The equation that the problem is built around $$cfrac{1}{1-rho}(w-pi)^{1-rho} = cfrac{1}{2} left( cfrac{1}{1-rho}(w-x)^{1-rho} right) + cfrac{1}{2} left( cfrac{1}{1-rho}(w+y)^{1-rho} right)$$

In these problems, typically $w > 0$, $x > 0$, $y > 0$, $pi geq 0$ (it’s not 3.14…)

The problem wants me to show that for $rho > 1$, $y$ can never be sufficiently large for $$cfrac{1}{1-rho}(w-pi)^{1-rho} leq cfrac{1}{2} left( cfrac{1}{1-rho}(w-x)^{1-rho} right) + cfrac{1}{2} left( cfrac{1}{1-rho}(w+y)^{1-rho} right)$$

when the following is satisfied.

$$cfrac{x}{w} geq 1 – 0.5^{1/(rho-1)} quad Leftrightarrow quad rho geq cfrac{log(0.5)+log(1- x/w)}{log(1-x/w)}$$

In a previous problem, I solved for $y$ in the first equation. Here’s my work so far.

We know from part (b) that the minimum value of $y$ is given by

begin{align*}

y &= left( 2(w-pi)^{1-rho} – (w-x)^{1-rho} right)^{1/(1-rho)} – w

end{align*}

Which implies that the agent will accept a gamble that satisfies

begin{align*}

y &geq left( 2(w-pi)^{1-rho} – (w-x)^{1-rho} right)^{1/(1-rho)} – w

end{align*}

Define $-gamma = 1-rho$ where $gamma geq 0$

begin{align*}

y &geq left( 2(w-pi)^{-gamma} – (w-x)^{-gamma} right)^{-(1/gamma)} – w \

y &geq left(frac{1}{ left( cfrac{2}{(w-pi)^gamma} – cfrac{1}{(w-x)^gamma} right)}right)^{1/gamma} \

end{align*}

For fixed $gamma$ we see that $y$ approaches $infty$ as

begin{align*}

cfrac{2}{(w-pi)^gamma} &= cfrac{1}{(w-x)^gamma} \

2(w-x)^gamma &= (w-pi)^gamma \

2^{1/gamma}(w-x) &= w – pi \

(w-x) &= cfrac{w-pi}{2^{1/gamma}} \

x &= w – cfrac{w-pi}{2^{1/gamma}} \

x &= cfrac{2^{1/gamma}w – w – pi}{2^{1/gamma}} \

x &= cfrac{w(2^{1/gamma} – 1) – pi}{2^{1/gamma}}

end{align*}

Now let $pi = 0$ (no risk premium)

begin{align*}

x &= cfrac{w(2^{1/gamma} – 1)}{2^{1/gamma}} \

cfrac{x}{w} &= 1 – left(cfrac{1}{2}right)^{1/gamma} = 1 – left(cfrac{1}{2}right)^{1/(1-rho)}

end{align*}

I don’t feel comfortable about setting $pi=0$. While it feels intuitive to set it equal to zero (have the gamble be a “free insurance policy”), if $pi$ were equal to $0$, the problem should have specified it. Is there a way I can show that $pi$ disappears in the solution without setting it equal to 0?