<p>We are given</p>
<p><span class="math-container">$$\pi_u : E[u(\tilde{\omega}-\pi_u)]= E[u(\tilde{\omega}+\tilde \epsilon)]$$</span></p>
<p>The left-hand-side is</p>
<p><span class="math-container">$$LHS: p u(\omega_1 + \tilde \epsilon -\pi_u) + (1-p)u(\omega_2 -\pi_u).$$</span></p>
<p>The right-hand-side is</p>
<p><span class="math-container">$$RHS: p u(\omega_1 + \tilde \epsilon) + (1-p)u(\omega_2).$$</span></p>
<ol>
<li><p>Apply A 1st-order Taylor expansion of LHS around with respect to <span class="math-container">$(\pi_u, \tilde \epsilon)$</span> around <span class="math-container">$(0,0)$</span>, and don't forget to take expectations, taking into account the assumed distribution of <span class="math-container">$\tilde \epsilon$</span></p>
</li>
<li><p>Apply a 2nd-order Taylor expansion of the RHS with respect to <span class="math-container">$\tilde \epsilon$</span> around <span class="math-container">$0$</span>, and don't forget to take expectations, taking into account the assumed distribution of <span class="math-container">$\tilde \epsilon$</span>.</p>
</li>
<li><p>Equate, cancel out and re-arrange.</p>
</li>
</ol>

CC BY-SA 4.0
2024-02-27T17:54:29.393Z
Thank you! I have another question on the expected utility. Could you have a chance to look over that for me? Any hint is welcome! https://economics.stackexchange.com/questions/57776/which-of-these-two-lotteries-a-consumer-with-von-neumann-morgenstern-preference
Jonathen
CC BY-SA 4.0
2024-02-27T17:46:25.260Z
2024-02-27T17:54:17.970Z
2024-02-27T17:54:17.970Z
Alecos Papadopoulos
<p>Suppose a consumer with Von-Neumann Morgenstern preferences is faced with two risks and can only eliminate one. Let <span class="math-container">$\tilde{\omega}= \omega_1$</span> with probability p and <span class="math-container">$\tilde{\omega}= \omega_2$</span> with probability (1 − p).Suppose <span class="math-container">$\tilde \epsilon = 0 $</span> if <span class="math-container">$\tilde{\omega}=\omega_2$</span>. If <span class="math-container">$\tilde{\omega}=\omega_1, \tilde \epsilon = -\epsilon $</span>  with probability 0.5 and <span class="math-container">$\tilde \epsilon =  \epsilon$</span> with probability 0.5. Now let's define a premium
risk <span class="math-container">$\pi_u$</span> for <span class="math-container">$\tilde \epsilon$</span> as follows:</p>
<p><span class="math-container">$$\pi_u : E[u(\tilde{\omega}-\pi_u)]= E[u(\tilde{\omega}+\tilde \epsilon)]$$</span></p>
<p>Show that it is possible to approximate this risk premium by</p>
<p><span class="math-container">$$\pi_u \approx \frac{-0.5p \cdot u^{''}(\omega_1)\cdot \epsilon^2}{p \cdot u'(\omega_1)+(1-p)\cdot u' (\omega_2)}$$</span></p>

<microeconomics>
How to show that it is possible to approximate this risk premium?
Jonathen
# How to show that it is possible to approximate this risk premium?

(Asked by: username_0 on 2024-02-27)

Suppose a consumer with Von-Neumann Morgenstern preferences is faced with two risks and can only eliminate one. Let $\tilde{\omega}= \omega\_1$ with probability p and $\tilde{\omega}= \omega\_2$ with probability (1 − p).Suppose $\tilde \epsilon = 0 $ if $\tilde{\omega}=\omega\_2$. If $\tilde{\omega}=\omega\_1, \tilde \epsilon = -\epsilon $ with probability 0.5 and $\tilde \epsilon = \epsilon$ with probability 0.5. Now let's define a premium
risk $\pi\_u$ for $\tilde \epsilon$ as follows:

$$\pi\_u : E[u(\tilde{\omega}-\pi\_u)]= E[u(\tilde{\omega}+\tilde \epsilon)]$$

Show that it is possible to approximate this risk premium by

$$\pi\_u \approx \frac{-0.5p \cdot u^{''}(\omega\_1)\cdot \epsilon^2}{p \cdot u'(\omega\_1)+(1-p)\cdot u' (\omega\_2)}$$

**Tags:** microeconomics

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## 1 answers

### Answer by username_1 on 2024-02-27. Score: 2 | Accepted answer

We are given

$$\pi\_u : E[u(\tilde{\omega}-\pi\_u)]= E[u(\tilde{\omega}+\tilde \epsilon)]$$

The left-hand-side is

$$LHS: p u(\omega\_1 + \tilde \epsilon -\pi\_u) + (1-p)u(\omega\_2 -\pi\_u).$$

The right-hand-side is

$$RHS: p u(\omega\_1 + \tilde \epsilon) + (1-p)u(\omega\_2).$$

1. Apply A 1st-order Taylor expansion of LHS around with respect to $(\pi\_u, \tilde \epsilon)$ around $(0,0)$, and don't forget to take expectations, taking into account the assumed distribution of $\tilde \epsilon$
2. Apply a 2nd-order Taylor expansion of the RHS with respect to $\tilde \epsilon$ around $0$, and don't forget to take expectations, taking into account the assumed distribution of $\tilde \epsilon$.
3. Equate, cancel out and re-arrange.

> **Comments:**
> **username_0**: Thank you! I have another question on the expected utility. Could you have a chance to look over that for me? Any hint is welcome! https://economics.stackexchange.com/questions/57776/which-of-these-two-lotteries-a-consumer-with-von-neumann-morgenstern-preference  

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