The delta method or variance-stabilizing transformation can be applied to make the variance be "nearly constant" (https://en.wikipedia.org/wiki/Variance-stabilizing_transformation). They use the Poisson distribution as an example and argue that the transformation $x \mapsto \sqrt x$ is such a function. They appeal to using a first-order Taylor series approximation of a transformation $Y = g(X)$ which results in $\sigma^2_Y \approx \sigma_X^2 g'(\mu_X)^2$ where $\mu,\sigma^2$ is the mean and variance, respectively.

My question, why did they stop at first order? Indeed, squaring and taking expectation of $g(x) = g(\mu_X) + (x - \mu) g'(\mu_X) + \frac{1}{2} (x - \mu)^2 g''(\mu_X)$ leads to $E[g(x)^2] = g(\mu_X)^2 + \sigma^2 ((g')^2 + gg'') + O((x-\mu)^3)$.

Noting that $(g')^2 + g g'' = (gg')'$, it is clear that a transformation that removes this term entirely is $g(x) = C \sqrt x$. Why can we neglect this second order term that seems to be relevant in terms of magnitude (especially since it cancels the term typically included).

Have I made a mistake somewhere? Thanks!

  • 2
    $\begingroup$ Why carry out the more complex calculations (to higher order) in the first place? What do you gain? $\endgroup$ – whuber Jul 13 at 19:31
  • $\begingroup$ You haven't really addressed the question. The question is why is the second-order term neglected when it seems to not only be the same magnitude, but precisely cancel in the case of the Poisson distribution. Often series approximations are truncated on the basis that higher order terms can be neglected. This seems to not be the case here. $\endgroup$ – Gregory Jul 13 at 21:14
  • $\begingroup$ I'm not trying to address the question--I'm trying to understand it. I see you write an expression for the raw second moment of $g(x),$ but that's not part of this method, which is based on the central second moment. Your question also seems to focus on a special property of the Poisson distribution, whereas this is a general method. I'm puzzled by what you are trying to get at. $\endgroup$ – whuber Jul 13 at 21:29
  • 1
    $\begingroup$ The problem is that when you expand the terms for the central second moment, no such terms appear. $\endgroup$ – whuber Jul 14 at 12:22
  • 1
    $\begingroup$ @Gregory please check the second moment of $Y=g(x)$. This should be $\mathbb{E}[(g(x)-\mathbb{E}[g(x)])^2]$, where $\mathbb{E}[g(x)]$ and $\mathbb{E}[g(x)^2]$ come from the Taylor expansion of $g(x)$ and $g(x)^2$ at the second order. $\endgroup$ – user289381 Jul 15 at 23:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.