 Solution

Write $c_1=X_1-Y$, $c_2=X_2+Y$, both positive, and $W_j=U_j^{\prime }(c_j)>0$. As usual the utility expectations must be meaningful; the improvement below is supported where both consumptions and derivatives are bounded, so it produces finite strictly positive expected gains without a tail differentiation problem.

Suppose $R=W_1/W_2$ is not almost surely constant. Choose $0<l<h$ and positive-probability sets $A$ and $B$ on which respectively $R\ge h$ and $R\le l$. By restricting these sets if necessary, keep both consumptions bounded and bounded away from zero, and their marginal utilities bounded. Put $a=\mathbb{E}(W_2{1}_A)>0$, $b=\mathbb{E}(W_2{1}_B)>0$, and choose $d$ strictly between $lb/(ha)$ and $b/a$. For $Z=1_B-d{1}_A$, we haveReplace $Y$ by $Y+\delta Z$ for small positive $\delta$. Agent one's utility has first derivative $-\mathbb{E}(W_{1}Z)>0$, and agent two's has derivative $\mathbb{E}(W_{2}Z)>0$. Positivity of consumptions is preserved on the bounded support, and uniform Taylor estimates there make both actual gains positive for sufficiently small $\delta$. Thus a Pareto-efficient transfer must satisfy

For given total wealth $s=X_1+X_2>0$, let $c$ be agent two's consumption. The ratio $R_s(c)=U_1^{\prime }(s-c)/U_2^{\prime }(c)$ is continuous and strictly increasing for $0<c<s$, by strict concavity. The endpoint marginal-utility assumptions give $R_s(c)\to 0$ as $c\downarrow 0$ and $R_s(c)\to \infty$ as $c\uparrow s$. Thus it has a unique solution $c=h_{\lambda }(s)$ to $R_s(c)=\lambda$. Continuous dependence on $s$ follows from strict monotonicity and continuity, so this solution is a measurable wealth-sharing rule. ThereforeFor a fixed weight, it is also the pointwise maximizer of $U_1(s-c)+\lambda U_2(c)$, whose strict concavity confirms uniqueness.