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I am reading this problem from DeGroot's "Probability & Statistics" 2nd edition.

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I can't understand why $$\Pr[G^{-1}[F(X)]\leq z]=\Pr[F(X)\leq G(z)]$$, as nowhere is strict monotonicity assumed. I don't get why $G^{-1}[F(X)]\leq z\iff F(X)\leq G(z)$.

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    $\begingroup$ The inverse of a strictly increasing function is also strictly increasing! $\endgroup$ – kjetil b halvorsen May 27 '14 at 12:18
  • $\begingroup$ Yes, but nowhere "strictly increasing" is assumed. $\endgroup$ – Silent May 27 '14 at 13:05
  • $\begingroup$ OK, but if the distribution function should be constant on some interval, there is zero probability on that interval, so can just be ignored! $\endgroup$ – kjetil b halvorsen May 27 '14 at 13:07
  • $\begingroup$ @kjetilbhalvorsen, will you please elaborate that with answer? I will be obliged. $\endgroup$ – Silent May 27 '14 at 13:08
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You are correct that $G$ might not be strictly increasing, and thus the equivalence $G^{-1}[F(X)]\leq z\iff F(X)\leq G(z)$ cannot be deduced. However, it holds in the probabilistic sense the probability of one of these conditions being true and another false is 0.

Let us decompose the event $F(X)\leq G(z)$ into two cases depending on whether $G^{-1}[F(X)]\leq z$: \begin{equation} P( F(X) \leq G(z) ) = P\left([F(X)\leq G(z)] \cap [G^{-1}[F(X)]\leq z]\right) + P\left([F(X)\leq G(z)] \cap [G^{-1}[F(X)] > z] \right). \end{equation} Due to (not necessarily strict) monotonicity of $G$, we have $G^{-1}[F(X)]\leq z \Rightarrow F(X) \leq G(z)$. This implies that the first term of the decomposition is actually $P(G^{-1}[F(X)]\leq z)$. It remains to be shown that the second term is 0.

Note that by monotonicity of $G$, $F(X)<G(z)$ implies $G^{-1}[F(X)]\leq z$ and thus the event in question, $[F(X)\leq G(z)] \cap [G^{-1}[F(X)] > z]$, is a subset of $F(X)=G(z)$. Hence, \begin{equation} P\left([F(X)\leq G(z)] \cap [G^{-1}[F(X)] > z] \right)\leq P(F(X)=G(z)). \end{equation} But, $X$ was assumed to have a continuous d.f., for which it was shown that $F(X) \sim U(0,1)$. In particular, the probability of $F(X)$ being equal to any particular value is 0. Thus, \begin{equation} P\left([F(X)\leq G(z)] \cap [G^{-1}[F(X)] > z] \right) = 0. \end{equation} Substituting this into the second term of the first equation, we conclude that \begin{equation} P\left( F(X) \leq G(z) \right) = P\left(G^{-1}[F(X)]\leq z\right). \end{equation}

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  • $\begingroup$ Thank you so much, Sir, for so elaborate and beautiful answer. Sir, if you will clear these following doubts, I will be obliged. $\endgroup$ – Silent May 29 '14 at 4:40
  • $\begingroup$ 1. How due to (not necessarily strict) monotonicity of $G$, we have $G^{-1}[F(X)]\leq z \Rightarrow F(X) \leq G(z)$? I know that for strict monotonicity that argument is correct. But, how does it follow for nonstrict case? 2. Why due to monotonicity of $G$, $F(X)<G(z)\implies G^{-1}[F(X)]\leq z$? 3. Why $G^{-1}[F(X)]\leq z$ implies that $[F(X)\leq G(z)] \cap [G^{-1}[F(X)] > z]$ is a subset of $F(X)=G(z)$? 4. How $F(X) \sim U(0,1)$? $\endgroup$ – Silent May 29 '14 at 4:50
  • $\begingroup$ OK, Sir, I got why $F(X) \sim U(0,1)$, it is just because probability integral transformation. Please just solve other three doubts, please! $\endgroup$ – Silent May 30 '14 at 3:41
  • $\begingroup$ For curious people and for my further reference, see 1., 2. and 3. as well as 3.. These are answers from brilliant people. $\endgroup$ – Silent May 30 '14 at 15:16

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