# Anscombe transform and normal approximation

The Anscombe transform is $a(x) = 2\sqrt{x+3/8}$.

Can anyone show me how to prove that an Anscombe-transformed version $Y = a(X)$ of a Poisson distributed random variable $X$ is approximately normal distributed (when $\lambda>4$)?

• Hint: Delta method. (Also, look up variance-stabilizing transforms, which is part of the motivation.) Commented May 10, 2011 at 12:03
• Thanks Mpikts! I'll be quite honest, I don't really get how to start. What are the main tools and the "start" that I'll need to proof this? Commented May 11, 2011 at 7:43

Here is a sketch of a proof which combines three ideas: (a) the delta method, (b) variance-stabilization transformations and (c) the closure of the Poisson distribution under independent sums.

First, let's consider a sequence of iid Poisson random variables $X_1, X_2, \ldots$ with mean $\lambda > 0$. Then, the Central Limit Theorem asserts that $$\newcommand{\barX}{\bar{X}_n}\newcommand{\convd}{\,\xrightarrow{\,d\,}\,}\newcommand{\Nml}{\mathcal{N}} \sqrt{n} (\barX - \lambda) \convd \Nml(0,\lambda) \> .$$

Notice that the asymptotic variance depends on the (presumably unknown) parameter $\lambda$. It would be nice if we could find some function of the data other than $\bar{X}_n$ such that, after centering and rescaling, it had the same asymptotic variance no matter what the parameter $\lambda$ was.

The delta method provides a handy way for determining the distribution of smooth functions of some statistic whose limiting distribution is already known. Let $g$ be a function with continuous first derivative such that $g'(\lambda) \neq 0$. Then, by the delta method (specialized to our particular case of interest), $$\sqrt{n}\big(g(\barX) - g(\lambda)\big) \convd \Nml(0, \lambda g'(\lambda)^2) \>.$$

So, how can we make the asymptotic variance constant (say, the value $1$) for all possible $\lambda$? From the expression above, we know we need to solve

$$g'(\lambda) = \lambda^{-1/2} \>.$$

It is not hard to see that the general antiderivative is $g(\lambda) = 2 \sqrt{\lambda} + c$ for any $c$, and the limiting distribution is invariant to the choice of $c$ (by subtraction), so we can set $c = 0$ without loss of generality. Such a function $g$ is called a variance-stabilizing transformation.

Hence, by the delta method and our choice of $g$, we conclude that $$\sqrt{n}\Big(2\sqrt{\barX} - 2\sqrt{\lambda}\Big) \convd \Nml(0, 1) \>.$$

Now, the Poisson distribution is closed under independent sums. So, if $X$ is Poisson with mean $\lambda$, then there exist random variables $Z_1, \ldots, Z_n$ that are iid Poisson with mean $\lambda/n$ such that $\sum_{i=1}^n Z_i$ has the same distribution as $X$. This motivates the approximation in the case of a single Poisson random variable.

What Anscombe (1948) found was that modifying the transformation $g$ (slightly) to $\tilde{g}(\lambda) = 2\sqrt{\lambda + b}$ for some constant $b$ actually worked better for smaller $\lambda$. In this case, $b = 3/8$ is about optimal.

Note that this modification "destroys" the true variance-stabilizing property of $g$, i.e., $\tilde{g}$ is not variance-stabilizing in the strict sense. But, it is close and gives better results for smaller $\lambda$.