tau(\mu) d \mu\right) \\
& \quad+(1-\alpha)\left(\int_{0}^{b(\underline{\mu}, \bar{\mu})}\left(U(\bar{\mu})+U^{\prime}(\bar{\mu})(\mu-\bar{\mu})\right) \tau(\mu) d \mu+\int_{b(\underline{\mu}, \bar{\mu})}^{1}\left(U(\underline{\mu})+U^{\prime}(\underline{\mu})(\mu-\underline{\mu})\right) \tau(\mu) d \mu\right)
\end{aligned}
$$

The function $V(\underline{\mu}, \bar{\mu})$ is continuously differentiable with partial derivatives (whenever

$$
\underline{\mu}<\bar{\mu}):
$$

$$
\begin{aligned}
& \frac{\partial V}{\partial \underline{\mu}}=U^{\prime \prime}(\underline{\mu})\left(\alpha \int_{0}^{\underline{\mu}}(\mu-\underline{\mu}) \tau(\mu) d \mu+(1-\alpha) \int_{b(\underline{\mu}, \bar{\mu})}^{1}(\mu-\underline{\mu}) \tau(\mu) d \mu\right) \\
& \frac{\partial V}{\partial \bar{\mu}}=U^{\prime \prime}(\bar{\mu})\left(\alpha \int_{\bar{\mu}}^{1}(\mu-\bar{\mu}) \tau(\mu) d \mu+(1-\alpha) \int_{0}^{b(\underline{\mu}, \bar{\mu})}(\mu-\bar{\mu}) \tau(\mu) d \mu\right)
\end{aligned}
$$

Intuitively, the first-order impact of a change in the trust boundary equals the change in the action played at that boundary, measured by $U^{\prime \prime}(\cdot)$, integrated over the belief regions in which the aligned and misaligned advisers induce that action, weighted by the alignment parameter. (Terms involving $\partial b / \partial \underline{\mu}$ and $\partial b / \partial \bar{\mu}$ vanish because at $\mu=b(\underline{\mu}, \bar{\mu})$ the misaligned adviser is indifferent between the two messages.)

Whenever the trust region is non-singleton, $\underline{\mu}<\bar{\mu}$, at the optimal choice of $\underline{\mu}$ and $\bar{\mu}$ these partial derivatives must equal zero, $\partial V / \partial \underline{\mu}=0$ and $\partial V / \partial \bar{\mu}=0$. Since $U^{\prime \prime}(\cdot)>0$, these first-order conditions can be rearranged as follows. Define functions $\Psi_{1}$ and $\Psi_{2}$ as

$$
\begin{aligned}
& \Psi_{1}(\underline{\mu}, \bar{\mu}, \alpha) \triangleq \alpha \int_{0}^{\underline{\mu}}(\mu-\underline{\mu}) \tau(\mu) d \mu+(1-\alpha) \int_{b(\underline{\mu}, \bar{\mu})}^{1}(\mu-\underline{\mu}) \tau(\mu) d \mu \\
& \Psi_{2}(\underline{\mu}, \bar{\mu}, \alpha) \triangleq \alpha \int_{\bar{\mu}}^{1}(\mu-\bar{\mu}) \tau(\mu) d \mu+(1-\alpha) \int_{0}^{b(\underline{\mu}, \bar{\mu})}(\mu-\bar{\mu}) \tau(\mu) d \mu
\end{aligned}
$$

Then, $\Psi_{1}(\underline{\mu}, \bar{\mu}, \alpha)=\Psi_{2}(\underline{\mu}, \bar{\mu}, \alpha)=0$ is equivalent to conditions (6) and (7).
First, we show that conditions (6) and (7) are incompatible with $\alpha<1 / 2$. (If $M$ were finite, this would follow directly from Theorem 3.) Indeed, if those conditions hold then (for the rest of the proof, we will often omit the arguments of the function $b$ for brevity):

$$
\begin{aligned}
\alpha\left(\int_{0}^{b}(b-\mu) \tau(\mu) d \mu+\int_{b}^{1}(\mu-b) \tau(\mu) d \mu\right) & \geq \alpha\left(\int_{0}^{\underline{\mu}}(\underline{\mu}-\mu) \tau(\mu) d \mu+\int_{\bar{\mu}}^{1}(\mu-\bar{\mu}) \tau(\mu) d \mu\right) \\
& =(1-\alpha)\left(\int_{0}^{b}(\bar{\mu}-\mu) \tau(\mu) d \mu+\int_{b}^{1}(\mu-\underline{\mu}) \tau(\mu) d \mu\right) \\
& \geq(1-\alpha)\left(\int_{0}^{b}(b-\mu) \tau(\mu) d \mu+\int_{b}^{1}(\mu-b) \tau(\mu) d \mu\right)
\end{aligned}
$$

where the inequalities hold because $\underline{\mu} \leq b(\underline{\mu}, \bar{\mu}) \leq \bar{\mu}$ and the equality is a consequence of (6) and (7). Because $\tau$ has full support, the multipliers on both sides of the inequality are
strictly positive, and thus $\alpha \geq 1-\alpha$, i.e., $\alpha \geq 1 / 2$.
Now we argue that for $\alpha \geq 1 / 2$ the solution to (6) and (7) such that $\underline{\mu} \leq \bar{\mu}$ exists. Note that at $\alpha=1 / 2,[\underline{\mu}, \bar{\mu}]=\left[\mu_{0}, \mu_{0}\right]$ is a solution. For the rest of this proof, we omit the dependence of $\Psi_{i}$ on $\alpha$. By Lemma 1 and direct inspection, $b(\underline{\mu}, \bar{\mu})$ is strictly and continuously increasing in its arguments, so $\Psi_{1}(\underline{\mu}, \bar{\mu})$ is strictly and continuously decreasing in $\underline{\mu}$ for each $\bar{\mu}$. Furthermore,

$$
\begin{aligned}
& \Psi_{1}(0, \bar{\mu})=(1-\alpha) \int_{b(0, \bar{\mu})}^{1} \mu \tau(\mu) d \mu \geq 0 \\
& \Psi_{1}(1, \bar{\mu})=\alpha \int_{0}^{1}(\mu-1) \tau(\mu) d \mu<0
\end{aligned}
$$

