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

enter image description here enter image description here

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 $G^{-1}[F(X)]\leq z\iff F(X)\leq G(z)$.

  • 1
    $\begingroup$ 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

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}

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.