# Square root transformation of Poisson process. How $\small Var[\sqrt{P(\lambda)}] \approx \frac{1}{4}$

I am working on Kaggle Neural data challenge. I am trying to understand the transformation applied on the neural spiking data. A number of spikes given a stimulus are Poisson distributed as

$$Y_i \sim P(\lambda_i)$$

The mean and variance of any Poisson process is given as

$$E[P(\lambda_i)] = Var[P(\lambda_i)] = \lambda_i$$

In order to normalize the data, the square root transformation is applied on the spike counts. Then the mean and variance becomes,

$$E[\sqrt{P(\lambda)}] \approx \sqrt{\lambda}$$

$$Var[\sqrt{P(\lambda)}] \approx \frac{1}{4}$$

I do not understand how the variance becomes constant by square root transformation on the $$P(\lambda_i)$$?

## 1 Answer

$$\operatorname{var}(\sqrt{P(\lambda)})=E[P(\lambda)]-E[\sqrt{P(\lambda)}]^2=\lambda-E[\sqrt{P(\lambda)}]^2$$

The second term can be approximated better as follows, see this post:

$$E[\sqrt{P(\lambda)}]\approx \sqrt{\lambda}-\frac{\lambda^{-1/2}}{8}+\frac{\lambda^{-3/2}}{16}+...$$

Which is why the expected value is also approximated by $$\sqrt \lambda$$. Square of it will be $$E[\sqrt{P(\lambda)}]^2\approx \lambda -\frac{1}{4}+\frac{9}{64\lambda}+...$$

So, the variance will be approximately $$\operatorname{var}(\sqrt{P(\lambda)})\approx \frac{1}{4}-\frac{9}{64\lambda}+...$$

which is approximately $$1/4$$ for large $$\lambda$$.