Therefore, for each $\bar{\mu}$, a best-response $b_{1}(\bar{\mu})$ such that $\Psi_{1}\left(b_{1}(\bar{\mu}), \bar{\mu}\right)=0$ exists and is unique. Since $\Psi_{1}(\underline{\mu}, \bar{\mu})$ strictly decreases in $\bar{\mu}, b_{1}(\bar{\mu})$ strictly decreases in $\bar{\mu}$. Finally, for any $\bar{\mu}$,

$$
\int_{b_{1}(\bar{\mu})}^{1}\left(\mu-b_{1}(\bar{\mu})\right) \tau(\mu) d \mu \geq \int_{b\left(b_{1}(\bar{\mu}), \bar{\mu}\right)}^{1}\left(\mu-b_{1}(\bar{\mu})\right) \tau(\mu) d \mu \geq \int_{0}^{b_{1}(\bar{\mu})}\left(b_{1}(\bar{\mu})-\mu\right) \tau(\mu) d \mu
$$

where the second inequality holds because $\alpha \geq 1 / 2$ and $\Psi_{1}\left(b_{1}(\bar{\mu}), \bar{\mu}\right)=0$. Thus, for any $\bar{\mu}$, $b_{1}(\bar{\mu}) \leq \mu_{0}$.

Analogously, for each $\underline{\mu}$, a best-response $b_{2}(\underline{\mu})$ such that $\Psi_{2}\left(\underline{\mu}, b_{2}\right)=0$, exists, is unique, strictly decreases in $\underline{\mu}$, and is everywhere greater than $\mu_{0}$.


Robust Trust

Dworczak, Piotr, and Alex Smolin. “Robust Trust.” Working paper, 2026.

Preserve the supplied title, complete author list, citation, and canonical URL when technically practicable.

Therefore, a solution to (6) and (7) is any $\underline{\mu} \in\left[0, \mu_{0}\right]$ and $\bar{\mu}=b_{2}(\underline{\mu}) \in\left[\mu_{0}, 1\right]$ such that $b_{1}\left(b_{2}(\underline{\mu})\right)=\underline{\mu}$. By the established properties of $b_{1}$ and $b_{2}, b_{1}\left(b_{2}(\underline{\mu})\right)$ is continuous in $\underline{\mu}$ with $b_{1}\left(b_{2}(\underline{\mu})\right) \in\left[0, \mu_{0}\right]$ for all $\underline{\mu} \in\left[0, \mu_{0}\right]$; hence, $b_{1}\left(b_{2}(0)\right)-0 \geq 0$ and $b_{1}\left(b_{2}\left(\mu_{0}\right)\right)-\mu_{0} \leq 0$. By the intermediate value theorem, there exists $\underline{\mu} \in\left[0, \mu_{0}\right]$ such that $b_{1}\left(b_{2}(\underline{\mu})\right)=\underline{\mu}$.

So far, we showed that for $\alpha \geq 1 / 2$, a solution exists and belongs to a closed rectangular set $D=\left\{(\underline{\mu}, \bar{\mu}): \underline{\mu} \in\left[0, \mu_{0}\right], \bar{\mu} \in\left[\mu_{0}, 1\right]\right\}$. To establish uniqueness, consider the function $-\Psi=\left(-\Psi_{1},-\Psi_{2}\right)$ on $D$. Any solution must satisfy $-\Psi(\underline{\mu}, \bar{\mu})=(0,0)$. Observe that for any
$(\underline{\mu}, \bar{\mu}) \in D$,

$$
\begin{aligned}
& \frac{\partial\left[-\Psi_{1}\right]}{\partial \underline{\mu}}=\alpha \int_{0}^{\underline{\mu}} \tau(\mu) d \mu+(1-\alpha) \frac{\partial b}{\partial \underline{\mu}} \tau(b)(b-\underline{\mu})+(1-\alpha) \int_{b}^{1} \tau(\mu) d \mu>0, \\
& \frac{\partial\left[-\Psi_{1}\right]}{\partial \bar{\mu}}=(1-\alpha) \frac{\partial b}{\partial \bar{\mu}} \tau(b)(b-\underline{\mu}) \geq 0, \\
& \frac{\partial\left[-\Psi_{2}\right]}{\partial \underline{\mu}}=-(1-\alpha) \frac{\partial b}{\partial \underline{\mu}} \tau(b)(b-\bar{\mu}) \geq 0, \\
& \frac{\partial\left[-\Psi_{2}\right]}{\partial \bar{\mu}}=\alpha \int_{\bar{\mu}}^{1} \tau(\mu) d \mu+(1-\alpha) \frac{\partial b}{\partial \bar{\mu}} \tau(b)(\bar{\mu}-b)+(1-\alpha) \int_{0}^{b} \tau(\mu) d \mu>0 .
\end{aligned}
$$

Moreover, for all $(\underline{\mu}, \bar{\mu}) \in D$, the Jacobian of $[-\Psi]$ is a P-matrix, i.e., it has strictly positive principal minors:

$$
\frac{\partial\left[-\Psi_{1}\right]}{\partial \underline{\mu}}>0, \quad \frac{\partial\left[-\Psi_{1}\right]}{\partial \underline{\mu}} \frac{\partial\left[-\Psi_{2}\right]}{\partial \bar{\mu}}-\frac{\partial\left[-\Psi_{1}\right]}{\partial \bar{\mu}} \frac{\partial\left[-\Psi_{2}\right]}{\partial \underline{\mu}}>0 .
$$

By the Gale-Nikaido Theorem (Gale and Nikaido (1965), Theorem 4), it follows that $[-\Psi]$ is injective on $D$, and thus there exists at most one solution to the equation $-\Psi(\underline{\mu}, \bar{\mu})=(0,0)$.

