I have searched google, and wikipedia, and have come up with nothing. If there are links that you could provide to help me figure out how to prove this, that would be very beneficial.

Is it possible to do this by checking the approximated variance? If so how?

I know that E() = V() = lambda which is why there is a transformation.


1 Answer 1


Up to a certain degree of approximation you can prove this.

If you have iid $X_k \sim P(\lambda)$ then $\overline{X} \sim P(n \lambda)$ so $E \overline{X} = \lambda$ and $Var \overline{X} = \frac{\lambda}{n}$. But if you use the central limit theorem and apply the delta method with $g(u) = 2 \sqrt{u}, g'(u) = \frac{1}{\sqrt{u}}$ therefore $Y = g(\overline{X}) \sim N(g(\lambda), g'(\lambda) \frac{\lambda}{n^2}) = N(2 \sqrt{\lambda},g'(\lambda)^2 \frac{1}{n^2})$ as $n \to \infty$.

So the variance of $Y$ does not depend on $\lambda$ at least to this order of approximation and therefore the square root transformation is variance-stabilizing for the Poisson distribution.


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