I was taught that a transformation f(X) is said to be a variance-stabilizing transformation if $[f'(E(X))]^2*Var(X)$ is independent of E(X).
For a Poisson-distributed random variable X, E(X) = Var(X) = $\lambda$.
The Anscombe transformation is $f(X) = 2*\sqrt(X + 3/8)$, but $[f'(E(X))]^2*Var(X)$ = $\lambda/(\lambda + 3/8)$, which is not independent of E(X).
Was I taught incorrectly about what makes a transformation variance-stabilizing? Or are there limitations on the above definition of which I am not aware?