Finally, we show that the proposed trust region strategy is robustly rationalizable by explicitly constructing a TRE. For $\alpha \geq 1 / 2$, we need to construct a measurable strategy of the misaligned adviser $\beta:[0, b] \rightarrow[\bar{\mu}, 1]$ such that for every set $X \subseteq[\bar{\mu}, 1]$ with $\alpha \tau(X)+(1-$ $\alpha) \tau\left(\beta^{-1}(X)\right)>0$,

$$
\frac{\alpha \int_{X} \mu \tau(\mu) d \mu+(1-\alpha) \int_{\beta^{-1}(X)} \mu \tau(\mu) d \mu}{\alpha \int_{X} \tau(\mu) d \mu+(1-\alpha) \int_{\beta^{-1}(X)} \tau(\mu) d \mu}=\bar{\mu}
$$

(The construction of $\beta:(b, 1] \rightarrow[0, \underline{\mu}]$ is analogous.) To this end, define two finite atomless nonnegative measures:

$$
\begin{aligned}
& \nu(Y) \triangleq(1-\alpha) \int_{Y}(\bar{\mu}-\mu) \tau(\mu) d \mu, \quad Y \subseteq[0, b] \\
& \eta(X) \triangleq \alpha \int_{X}(\mu-\bar{\mu}) \tau(\mu) d \mu, \quad X \subseteq[\bar{\mu}, 1]
\end{aligned}
$$

Observe that condition (7) is precisely $\eta([\bar{\mu}, 1])=\nu([0, b])$ whereas condition (27) is the
pushforward identity:

$$
\eta(X)=\nu\left(\beta^{-1}(X)\right), \quad X \subseteq[\bar{\mu}, 1] .
$$

In other words, we need to find $\beta$ that transports $\nu$ to $\eta$. It is always possible. For a canonical quantile construction, define the cumulative mass functions $F_{\nu}(\mu) \triangleq \nu([0, \mu])$ for $\mu \in[0, b]$ and $F_{\eta}(\mu) \triangleq \eta([\bar{\mu}, \mu])$ for $\mu \in[\bar{\mu}, 1]$. The transport map can then be set:

$$
\beta(\mu)=F_{\eta}^{-1}\left(F_{\nu}(\mu)\right), \quad \mu \in[0, b],
$$

where $F_{\eta}^{-1}(\cdot)$ is the generalized inverse: $F_{\eta}^{-1}(q)=\inf \left\{\mu \in[\bar{\mu}, 1]: F_{\eta}(\mu) \geq q\right\}$.
For $\alpha<1 / 2, T=\left\{\mu_{0}\right\}$, so the misaligned adviser is indifferent between all messages and it suffices to construct a strategy $\beta:[0,1] \rightarrow[0,1]$ such that for all $X \subseteq[0,1]$ with $\alpha \int_{X} \tau(\mu) d \mu+(1-\alpha) \int_{\beta^{-1}(X)} \tau(\mu) d \mu>0$,

$$
\frac{\alpha \int_{X} \mu \tau(\mu) d \mu+(1-\alpha) \int_{\beta^{-1}(X)} \mu \tau(\mu) d \mu}{\alpha \int_{X} \tau(\mu) d \mu+(1-\alpha) \int_{\beta^{-1}(X)} \tau(\mu) d \mu}=\mu_{0}
$$

which is equivalent to:

$$
\alpha \int_{X}\left(\mu-\mu_{0}\right) \tau(\mu) d \mu+(1-\alpha) \int_{\beta^{-1}(X)}\left(\mu-\mu_{0}\right) \tau(\mu) d \mu=0
$$

To do that, observe that $\int_{0}^{1}\left(\mu-\mu_{0}\right) \tau(\mu) d \mu=0$ and $\int_{0}^{\mu_{0}}\left(\mu_{0}-\mu\right) \tau(\mu) d \mu=\int_{\mu_{0}}^{1}(\mu-$ $\left.\mu_{0}\right) \tau(\mu) d \mu>0$. Since $\alpha \in(0,1 / 2), \alpha /(1-\alpha) \in(0,1)$ and by the intermediate value theorem, there exist $\mu_{L} \in\left(0, \mu_{0}\right)$ and $\mu_{H} \in\left(\mu_{0}, 1\right)$ such that

$$
\begin{aligned}
& \int_{0}^{\mu_{L}}\left(\mu_{0}-\mu\right) \tau(\mu) d \mu=\frac{\alpha}{1-\alpha} \int_{0}^{\mu_{0}}\left(\mu_{0}-\mu\right) \tau(\mu) d \mu \\
& \int_{\mu_{H}}^{1}\left(\mu-\mu_{0}\right) \tau(\mu) d \mu=\frac{\alpha}{1-\alpha} \int_{\mu_{0}}^{1}\left(\mu-\mu_{0}\right) \tau(\mu) d \mu
\end{aligned}
$$

By construction,

$$
\begin{aligned}
& \int_{0}^{\mu_{L}}\left(\mu_{0}-\mu\right) \tau(\mu) d \mu=\frac{\alpha}{1-\alpha} \int_{\mu_{0}}^{1}\left(\mu-\mu_{0}\right) \tau(\mu) d \mu \\
& \int_{\mu_{H}}^{1}\left(\mu-\mu_{0}\right) \tau(\mu) d \mu=\frac{\alpha}{1-\alpha} \int_{0}^{\mu_{0}}\left(\mu_{0}-\mu\right) \tau(\mu) d \mu \\
& \int_{\mu_{L}}^{\mu_{H}}\left(\mu-\mu_{0}\right) \tau(\mu) d \mu=0
\end{aligned}
$$


Robust Trust

We can set $\beta(\mu)=\beta_{L}(\mu)$ when $\mu \in\left[0, \mu_{L}\right], \beta(\mu)=\mu_{0}$, when $\mu \in\left(\mu_{L}, \mu_{H}\right)$, and $\beta(\mu)=\beta_{H}(\mu)$, when $\mu \in\left[\mu_{H}, 1\right]$. Here, $\beta_{L}$ is a quantile transport map that transports measure $\nu_{L}(Y)=(1-\alpha) \int_{Y}\left(\mu_{0}-\mu\right) \tau(\mu) d \mu$ on $\left[0, \mu_{L}\right]$ to measure $\eta_{L}(X)=\alpha \int_{X}\left(\mu-\mu_{0}\right) \tau(\mu) d \mu$ on [ $\mu_{0}, 1$ ], just like in the case of $\alpha \geq 1 / 2$; it ensures that (28) holds for all $X \subseteq\left(\mu_{0}, 1\right]$. Similarly, $\beta_{H}$ is a quantile transport map that transports measure $\nu_{H}(Y)=(1-\alpha) \int_{Y}\left(\mu-\mu_{0}\right) \tau(\mu) d \mu$ on $\left[\mu_{H}, 1\right]$ to measure $\eta_{H}(X)=\alpha \int_{X}\left(\mu_{0}-\mu\right) \tau(\mu) d \mu$ on $\left[0, \mu_{0}\right]$; it ensures that (28) holds for all $X \subseteq\left[0, \mu_{0}\right)$. (The transported masses match the targets by equations (29) and (30).) Finally, by equation (31), condition (28) holds for $\mu=\mu_{0}$. The result follows.


