# How does Anscombe transformation stabilize the variance of a Poisson R.V.?

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?

## 2 Answers

The definition is correct, but in practice it is applied to transforms that are only approximately variance stabilizing, largely because in many cases there either isn't one or it's annoyingly complex and buys little improvement over a simpler transform. The Anscombe transform is the "best" transform in the class of functions of the form $\sqrt(x+c)$, where $x \sim$ Poisson. but, as the Wikipedia page points out, there are other transforms, some of which are better. However, for $\lambda$ mildly far from 0, say $\geq 5$, the Anscombe transform does quite well.

Note that mild heteroskedasticity doesn't really cause any problems in the great bulk of applications, so getting an exact transform is typically of theoretical but not practical importance.

The definition of a variance-stabilizing transformation is simply that $\operatorname{Var}[f(X)] = \mbox{constant}$.

The expression $[f'(\operatorname{E}(X))]^2 \cdot \operatorname{Var}(X)$ is an approximation for $\operatorname{Var}[f(X)]$, as the latter is often hard to compute. Therefore, the criterion you cite is only approximate.

Indeed, for the Poisson case, $f(X) = 2 \sqrt{X}$ would have $[f'(\operatorname{E}(X))]^2 \cdot \operatorname{Var}(X) = 1$, but is not exactly variance-stabilizing (at $\operatorname{E}(X) = \lambda = 0$ the variance is still $0$, of course).

Anscombe's addition of the $3/8$ makes the variance closer to $1$ for large $\lambda$'s, or, equivalently, acceptably close to $1$ for lower values of $\lambda$.