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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?

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  • $\begingroup$ 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. $\endgroup$ – RRL Jun 10 at 5:08
  • $\begingroup$ 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 $u$) $\endgroup$ – wiwnes691 Jun 10 at 5:13
  • $\begingroup$ Can you clarify “where $w > 10 -$ wealth”? Is that 10 minus wealth? $\endgroup$ – RRL Jun 10 at 5:16
  • $\begingroup$ $w$ means wealth and $w$ is bigger than $10$. Sorry for misunderstanding🙂 $\endgroup$ – wiwnes691 Jun 10 at 5:21
  • $\begingroup$ 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. $\endgroup$ – Vizag Jun 10 at 6:19
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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?

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

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