Robust Trust

## A. 7 Proof of Proposition 2

At $\alpha=1 / 2,[\underline{\mu}, \bar{\mu}]=\left[\mu_{0}, \mu_{0}\right]$ satisfies conditions (6) and (7). At $\alpha=1,[\underline{\mu}, \bar{\mu}]=[0,1]$ satisfies conditions (6) and (7).

For $\alpha \in(1 / 2,1)$, denote by $\Psi_{i 1}, \Psi_{i 2}$, and $\Psi_{i \alpha}$ the partial derivatives of $\Psi_{i}$ with respect to $\underline{\mu}, \bar{\mu}$, and $\alpha$ respectively. Define the Jacobian:

$$
J(\underline{\mu}, \bar{\mu}, \alpha) \triangleq\left(\begin{array}{ll}
\Psi_{11} & \Psi_{12} \\
\Psi_{21} & \Psi_{22}
\end{array}\right)
$$

As we argued in the proof of Proposition 1, for all $\underline{\mu}, \bar{\mu}, \alpha>1 / 2, \operatorname{det} J(\underline{\mu}, \bar{\mu}, \alpha)>0$, and therefore, by the implicit function theorem, optimal $\underline{\mu}(\alpha)$ and $\bar{\mu}(\alpha)$ are continuously
differentiable and ${ }^{21}$

$$
\binom{d \underline{\mu} / d \alpha}{d \bar{\mu} / d \alpha}=-J(\underline{\mu}, \bar{\mu}, \alpha)^{-1}\binom{\Psi_{1 \alpha}}{\Psi_{2 \alpha}} .
$$

Consequently,

$$
\begin{aligned}
& \frac{d \underline{\mu}}{d \alpha}=-\frac{\Psi_{2 \bar{\mu}} \Psi_{1 \alpha}-\Psi_{1 \bar{\mu}} \Psi_{2 \alpha}}{\Psi_{1 \underline{\mu}} \Psi_{2 \bar{\mu}}-\Psi_{1 \bar{\mu}} \Psi_{2 \underline{\mu}}}<0, \\
& \frac{d \bar{\mu}}{d \alpha}=\frac{\Psi_{2 \underline{\mu}} \Psi_{1 \alpha}-\Psi_{1 \underline{\mu}} \Psi_{2 \alpha}}{\Psi_{1 \underline{\mu}} \Psi_{2 \bar{\mu}}-\Psi_{1 \bar{\mu}} \Psi_{2 \underline{\mu}}}>0,
\end{aligned}
$$

where the inequalities hold because, as we already showed, $\Psi_{1 \underline{\mu}}<0, \Psi_{1 \bar{\mu}} \leq 0, \Psi_{2 \underline{\mu}} \leq 0$, $\Psi_{2 \bar{\mu}}<0$, and

$$
\begin{aligned}
& \Psi_{1 \alpha}=\int_{0}^{\underline{\mu}}(\mu-\underline{\mu}) \tau(\mu) d \mu-\int_{b}^{1}(\mu-\underline{\mu}) \tau(\mu) d \mu<0 \\
& \Psi_{2 \alpha}=\int_{\bar{\mu}}^{1}(\mu-\bar{\mu}) \tau(\mu) d \mu-\int_{0}^{b}(\mu-\bar{\mu}) \tau(\mu) d \mu>0
\end{aligned}
$$

The result follows.


Robust Trust

## A. 8 Proof of Proposition 3

Throughout, fix $\alpha>1 / 2$; the proposition is trivially true otherwise. We begin with a simple lemma.

Lemma 7. Let $U_{1}, U_{2}$ be twice differentiable and strictly convex on $[0,1]$. Assume that the ratio $U_{1}^{\prime \prime}(\mu) / U_{2}^{\prime \prime}(\mu)$ is decreasing. Then, with $b_{i}$ defined for each $U_{i}, i \in\{1,2\}$, by equation (5), we have that $b_{1}(\underline{\mu}, \bar{\mu}) \leq b_{2}(\underline{\mu}, \bar{\mu})$ for all $\underline{\mu} \leq \bar{\mu}$.

Proof. By equation (5), for $i \in\{1,2\}, b_{i}(\underline{\mu}, \bar{\mu})=\mathbb{E}_{\mu \sim f_{i}}[\mu]$, where $f_{i}(\mu)$ is a density of a

[^17]probability measure on $[\underline{\mu}, \bar{\mu}]$ defined as
$$
f_{i}(\mu) \triangleq \frac{U_{i}^{\prime \prime}(\mu)}{\int_{\underline{\mu}}^{\bar{\mu}} U_{i}^{\prime \prime}(\mu) d \mu} .
$$
By assumption, $f_{1}(\mu) / f_{2}(\mu)$ is decreasing, and hence $f_{2}$ likelihood-ratio dominates $f_{1}$. This implies that the probability distribution $f_{1}$ is first-order stochastically dominated by $f_{2}$; in particular, it has a lower mean. The result follows. $\square$

We now take the second step by showing a monotone relationship between the cutoff function $b$ and the trust region.

