# Utility function and Insurance premium

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:

1. If he has $$0,95w$$ he also buy insurance for $$P$$

2. If he has $$1,05w$$ he also buy insurance for $$P$$

3. 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?

• Why can’t you solve for $P$ in terms of other parameters using the formula? What does it mean that you “failed”? Also you write an expression for $u(x)$ at the beginning with no $x$ appearing on the RHS. Surely this is a typo. – RRL Jun 10 at 5:08
• I’ve calculated $P$ but I cant get to the right conclusion - I cant see why $2$ is true and the rest not. (Thanks I corrected $u$) – wiwnes691 Jun 10 at 5:13
• Can you clarify “where $w > 10 -$ wealth”? Is that 10 minus wealth? – RRL Jun 10 at 5:16
• $w$ means wealth and $w$ is bigger than $10$. Sorry for misunderstanding🙂 – wiwnes691 Jun 10 at 5:21
• The question is incomplete. It's missing the payout he'll receive if he buys the insurance. It's also missing how much wealth he'll lose if with probability $0.5$ he gets into the accident. – Vizag Jun 10 at 6:19

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?

• Yes, I’ve tried exactly the same way. But are you sure that there is $$\tag{2} P^2 - 2(b-w)P +0.1(b-w) - 0.05 = 0$$ Shoudnt it be ?$$\tag{2} P^2 + 2(b-w)P -0.1(b-w) - 0.05 = 0$$ – wiwnes691 Jun 10 at 13:41
• @wiwnes691: Yes you are correct. – RRL Jun 10 at 17:05
• I have stucked at this point. I cant see how use this information to solve the task – wiwnes691 Jun 10 at 19:48
• Are you able to help me? – wiwnes691 Jun 12 at 3:28
• @wiwnes691: I’m still thinking about it. Are you sure you have all the correct information. Too much seems to be unspecified. – RRL Jun 12 at 5:06