Derivation for an Arrow-Pratt risk premium is given as
$U(W_0-\pi)=E[U(W_0+X)]=U(C)$; We assume $E(X)=0$
Taylor series expansion on both sides:
$U(W_0)-\pi U'(w_0)+\frac{\pi^2}{2}U''(W_0)+...=E[U(W_0)+XU'(W_0)+\frac{X^2}{2}U''(W_0)]$
Introducing $E(X)=0$
$U(W_0)-\pi U'(w_0)+\frac{\pi^2}{2}U''(W_0)+...= U(W_0)+\frac{1}{2}U''(W_0).Var(X)+...$
since $E(X)=0$ and $Var$ (X)$=E(X^2)-[E(X)]^2=E(X^2)$
If higher order terms are negligible
$-\pi U'(W_0) \approx \frac{1}{2}U''(W_0)Var(X)$
$\rightarrow \pi^*=-\frac{1}{2} \sigma^2_x \frac{U''(W_0)}{U'(0)}$
Hence the absolute Arrow pratt local risk premium
Derivation for an Arrow Pratt Relative Risk premium
$U[E(W_0(1+Y))-\pi(W_O,Y)=E[U(W_0(1+Y))]$
We assume $E(Y)=0$ and that investment is actuarially fair. I need HELP to compute the Taylor series in a similar approach to obtain
$\pi^*=-\frac{1}{2} \sigma_y^2 \frac{U''(W_0)}{U'(W_0)}W_0$.
It looks like you don't have to re-do the Taylor series - you can plug $X=W_0Y$ into the absolute risk premium.
Using absolute risk premium $\pi_a^*=-\frac{1}{2} \sigma^2_x \frac{U''(W_0)}{U'(W_0)},$ equating $U[E(W_0(1+Y-\pi_r))]=E[U(W_0(1+Y))]$ and ignoring low order terms in $\pi_r$ is the same as setting $X=W_0Y$ in $\pi_a^*/W_0$. Using:
gives $\pi_r^*=\pi_a^*/W_0=-\frac{1}{2} \sigma_y^2 \frac{U''(W_0)}{U'(W_0)}W_0$.