Lemma 8. Consider two decision problems $U_{i}, i \in\{1,2\}$, and let $\left(\underline{\mu}_{i}, \bar{\mu}_{i}\right) \in D$ denote the unique solution to the system

$$
\Psi_{1}^{i}(\underline{\mu}, \bar{\mu}, \alpha)=0, \quad \Psi_{2}^{i}(\underline{\mu}, \bar{\mu}, \alpha)=0,
$$

where $\Psi_{1}^{i}, \Psi_{2}^{i}$ and $D$ are defined as in the proof of Proposition 1 for each $U_{i}$. If $b_{1}(\underline{\mu}, \bar{\mu}) \leq$ $b_{2}(\underline{\mu}, \bar{\mu})$ for all $\underline{\mu} \leq \bar{\mu}$, then $\underline{\mu}_{2} \leq \underline{\mu}_{1}$ and $\bar{\mu}_{2} \leq \bar{\mu}_{1}$.

Proof. For any $h \in[0,1]$, define the auxiliary functions

$$
\begin{aligned}
& \widetilde{\Psi}_{1}(\underline{\mu}, h) \triangleq \alpha \int_{0}^{\underline{\mu}}(\mu-\underline{\mu}) \tau(\mu) d \mu+(1-\alpha) \int_{h}^{1}(\mu-\underline{\mu}) \tau(\mu) d \mu \\
& \widetilde{\Psi}_{2}(\bar{\mu}, h) \triangleq \alpha \int_{\bar{\mu}}^{1}(\mu-\bar{\mu}) \tau(\mu) d \mu+(1-\alpha) \int_{0}^{h}(\mu-\bar{\mu}) \tau(\mu) d \mu
\end{aligned}
$$

For each $h$, let $\underline{\mu}(h)$ be the unique solution to $\widetilde{\Psi}_{1}(\underline{\mu}, h)=0$ in $\left[0, \mu_{0}\right]$, and let $\bar{\mu}(h)$ be the unique solution to $\widetilde{\Psi}_{2}(\bar{\mu}, h)=0$ in $\left[\mu_{0}, 1\right]$, with the existence and uniqueness following from an argument analogous to that used in the proof of Proposition 1.

For each $i \in\{1,2\}$ define the scalar map

$$
\varphi_{i}(h) \triangleq b_{i}(\underline{\mu}(h), \bar{\mu}(h)) .
$$

Let $h_{i} \triangleq b_{i}\left(\underline{\mu}_{i}, \bar{\mu}_{i}\right)$ be the cutoff evaluated at the optimal endpoints of problem $i$. Then $\left(\underline{\mu}_{i}, \bar{\mu}_{i}\right)$ solves $\Psi_{1}^{i}=\Psi_{2}^{i}=0$ if and only if $\underline{\mu}_{i}=\underline{\mu}\left(h_{i}\right), \bar{\mu}_{i}=\bar{\mu}\left(h_{i}\right)$, and $h_{i}=\varphi_{i}\left(h_{i}\right)$. By the assumption
$b_{1} \leq b_{2}$ pointwise, for every $h$,

$$
\varphi_{1}(h)=b_{1}(\underline{\mu}(h), \bar{\mu}(h)) \leq b_{2}(\underline{\mu}(h), \bar{\mu}(h))=\varphi_{2}(h) .
$$

Define a region $H \triangleq\{h \in[0,1]: \underline{\mu}(h) \leq h \leq \bar{\mu}(h)\}$. Note that $H$ is an interval and $h_{1}, h_{2} \in H$. For $h \in H$, implicit differentiation yields

$$
\underline{\mu}^{\prime}(h)=-\frac{\partial \widetilde{\Psi}_{1} / \partial h}{\partial \widetilde{\Psi}_{1} / \partial \underline{\mu}} \leq 0, \quad \bar{\mu}^{\prime}(h)=-\frac{\partial \widetilde{\Psi}_{2} / \partial h}{\partial \widetilde{\Psi}_{2} / \partial \bar{\mu}} \leq 0 .
$$

Thus, $\varphi_{i}$ is weakly decreasing in $h$ on $H$ : both $\underline{\mu}(h)$ and $\bar{\mu}(h)$ are weakly decreasing in $h$, while $b_{i}(\underline{\mu}, \bar{\mu})$ is weakly increasing in each endpoint, so the composition $h \mapsto \varphi_{i}(h)$ is weakly decreasing.

It follows that $h_{2} \geq h_{1}$ : if $h_{2}<h_{1}$, then, since $\varphi_{2}$ is decreasing on $H$,

$$
h_{2}=\varphi_{2}\left(h_{2}\right) \geq \varphi_{2}\left(h_{1}\right) \geq \varphi_{1}\left(h_{1}\right)=h_{1},
$$

which is a contradiction. Since $\underline{\mu}(\cdot)$ and $\bar{\mu}(\cdot)$ are weakly decreasing,

$$
\underline{\mu}_{2}=\underline{\mu}\left(h_{2}\right) \leq \underline{\mu}\left(h_{1}\right)=\underline{\mu}_{1}, \quad \bar{\mu}_{2}=\bar{\mu}\left(h_{2}\right) \leq \bar{\mu}\left(h_{1}\right)=\bar{\mu}_{1},
$$

completing the proof. $\square$

By Lemma 7, $U_{1}^{\prime \prime} / U_{2}^{\prime \prime}$ decreasing implies $b_{1}(\underline{\mu}, \bar{\mu}) \leq b_{2}(\underline{\mu}, \bar{\mu})$ for all $\underline{\mu} \leq \bar{\mu}$. Lemma 8 then yields $\underline{\mu}_{2} \leq \underline{\mu}_{1}$ and $\bar{\mu}_{2} \leq \bar{\mu}_{1}$. This proves Proposition 3.


