Let $u_1$ and $g$ be increasing strictly concave functions from $\mathbb{R}$ to $\mathbb{R}$. Let $u_2:=g\circ u_1$. If we regard $u_1$ and $u_2$ as utility functions of two players, this is saying that player $2$ has Arrow-Pratt coefficient of absolute risk aversion greater than that of player $1$.
Suppose now $X$ is a risky asset, i.e. a (non-constant) random variable. Let $t_1^*=argmax_{t\in [0,1]} \mathbb{E}(u_1(1-t+tX))$ and $t_2^*=argmax_{t\in [0,1]} \mathbb{E}(u_2(1-t+tX))$.
How do I prove that $t_1^*>t_2^*$? Intuitively I expect this to be true since it means that the more risk averse player allocates less money to the risky asset.
Maybe it's easier than I thought. We know that $$\mathbb{E}[u'(1-t^*+t^*X)(X-1)]=0$$ and want to prove that for $g$ concave $$\mathbb{E}[g'(u(1-t^*+t^*X))u'(1-t^*+t^*X)(X-1)]<0.$$ This follows simply from the fact that if $X(\omega)>1$ then by concavity of $g$ and monotonicity of $u$ $$g'(u(1-t^*+t^*X(\omega)))<g'(u (1)),$$ so that $$\mathbb{E}[g'(u(1-t^*+t^*X))u'(1-t^*+t^*X)(X-1)\mathbb{1}_{(\ X>1)}]<g'(u(1))\mathbb{E}[u'(1-t^*+t^*X)(X-1)\mathbb{1}_{(X>1)}].$$ Similarly, $$\mathbb{E}[g'(u(1-t^*+t^*X))u'(1-t^*+t^*X)(X-1)\mathbb{1}_{(X<1)}]<g'(u(1))E[u'(1-t^*+t^*X)(X-1)\mathbb{1}_{(X<1)}].$$ Adding up we get the final inequality.