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.

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    $\begingroup$ 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? $\endgroup$ – whuber Apr 24 at 3:14
  • $\begingroup$ I don't have any criteria for "good". I'm eager to learn all the possibilities. $\endgroup$ – MSE Apr 24 at 3:17
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    $\begingroup$ 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? $\endgroup$ – whuber Apr 24 at 3:29
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    $\begingroup$ 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. $\endgroup$ – BruceET Apr 24 at 7:15
  • $\begingroup$ 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. $\endgroup$ – 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 $\chi^2_{200}$.

  • $\begingroup$ 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! $\endgroup$ – MSE May 6 at 17:06

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