Robust Trust

## A. 9 Proof of Proposition 4

With a small abuse of notation, we can parameterize each private strategy by $\hat{\sigma}=\operatorname{Pr}\left(a=a_{2}\right)$. We also drop the dependence of $G$ and $L$ on $\hat{\tau}$ in the notation. Then, by the arguments behind Theorem 1, if the agent employs the set of private strategies $\hat{\Sigma}_{0}=\{\hat{\sigma}(m)\}_{m \in \Delta(\Omega)}$, the payoffs coming from both aligned and misaligned adviser depend only on $\hat{\sigma}_{L} \triangleq \inf \hat{\Sigma}_{0}$ and $\hat{\sigma}_{H} \triangleq \sup \hat{\Sigma}_{0}$, and the optimal payoffs from using $\hat{\Sigma}_{0}$ are the same as if the agent plays
$\hat{\sigma}(m)=\hat{\sigma}_{L}$ when $v(m)<0$ and $\hat{\sigma}(m)=\hat{\sigma}_{H}$ when $v(m) \geq 0$. This payoff is:

$$
\begin{aligned}
& \int_{-\infty}^{0}\left(\alpha \hat{\sigma}_{L}+(1-\alpha) \hat{\sigma}_{H}\right) v \hat{\tau}(d v)+\int_{0}^{+\infty}\left(\alpha \hat{\sigma}_{H}+(1-\alpha) \hat{\sigma}_{L}\right) v \hat{\tau}(d v) \\
& =\hat{\sigma}_{L}((1-\alpha) G-\alpha L)+\hat{\sigma}_{H}(\alpha G-(1-\alpha) L)
\end{aligned}
$$

The optimal choice of $\hat{\sigma}_{L}$ and $\hat{\sigma}_{H}$ must maximize (32) subject to $\hat{\sigma}_{L}, \hat{\sigma}_{H} \in[0,1]$ and $\hat{\sigma}_{L} \leq \hat{\sigma}_{H}$. This is a linear optimization subject to $\left(\hat{\sigma}_{L}, \hat{\sigma}_{H}\right)$ being in a triangle with vertices (0, 0), (0, 1), and $(1,1)$. A straightforward calculation gives the following solution:

If $G=L:$ if $\alpha<\hat{\alpha} \triangleq 1 / 2$, then any $\hat{\sigma}_{L}=\hat{\sigma}_{H}$ is optimal; if $\alpha>\hat{\alpha}$, then $\hat{\sigma}_{L}=0$ and $\hat{\sigma}_{H}=1$; if $\alpha=\hat{\alpha}$, then any $\left(\hat{\sigma}_{L}, \hat{\sigma}_{H}\right)$ is optimal. If $G>L$ : if $\alpha<\hat{\alpha}$, then $\hat{\sigma}_{L}=\hat{\sigma}_{H}=1$; if $\alpha>\hat{\alpha}$, then $\hat{\sigma}_{L}=0$ and $\hat{\sigma}_{H}=1$; if $\alpha=\hat{\alpha}$, then $\hat{\sigma}_{H}=1$ and any $\hat{\sigma}_{L}$ is optimal. If $G<L$ : if $\alpha<\hat{\alpha}$, then $\hat{\sigma}_{L}=\hat{\sigma}_{H}=0$; if $\alpha>\hat{\alpha}$, then $\hat{\sigma}_{L}=0$ and $\hat{\sigma}_{H}=1$; if $\alpha=\hat{\alpha}$, then $\hat{\sigma}_{L}=0$ and any $\hat{\sigma}_{H}$ is optimal.

The cases $\hat{\sigma}_{L}=\hat{\sigma}_{H}$ correspond to not trusting any message and always acting in the same way, optimal at the prior belief, so $T=\left\{\mu_{0}\right\}$. The cases $\hat{\sigma}_{L}=0$ and $\hat{\sigma}_{H}=1$ correspond to trusting all messages, so $T=\Delta(\Omega)$. Since we assumed that the probability of $v(\mu)=0$ is 0 and $G \neq L$, the corresponding optimal strategy is uniquely determined.

It is left to show that those strategies are robustly rationalizable. Define $M_{0}=\{\mu:$ $v(\mu)=0\}, M_{-}=\{\mu: v(\mu)<0\}$, and $M_{+}=\{\mu: v(\mu)>0\}$. Define probability measures $q_{+}(X)=\int_{X} v(\mu) \tau(d \mu) / G$ for $X \subseteq M_{+}, q_{-}(Y)=\int_{Y}(-v(\mu)) \tau(d \mu) / L$ for $Y \subseteq M_{-}$.

For $\alpha>\hat{\alpha}$, the agent fully trusts the adviser. Consider the following strategy of the misaligned adviser. If $\mu \in M_{0}$, then $\beta(\mu)=\mu$. If $\mu \in M_{-}$, then $\beta$ randomizes over messages $m \in M_{+}$according to $q_{+}$. If $\mu \in M_{+}$, then $\beta$ randomizes over messages $m \in M_{-}$according to $q_{-}$. This strategy is clearly adversarial. Furthermore, since $\alpha>\hat{\alpha}$, after any message $m \in M_{+}$, the posterior expected payoff from action $a_{2}$ is strictly positive: for any $X \subseteq M_{+}$ with $\tau(X)>0$,

$$
\alpha \int_{X} v(m) \tau(d m)+(1-\alpha) \int_{\Delta(\Omega)} v(\mu) \beta(X \mid \mu) \tau(d \mu)=\alpha G q_{+}(X)-(1-\alpha) L q_{+}(X)>0 .
$$

Analogously, after any message $m \in M_{-}$, the posterior expected payoff from action $a_{2}$ is
strictly negative: for any $Y \subseteq M_{-}$with $\tau(Y)>0$,

$$
\alpha \int_{Y} v(m) \tau(d m)+(1-\alpha) \int_{\Delta(\Omega)} v(\mu) \beta(Y \mid \mu) \tau(d \mu)=-\alpha L q_{-}(Y)+(1-\alpha) G q_{-}(Y)<0 .
$$

And by construction, after any message $m \in M_{0}$, the posterior expected payoff from action $a_{2}$ is zero.

