Let u be the utility function $u(x)=-\frac{x^{-\eta}-1}{\eta},$ with $x,\eta>0$
Assume that an investor is indifferent between an investment with riskless outcome of 101.005 and a stochastic outcome of K, where $\log(K/100)\sim N(0,05 ;0,16)$. Determine $\eta>0$.
As someone asked for in the comments, my thought process was the following(not sure if any of it is correct): By certainty equivalent $$u(101,005)=E[u(K)]$$ $$-\frac{101,005^{-\eta}-1}{\eta}=E[-\frac{K^{-\eta}-1}{\eta}]$$ due to linearity $$101,005^{-\eta}=E[K^{-\eta}]$$ $$101,005^{-\eta}=E[\exp(\log(K^{-\eta}))]$$
$$101,005^{-\eta}=E[\exp(-\eta\log(K))]$$ $$101,005^{-\eta}=E[\exp(-\eta\log(\frac{K}{100}100))]$$ $$101,005^{-\eta}=E[\exp(-\eta\log(\frac{K}{100}100))]$$ $$101,005^{-\eta}=E[\exp(-\eta\log(\frac{K}{100})+\eta\log(100))]$$ $$101,005^{-\eta}=E[\exp(-\eta\log(\frac{K}{100}))]100^{-\eta}$$ using the fact that log(K/100) is normally distributied $$(\frac{101,005}{100})^{-\eta}=\exp((\eta^2*0,16/2)-\eta*0,05 )$$ $$-\eta \log(1,01005)= \eta^2*0,08-\eta*0,05$$ and with this one has $$\eta= \frac{0,05-ln(1,01005)}{0,08}$$ which more or less 0,5. I'm honestly not sure though if this is correct. In fact I think that what I did is wrong...