Variance function is variance stabilising

Question

Y has mean u and variance function V(u).

If $V(u) = \alpha.u^v$ then $h(y) = y^{(2-v)/2}$ is variance stabilising which means that Var(h(Y)) is approximately constant.

I tried to prove it computing $Var(h(Y)) = E(h(Y)^2)-E(h(Y))^2$ but I'm completely stuck. Can you help ?

Answer 1

Use the Taylor series approximation $$ h(Y) - h(\mu) \approx h'(\mu)(Y - \mu), $$ where $\mu = E(Y)$. Then, we have $$ [h(Y) - h(\mu)]^{2} \approx [h'(\mu)]^{2}(Y - \mu)^{2}. $$ Hence, $$ Var(h(Y)) = E\{ [h(Y) - h(\mu)]^{2} \} \approx [h'(\mu)]^{2}E\{ (Y - \mu)^{2} \} = [h'(\mu)]^{2}Var(Y). $$ Now, since $h'(\mu) = \frac{2 - v}{2}\mu^{-v/2}$ and $Var(Y) = \alpha \mu^{v}$, it follows from the above that $$ Var(h(Y)) \approx \frac{(2 - v)^{2}\mu^{-v}}{4} Var(Y) = \frac{\alpha(2 - v)^{2}}{4}. $$ So, the approximate variance of $h(Y)$ does not depend on the mean.