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Chapter 5: Problem 8

Let \(Z_{n}\) be \(\chi^{2}(n)\) and let \(W_{n}=Z_{n} / n^{2}\). Find the limiting distribution of \(W_{n}\).

Short Answer

Expert verified
The limiting distribution of \(W_{n}\) when n approaches infinity is zero.

Step by step solution

01

Understanding the Variables

First understand the variables given. \(Z_{n}\) has a chi-square (\(\chi^{2}\)) distribution with n degrees of freedom. Additionally, it is specified that \(W_{n}=Z_{n} / n^{2}\).
02

Applying the Limit Theorem

Now, as per the limit theorem, if \(Z_{n}\) is a chi-square distribution with n degrees of freedom, \(Z_{n} / n\) will converge in distribution to a standard normal distribution as n approaches infinity. Therefore, \(Z_{n} / n\) has distribution which is approximately \(N(1, 2/n)\) for large n.
03

Applying the Given Relation

Now apply the relationship \(W_{n}=Z_{n} / n^{2}\), we find as n approaches infinity, the distribution \(W_{n}\) will tend towards zero because \(Z_{n} / n^{2}\) will approach zero (as n goes to infinity, the denominator increases much faster than the numerator).

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