How to show that it is possible to approximate this risk premium?

Question

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

$$\pi_u \approx \frac{-0.5p \cdot u^{''}(\omega_1)\cdot \epsilon^2}{p \cdot u'(\omega_1)+(1-p)\cdot u' (\omega_2)}$$

Answer 1

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.