I'm having trouble proving something I'm reading in a research paper, and wanted to get some other thoughts:

Suppose we define the transformation $\psi : [-\infty, \infty] \times [0,1] \rightarrow [0,1]$ given by $$ \psi(x,u) = P[X<x] + uP[X=x] $$ Consider also the lexicographical order $$ (x,u) \preceq (x^*,u^*) \Leftrightarrow (x<x^*) \cup (x=x^* \cap u\leq u^*) $$

It is proved in the paper that $$ P[\psi(X,U) \leq \psi(x,u)] = P[(X,U) \preceq (x,u)] $$

The part I don't understand is the following: If $T$ is strictly increasing and continuous on the range of $X$, then $$ \psi(T(X),U) = \psi(X,U) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [1] $$

What I have worked out so far is the following. Suppose we define $X^* := \psi(X,U)$ and $X_T^* := \psi(T(X),U)$. Then, the claim that the paper is making the transformed random variables are equivalent, meaning that the following statement should be true: $$P[X^* \leq z] \overset{?}{=} P[X_T^* \leq z] \ \ \ \ \forall \ z$$, where $z := \psi(x,u)$. This means, we can do the following simplifications: $$\begin{align} P[\psi(X,U) \leq \psi(x,u)] &\overset{?}{=} P[\psi(T(X),U) \leq \psi(x,u)] \\ = P[(X,U) \preceq (x,u)] &\overset{?}{=} P[(T(X),U) \preceq (x,u)] \\ = P[X<x] + P[X=x]P[U\leq u] &\overset{?}{=} P[T(X)<x] + P[T(X)=x]P[U\leq u] \\ = P[X<x] + P[X=x]P[U\leq u] &\overset{?}{=} P[X<T^{-1}(x)] + P[X=T^{-1}(x)]P[U\leq u] \end{align}$$

This is as far as I've gotten mathematically. If I assume that X is a continuous random variable, then we know that $P[X=x] = P[T(X)=x] = 0$, and the expression simplifies to $$P[X<x] \overset{?}{=} P[X<T^{-1}(x)] $$ I don't think the equality is satisfied here ... any ideas? Does Equation 1 become valid if X is a discrete random variable rather than continuous?

  • 1
    $\begingroup$ Could you tell us what "$U$" is? It seems it ought to be a uniform random variable on $[0,1]$, independent of $X$, but you haven't actually said anything about it. $\endgroup$ – whuber Sep 15 '15 at 20:09
  • 1
    $\begingroup$ @whuber , I'm sorry, yes you are correct, $U$ is a uniform random variable and independent of $X$. In the spirit of providing more information, it should also be noted that the transformation $\psi$ defined above is in the context of this paper applied to discrete random variables so that the transformed random variable $\psi(X,U)$ is uniformly distributed. However, the section I am referring to doesn't explicitly state that $X$ has to be discrete, and hence I used $X$ being continuous as what I think is a counter-example. $\endgroup$ – Kiran K. Sep 15 '15 at 20:13
  • $\begingroup$ If $X$ is continuous then $\Pr(X=x)=0$. How could that possibly lead to a counterexample? You are getting lost in a vague notation and winding up with something that makes no sense. $\endgroup$ – whuber Sep 15 '15 at 21:07
  • $\begingroup$ @whuber Right, if X is continuous, then as you and I both stated, $Pr(X=x)=0$, and thus we are left to see if $Pr(X<x)$ equals $P(X<T^{-1}(x))$ to make Equation 1 true. There must be some misunderstanding on my part here because the only way I see those two being equal is if $x$ equals $T^{-1}(x)$, but that is not true in general, correct? $\endgroup$ – Kiran K. Sep 15 '15 at 21:14
  • 2
    $\begingroup$ If that's the case, then the notation is utterly confused. After all, it is a simple matter to apply the definition of $\psi$, substituting $T(X)$ for $x$ and $U$ for $u$, to obtain $\psi(T(X),U)=\Pr(X\lt T(X))+U\Pr(X=T(X))$, which obviously is not what was intended. $\endgroup$ – whuber Sep 18 '15 at 14:16

Your Answer

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

Browse other questions tagged or ask your own question.