# calculus – Help incorporating a variable that I set to 0 even though the problem didn’t explicitly ask for it. 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? Posted on Categories Articles