Discussion
The \(p\)th quantile of a standard normal random variable is \(\xi_{p}\). Then, the \(p\)th quantile of a chi-square random variable with 1 degree of freedom is :
Answer: C
The \(p\)th quantile of \(\chi_{1}^{2}\) is \(\xi_{(1 + p)/2}^{2}\).
A \(\chi_{1}^{2}\) variable is \(Z^{2}\) with \(Z\) standard normal. For \(c \gt 0\),
\[ P(Z^{2} \le c^{2}) = P(-c \le Z \le c) = 2\Phi(c) - 1 \]
Setting this equal to \(p\) gives \(\Phi(c) = (1 + p)/2\), so \(c = \xi_{(1 + p)/2}\) and the quantile of \(Z^{2}\) is \(c^{2}\).
\(\xi_{p}^{2}\) ignores the two tails (and fails for \(p \lt 1/2\), where \(\xi_{p} \lt 0\)); \(\xi_{1 + p}\) is undefined since \(1 + p \gt 1\); and \(2\xi\) is not even a quantile expression.
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