A policy maker has utility function $u(w)=b^2-(b-w)^2$ where $w>10$ (wealth ) and $b>0$ constant such as $b \geq 3w$. The policy maker is exposed to risk of loss $X$. $X=1$ with probability $0.05$ and $X=0$ with probability $0.95$. We know that he got the full insurance for premium $P$. Is it possible that:
If he has $0,95w$ he also buy insurance for $P$
If he has $1,05w$ he also buy insurance for $P$
If he has $0,95w$ he also buy insurance for $0,95P$
So I’ve tried to calculate $P$ using formula $\mathbb{E} (u(w-X))=u(w-P)$ but I failed. Is there any other way to solve it?
Insurance with premium $P$ is justified if expected utility in the presence of risk equals utility of wealth minus the premium,
$$\tag{1}E[u(w -X)] = u(w -P)$$
By hypothesis, there is an acceptable premium $P$, i.e, $P < w$, that solves (1) given that $b> 3w$ and $w > 10$. These last two conditions also imply $(b-w) > 20$.
Substituting the given utility function in (1) we get
$$0.95[b^2 - (b-w)^2] + 0.05 [b^2 - (b-(w-1))^2] = b^2 - (b - (w - P))^2$$
This reduces to
$$\tag{2} P^2 + 2(b-w)P -0.1(b-w) - 0.05 = 0$$
It appears the only information we have is that $b-w > 20$ and there is an acceptable solution $P$ of equation (2).
Presumably you can evaluate the three proposals by replacing $w$ with $0.95 w$ for case (1), replacing $w$ with $1.05 w$ for case (2), and replacing $w$ with $0.95 w$ and $P$ with $0.95P$ for case (3) and determining if an acceptable solution for $P$ can be obtained as before given the information on hand.
Have you tried this?