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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.

Fire
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    https://stats.stackexchange.com/questions/46418/why-is-the-square-root-transformation-recommended-for-count-data?rq=1, and the link to Wikipedia in the answer, may help. – jbowman Feb 26 '20 at 03:55
  • [A list of some similar posts](https://stats.stackexchange.com/search?q=poisson+varian*+stabili*+) that could help. – kjetil b halvorsen Feb 26 '20 at 13:18
  • Emulate the analysis in https://stats.stackexchange.com/a/251661/919, which carries out the proof for a Binomial distribution. All you have to do is replace $p(1-p)/n$ there by the Poisson parameter $\lambda.$ – whuber Feb 26 '20 at 14:07

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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.

Rodrigo
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