2

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 $P(\lambda_i)$?

mhdadk
  • 2,582
  • 1
  • 4
  • 17
Saranraj Nambusubramaniyan
  • 123
  • 6
  • 1
    https://stats.stackexchange.com/questions/251449 shows how to find an appropriate transformation generally. https://stats.stackexchange.com/questions/46418 provides additional references. – whuber Jan 17 '21 at 18:50

1 Answers1

3

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

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 $\lambda$.

gunes
  • 49,700
  • 3
  • 39
  • 75