There is an economy, populated by a large number of agents. A first order condition common to all agents, is the following:
$$E[\exp^{(1-\theta)\eta_i}(r-R+\eta_i)]=0$$
the index $i$ indicates the individual, $\eta$ is a an exogenous random variable that each individual receives (idiosyncratic shock) it is assumed to be identically and independently distributed among them, and across time. It is also assumed that $\eta_i \sim N (0, \sigma^2)$. The variables, $\theta, r, R$ are time varying. I suppress the time subscript on them, as the problem is stationary and so can be thought as some constants/parameters. The scope, is to solve for $\theta$. Also, $r \neq R$.
$$ E[\exp^{(1-\theta)\eta_i}(r+\eta_i - R)]=0 \Rightarrow $$ $$ E[\exp^{(1-\theta)\eta_i}(r+\eta_i ) - \exp^{(1-\theta)\eta_i}R)]=0 \Rightarrow $$ $$ E[\exp^{(1-\theta)\eta_i}(r+\eta_i)] - E [\exp^{(1-\theta)\eta_i}R)]=0 \Rightarrow $$ $$ E[\exp^{(1-\theta)\eta_i}(r+\eta_i)] = E [\exp^{(1-\theta)\eta_i}]R $$ $$ E[\exp^{(1-\theta)\eta_i}r] + E[\exp^{(1-\theta)\eta_i}\eta_i] = E [\exp^{(1-\theta)\eta_i}]R $$ $$ \log \left(E[\exp^{(1-\theta)\eta_i}(r+\eta_i)]\right) = \log \left(E [\exp^{(1-\theta)\eta_i}]R\right) $$
I then use the following approximation: $\log E (X) = E (\log X) + \frac{Var(X)}{2}$
$$ \log \left(E[\exp^{(1-\theta)\eta_i}(r+\eta_i)]\right) = \log \left(E [\exp^{(1-\theta)\eta_i}]R\right) $$ $$ E[\log (\exp^{(1-\theta)\eta_i}(r+\eta_i))] + \frac{Var(\exp^{(1-\theta)\eta_i}(r+\eta_i))}{2} = E [\log \exp^{(1-\theta)\eta_i}] + \frac{Var(\exp^{(1-\theta)\eta_i})}{2} + \log R $$
$$ E[(1-\theta)\eta_i + \log (r+\eta_i)] + \frac{Var(\exp^{(1-\theta)\eta_i}(r+\eta_i))}{2} = E [{(1-\theta)\eta_i}] + \frac{Var(\exp^{(1-\theta)\eta_i})}{2} + \log R $$
$$ E [\log (r+\eta_i)] + \frac{Var(\exp^{(1-\theta)\eta_i}(r+\eta_i))}{2} = \frac{Var(\exp^{(1-\theta)\eta_i})}{2} + \log R $$
This is where I am not quite sure how to proceed. And please let me know whether you see something wrong so far.
Since $\eta_i=\sigma Z$ with $Z$ standard normal, the goal is to find $\theta$ such that $$E(Z\mathrm e^{bZ})=aE(\mathrm e^{bZ}),$$ for the parameters $a$ and $b$ defined as $$a=(R-r)/\sigma,\qquad b=(1-\theta)\sigma.$$ A basic computation anybody interested in normal random variables should master yields the...
The differentiation of this identity with respect to $c$ (or a direct computation) yields $$E(Z\mathrm e^{cZ})=c\,\mathrm e^{c^2/2}.$$ Applying this to $c=b$, one sees that the solution to the initial problem is unique and such that $a=b$, that is, $$\theta=1-\frac{R-r}{\sigma^2}.$$
Addendum: (Proof of the key formula) By definition, $$E(\mathrm e^{cZ})=\int_\mathbb R\mathrm e^{cz}\frac1{\sqrt{2\pi}}\mathrm e^{-z^2/2}\mathrm dz=\mathrm e^{c^2/2}\int_\mathbb R\frac1{\sqrt{2\pi}}\mathrm e^{-(z-c)^2/2}\mathrm dz=\mathrm e^{c^2/2}.$$