My question is about the following statement:
Consider a random walk whose step size has distribution Y, which is Poisson with mean 1 minus 1. Asymptotically the critical items are that Y has zero expectation and unit variance.
I am a little confused as to why the expectation and variance differ for this Poisson distribution. I assume this has to do with the statement " Asymptotically the critical items are ". Can someone point me in the right direction how to derive this?
We are subtracting $1$ from the Poisson distribution. We have a random variable $\mathbf X \sim \text{Poisson}(1)$, with mean $1$ and variance $1$; then, we define $\mathbf Y = \mathbf X - 1$, which has mean $0$ and variance $1$.
The next statement is just saying that to understand the asymptotic behavior of the random walk, the mean and variance of $\mathbf Y$ are all that we need to know.