For $\alpha<\hat{\alpha}$, the agent doesn't trust the adviser so any adviser's strategy is adversarial. Consider the case $G>L$; the complementary case is analogous. We need to construct the misaligned adviser strategy that makes communication not valuable. To do so, define $\gamma=\alpha L /((1-\alpha) G) \in[0,1]$. Consider the following strategy of the misaligned adviser. If $\mu \in M_{-}$, then $\beta$ randomizes over messages $m \in M_{+}$according to $q_{+}$. If $\mu \in M_{+}$, then with probability $\gamma, \beta$ randomizes over messages $m \in M_{-}$according to $q_{-}$and with probability $(1-\gamma), \beta$ randomizes over messages $m \in M_{+}$according to $q_{+}$. This strategy makes the posterior expected payoff from action $a_{2}$ zero after every message $m \in M_{-}$: for any $Y \subseteq M_{-}$ with $\tau(Y)>0$,

$$
\alpha \int_{Y} v(m) \tau(d m)+(1-\alpha) \int_{\Delta(\Omega)} v(\mu) \beta(Y \mid \mu) \tau(d \mu)=-\alpha L q_{-}(Y)+(1-\alpha) \gamma G q_{-}(Y)=0 .
$$

After any message $m \in M_{+}$, the posterior expected payoff from action $a_{2}$ is strictly positive: for any $X \subseteq M_{+}$with $\tau(X)>0$,

$$
\alpha \int_{X} v(m) \tau(d m)+(1-\alpha) \int_{\Delta(\Omega)} v(\mu) \beta(X \mid \mu) \tau(d \mu)=(G-L) q_{+}(X)>0 .
$$

Thus, this $\beta$ robustly rationalizes the strategy.
Finally, at $\alpha=\hat{\alpha}$, both full trust and no trust are optimal, along a continuum of other strategies, and robustly rationalizable by the same strategies $\beta$ as above.


Robust Trust

## A. 10 Supporting calculations for Example 1

The key step in the construction of the optimal trust region in Example 1 comes from the following simple lemma.

Lemma 9 (Spherical U). Let $U(\mu)=\tilde{U}(\|\mu-b\|)$ for some vector $b$ and function $\tilde{U}$. Let $\mu^{\prime}(r, n)=b+r n$, where $n \in \mathbb{R}^{N}$ with $\|n\|=1$ and $r \in \mathbb{R}$. Then, (i) $D_{U}\left(\mu, \mu^{\prime}(r, n)\right)$ is strictly increasing in $n \cdot(b-\mu)$ whenever $r \neq 0$, and (ii) $D_{U}\left(\mu, \mu^{\prime}(r, n)\right)$ is a unimodal function in $r$ with a minimum at $r=n \cdot(\mu-b)$.

Proof. In this case, $\nabla U\left(\mu^{\prime}(r, n)\right)=\tilde{U}^{\prime}(r) n$, and thus

$$
D_{U}\left(\mu, \mu^{\prime}(r, n)\right)=\tilde{U}(\|\mu-b\|)-\tilde{U}(r)-\tilde{U}^{\prime}(r)(n \cdot(\mu-b)-r) .
$$

Since $U(\mu)$ is strictly convex, $\tilde{U}^{\prime}(r)>0$ whenever $r \neq 0$ and $\tilde{U}^{\prime \prime}(r)>0$. Thus $D_{U}\left(\mu, \mu^{\prime}(r, n)\right)$ is strictly increasing in $n \cdot(b-\mu)$. Furthermore,

$$
\frac{\partial D_{U}\left(\mu, \mu^{\prime}(r, n)\right)}{\partial r}=-\tilde{U}^{\prime}(r)-\tilde{U}^{\prime \prime}(r)(n \cdot(\mu-b)-r)+\tilde{U}^{\prime}(r)=\tilde{U}^{\prime \prime}(r)(r-n \cdot(\mu-b)) .
$$

Because $\tilde{U}^{\prime \prime}(r)>0$, the unimodality follows. $\square$

Suppose that the misaligned adviser holds belief $\mu$. By Corollary 3, the adviser will try to maximize $D_{U}\left(\mu, \mu^{\prime}\right)$ over $\mu^{\prime}$ on the boundary of the trust region. Parameterizing $\mu^{\prime}=b+r n$ and applying Lemma 9 yields that $n$ is optimally chosen to be $(b-\mu) /\|b-\mu\|$, whereas $r$ is chosen to be maximal within the trust region, $r=r^{*}(\alpha)$. Thus, the optimal $\mu^{\prime}$ is given uniquely by $b+r^{*}(\alpha)(b-\mu) /\|b-\mu\|$.

[^0]:    *Dworczak: Department of Economics, Northwestern University and Group for Research in Applied Economics, piotr.dworczak@northwestern.edu. Smolin: Toulouse School of Economics, alexey.v.smolin@gmail.com. We thank Nageeb Ali, Ricardo Alonso, Ben Brooks, Laura Doval, Tan Gan, Alexis Ghersengorin, Marina Halac, Jason Hartline, Nicole Immorlica, Emir Kamenica, David Levine, Annie Liang, Stephen Morris, Jacopo Perego, Balázs Szentes, and Mark Whitmeyer for helpful conversations. Alex Smolin gratefully acknowledges funding from the French National Research Agency (ANR) under the Investments for the Future program (grant ANR-17-EURE-0010) and the AI Interdisciplinary Institute ANITI (grant ANR-23-IACL-0002). Part of the analysis in this paper was conducted while Alex Smolin was visiting Northwestern University and Columbia Business School, and we thank both institutions for their hospitality.

[^1]:    ${ }^{1}$ Strictly speaking, we assume that with probability $\alpha$ the Sender reveals his signal truthfully, but this can be shown to be optimal for the Receiver.

[^2]:    ${ }^{2}$ See also Levy and Szentes (2025) for a related model of AI under imperfect recall.

[^3]:    ${ }^{3}$ We allow $S$ to be infinite, which is useful for constructing tractable examples. Whenever we work with an infinite space, we endow it with the Borel $\sigma$-algebra, and require all sets and functions that we define to be measurable; statements involving "for all" should be interpreted as "for almost all" with respect to the underlying distributions.

[^4]:    ${ }^{4}$ It can be shown that the assumption of truthful reporting of the belief is equivalent (in terms of equilibrium payoff consequences) to assuming that the aligned adviser is attempting to maximize the agent's expected payoff. However, the assumption of truthful reporting is natural for an aligned AI system and useful, as it provides a natural meaning to each message (see Sobel (2020)).
    ${ }^{5}$ Formally, $\mathbb{E}_{\beta, \sigma}[u(a, \omega, \theta)]=\sum_{\omega \in \Omega} \mu_{0}(\omega) \int_{S} \int_{\Delta(\Omega)} \int_{\Theta} \int_{A} u(a, \omega, \theta) \sigma(d a \mid m, \theta) f(d \theta \mid \omega) \beta(d m \mid s) \pi(d s \mid \omega)$.

