# Approximate distribution for sum of squares of standardized Poisson random variables

Suppose that $$X_1, ..., X_n$$ are independent and identically distributed Poisson($$\lambda$$) random variables.

What is a good approximating distribution for $$\sum_{i = 1}^{200} \frac{(X_i - \lambda)^2}{\lambda}$$?

I think that it is $$\chi^2_{200}$$, because it is a sum of the squares of standardized normal random variables. However, I'm not sure if they really are standardized normal random variables.

• They (the $(X_i-\lambda)/\sqrt{\lambda}$) are not standardized normal random variables: they are standardized Poisson variables! There are many possible answers. Which are good ones depend on what you mean by a "good" approximation: could you tell us what aspects of the approximation need to be good and how you measure how good they are? – whuber Apr 24 at 3:14
• I don't have any criteria for "good". I'm eager to learn all the possibilities. – MSE Apr 24 at 3:17
• There are innumerable possibilities. Perhaps you could tell us the context in which this question arises or the actual statistical problem you are dealing with? – whuber Apr 24 at 3:29
• Two possibilities out of "innumerably many": If $\lambda$ is large then $X_i$ are nearly normal, $Z_i =(X_i - \lambda)/\sqrt{\lambda}$ are nearly standard normal, and the sum of 200 $Z_i^2$ is nearly CHISQ(200), which in turn is nearly normal. (Why and with what $\mu$ and $\sigma$ ?). // If $\lambda$ is very small, then the $X_i$ are mostly zeros and ones, and normal by CLT seems a better fit to the sum of 200. – BruceET Apr 24 at 7:15
• This question is from an exam in a second-year course in mathematical statistics. There is no more context. The question actually asks for the expected value of the approximate distribution. – MSE Apr 24 at 13:28

There are many possible approximations. But I guess the idea here is simply CLT.

Consider that, for large $$n$$, the sum $$Z= \sum_n Y_i$$ tends to a normal $$N(n\mu ,n \sigma^2 )$$ if $$Y_i$$ are iid variables of mean $$\mu$$ and variance $$\sigma^2$$.

In our case $$Y_i = \frac{(X_i-\lambda)^2}{\lambda}$$, with $$X_i$$ being a $$\lambda-$$Poisson.

We only need to calculate the expectation and variance of $$Y_i$$, I leave that to you.

Notice that the approximation is obviously wrong in some respects: in particular, $$Z$$ is non-negative and discrete.

I'm not sure if [it's the sum of the squares of] standardized normal random variables.

No, of course not. $$\frac{(X_i-\lambda)}{\sqrt{\lambda}}$$ is a "standardized" Poisson variable. Hence it not justified to assume that $$Z$$ is $$\chi^2_{200}$$.

• Ah - that makes sense. Thanks for providing a clear, concise, and straightforward answer to my question. I knew that it would not be so complicated! – MSE May 6 at 17:06