Discussion
Let \(\{X_{i}\}\) be a sequence of independent Poisson(\(\lambda\)) variables and let \(W_{n} = \frac{1}{n}\sum_{i=1}^{n} X_{i}\). Then the limiting distribution of \(\sqrt{n}(W_{n} - \lambda)\) is the normal distribution with zero mean and variance given by
Answer: C
The variance is \(\lambda\).
A Poisson(\(\lambda\)) variable has mean \(\lambda\) and variance \(\lambda\). \(W_{n}\) is the sample mean of independent, identically distributed variables with finite variance, so by the central limit theorem
\[ \begin{aligned} \sqrt{n}(W_{n} - \lambda) &\to N(0, \sigma^{2}) \\ &= N(0, \lambda) \end{aligned} \]
The scaling by \(\sqrt{n}\) is exactly what keeps the variance at \(\sigma^{2}\) rather than shrinking it to \(\sigma^{2}/n\). \(\sqrt{\lambda}\) is the standard deviation, not the variance; \(\lambda^{2}\) squares the variance again; and 1 is the variance only after dividing by \(\sqrt{\lambda}\) as well.
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