[^5]:    ${ }^{6}$ Formally, $\mathbb{E}_{\omega \sim \mu, \hat{\sigma}}[u(a, \omega, \theta)]=\sum_{\omega \in \Omega} \mu(\omega) \int_{\Theta} \int_{A} u(a, \omega, \theta) \hat{\sigma}(d a \mid \theta) f(d \theta \mid \omega)$.

[^6]:    ${ }^{7}$ Formally, to define the gradient, we extend the function $U$ beyond the probability simplex by assuming that, for any non-negative measure $\mu, U(\mu)=\mu(\Omega) U(\mu / \mu(\Omega))$.
    ${ }^{8}$ Point $m^{\prime} \in T$ is visible from $m$ if the line segment connecting $m^{\prime}$ and $m$ does not intersect $T \backslash\left\{m^{\prime}\right\}$.

[^7]:    ${ }^{9}$ However, it is not without loss of generality to assume that $T \subseteq M$.

[^8]:    ${ }^{10}$ Without loss of generality, we assume that $\beta$ uses only messages in $M$; any message $m \notin M$ cannot be sent by an aligned adviser and hence reveals that the adviser is misaligned.
    ${ }^{11}$ The assumption of finite $M$ and $\Theta$ is made for technical reasons; verifying the assumptions of Sion (1958)'s minimax theorem (in particular, its continuity requirements) is difficult for a cheap-talk-like game with infinite-dimensional strategy spaces since the impact of messages on payoffs is endogenous.

[^9]:    ${ }^{12}$ The strictly convex indirect payoff function can be a result of the agent having a continuum of actions or, as we show in Appendix A.4, finitely many actions and a continuum of private types.

[^10]:    ${ }^{13}$ One way to see that is to use our observation that in a TRE, the mapping from messages to belief defined by Bayes' rule must agree with the mapping defined by minimizing the Bregman distance to the trust region.

[^11]:    ${ }^{14}$ A recent systematic review of real-world AI deployment in medical imaging finds that AI most commonly serves as a secondary reader or a triage tool rather than as a fully autonomous reader; see Wenderott et al. (2024). For a regulatory example, the FDA-cleared DrAid chest X-ray system is indicated as a triage and prioritization aid and is not intended for stand-alone clinical decision-making; see FDA 510(k) summary K221241, released in 2024.
    ${ }^{15}$ See Dratsch et al. (2023). An RSNA summary of the study reports that incorrect AI suggestions reduced reader accuracy even for experienced radiologists.


Robust Trust

[^12]:    ${ }^{16}$ This conclusion is driven by the coarseness of the strategy space; we know from Theorem 3 that, for any $\alpha>0$, the agent would optimally use a non-trivial trust region if the state space and her action space were sufficiently rich.

[^13]:    ${ }^{17}$ To see why, note that any trust region $T$ can be convexified by replacing it with the intersection of sets $T_{\mu}$ over all $\mu \in \operatorname{supp}(\tau)$, where $T_{\mu}$ is the (convex) set of all points that are not further away from $\mu$ than any point in $T$.

[^14]:    ${ }^{18}$ It is equivalent to $\sigma^{*}$ because it is weakly better than $\sigma^{*}$ and $\sigma^{*}$ was optimal.

[^15]:    ${ }^{19}$ A matrix of adviser's posteriors can be computed by Bayes' rule as $(\mu(s))_{s \in S}=$ $\left(\operatorname{diag}\left(\mu_{0}(\omega)\right)_{\omega \in \Omega}\right) \Pi\left(\operatorname{diag}(\tau(s))_{s \in S}\right)^{-1}$. The diagonal matrices are invertible and the multiplication by an invertible matrix preserves the rank.

[^16]:    ${ }^{20}$ By the Auerbach theorem, there exist $v_{1}, \ldots, v_{R-1} \in \mathbb{W}$ and $\phi_{1}, \ldots, \phi_{R-1} \in \mathbb{W}^{*}$ such that $\left\|v_{i}\right\|_{1}=1$, $\left\|\phi_{i}\right\|_{\mathbb{W}^{*}}=1$, and $\phi_{i}\left(v_{j}\right)=\delta_{i j}$ (Section II.E, Lemma 11 in Wojtaszczyk (1991); see also Gershkov et al. (2025) for another recent application). By Hahn-Banach theorem, these $\phi_{i}$, operating on $\mathbb{W}$, can be extended to $\tilde{\phi}_{i}$, operating on $\mathbb{R}^{K}$, without a change in their norm. By the duality between spaces $l_{1}$ and $l_{\infty}$, for each $i$, there exists $x_{i} \in \mathbb{R}^{K}$ such that $\tilde{\phi}_{i}(z)=z^{\top} x_{i}$ and $\left\|x_{i}\right\|_{\infty}=\left\|\tilde{\phi}_{i}\right\|=1$. Then, for $w \in \mathbb{W}$, $w_{i}^{\top} x_{j}=\tilde{\phi}_{j}\left(w_{i}\right)=\phi_{j}\left(w_{i}\right)=\delta_{i j}$.

[^17]:    ${ }^{21}$ Differentiating the optimality conditions with respect to $\alpha$ we obtain $\Psi_{11} \frac{d \underline{\underline{\mu}}}{d \alpha}+\Psi_{12} \frac{d \bar{\mu}}{d \alpha}+\Psi_{1 \alpha}=0, \Psi_{21} \frac{d \underline{\underline{\mu}}}{d \alpha}+$ $\Psi_{22} \frac{d \bar{\mu}}{d \alpha}+\Psi_{2 \alpha}=0$.


Robust Trust

## Citation and provenance

**Authors:** Piotr Dworczak; Alex Smolin

**Canonical citation:** Dworczak, Piotr, and Alex Smolin. “Robust Trust.” Working paper, 2026.

**Canonical machine-readable version:** https://alexsmolin.com/corpus/papers/robust-trust.md

**Source record:** https://arxiv.org/abs/2602.09490

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