Discussion
Suppose that \(G_{1}\), \(G_{2}\) and \(G_{3}\) are functions defined by :
\[ G_{1}(x) = \frac{pF(x)}{1 - qF(x)}; \]
\[ G_{2}(x) = \frac{F(x)}{p + qF(x)}; \]
\[ G_{3}(x) = p + qF(2), \]
where \(0 \lt p \lt 1\), \(0 \lt q \lt 1\) and \(p + q = 1\) and F is a probability distribution function then, which of the following is true ?
Answer: B
Only \(G_{1}\) and \(G_{2}\) are distribution functions.
Take F as u running from 0 to 1. \(G_{1} = \frac{pu}{1 - qu}\) and \(G_{2} = \frac{u}{p + qu}\) are increasing in u (their derivatives are \(\frac{p}{(1 - qu)^{2}}\) and \(\frac{p}{(p + qu)^{2}}\), both positive), equal 0 at \(u = 0\), and equal \(\frac{p}{1 - q} = 1\) and \(\frac{1}{p + q} = 1\) at \(u = 1\). Composed with F they are non-decreasing, right-continuous, and tend to 0 and 1 at \(\mp\infty\).
\(G_{3}(x) = p + qF(2)\) is a constant, not depending on x. It cannot tend to 0 at \(-\infty\) (it is at least \(p \gt 0\)), so it is not a distribution function. That rules out every option that includes \(G_{3}\), and 'none' is wrong since \(G_{1}\) and \(G_{2}\) qualify.
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