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For a statistics class, I have to prove a result which leads me to the following question. If I can show that it is true, my proof is done. So here is the question.

Suppose that $U: \Omega \rightarrow [0,1]$ is a random variable uniformly distributed on $[0, 1]$ and $X: \Omega \rightarrow \mathbb{R}$ is continuously distributed as some distribution $F_X(·)$ (where $\Omega$ is the sample space). Is it true that

$F_U(U(\bar{\omega})) = F_X(z_{\bar{\omega}})$ for all $\bar{\omega} \in \Omega \implies z_{\bar{\omega}}=X(\bar{\omega})$

or equivalently,

$P \{\omega \in \Omega~|~ U(\omega) \leq U(\bar{\omega})\} = P \{\omega \in \Omega~|~ X(\omega) \leq z_{\bar{\omega}}\}$ for all $\bar{\omega} \in \Omega \implies z_{\bar{\omega}}=X(\bar{\omega})$

? I suspect that it is true, but I haven't been able to prove or to disprove it.

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For any random variable $Y$ with the distribution function $F_Y$ the random variable $F_Y(Y)$ has a uniform distribution in the interval $[0,1]$. Hence your first statement should be wrong, if we ignore the fact that it is not mathematically correct.

You equate two different mathematical objects. $F_U(U(\omega))$ is a random variable, $F_X(z)$ is a real function. They cannot be compared without giving precise definition of what you mean by $=$ in this case.

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  • $\begingroup$ @ mpiktas: I am not sure I understand your second comment. I understand that $F_U(U({\omega}))$ is a function and not a real number. But in the statement, I consider a particular $\bar{\omega} \in \Omega$, so $U(\bar{\omega})$ is a real number right? Then isn't $F_U(U(\bar{\omega})) \in \mathbb{R}$, making it clear what the equality sign means? $\endgroup$ – Martin Van der Linden Sep 26 '13 at 14:39
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    $\begingroup$ @ mpiktas : I edited my post following your comment to make things clearer. $\endgroup$ – Martin Van der Linden Sep 26 '13 at 15:03
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Following mpiktas answer, things became clearer to me and I think I can know answer the question. The first statement

$F_U(U(\bar{\omega})) = F_X(z_{\bar{w}})$ for all $\bar{\omega} \in \Omega \implies z_{\bar{\omega}}=X(\bar{\omega})$

Is in fact equivalent to (and better understood as)

$F_U(U({\omega})) \sim F_X(Z(\omega)) \implies Z(\omega)=X({\omega})$, where $\sim$ means "distributed as".

Now, this turns out not to be quite true (see mpiktas comment below). What is true however is

$F_U(U({\omega})) \sim F_X(Z(\omega)) \implies Z(\omega) =^d X({\omega})$, where $=^d$ means that the two random variables have the same distribution.

As noticed by mpiktas, $F_U(U({\omega})) \sim U(0,1)$ ( a proof can be found here). Hence we must have $F_X(Z(\omega)) \sim U(0,1)$. But then we must have $Z(\omega) =^d X(\omega)$. All that is needed for the proof of this last fact can also be found here. But in the link, the statement is reversed ($F_X(X(\omega)) \sim U(0,1)$ instead of $F_X(Z(\omega)) \sim U(0,1) \Rightarrow Z(\omega) =^d X(\omega)$ ) so I figured it might be helpful to reformulate it here.

  • Assume $F_X(Z(\omega)) \sim U(0,1)$.
  • Then $F_{F_X(Z(\omega))}(a) = a$
  • Notice that \begin{align} a &= F_X(F^{-1}_X(a))\\ &= P(X\leq F^{-1}_X(a))\\ &= P(F_X(X) \leq a)\\ &= F_{ F_X(X)} (a) \end{align} where I omitted the $\omega$'s for notational convenience.
  • So we have $F_{F_X(Z(\omega))}(a) = a = F_{ F_X(X(\omega))} (a)$, which in turn implies $Z(\omega) =^d X(\omega)$.
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    $\begingroup$ In fact your statement can be simplified to this one: if $F_X(Z)$ is distributed as $U(0,1)$ then $Z$ must be distributed as $F_X$, where $F_X$ is a distribution function of random variable $X$. I would not write $X(\omega)=Z(\omega)$ since this means that $P(X=Z)=1$, which is too strong, since if we take two independent copies of $X$, $X_1$ and $X_2$ then $F_X(X_1)\sim U(0,1)$ and $F_X(X_2)\sim U(0,1)$, but $P(X_1=X_2)\neq 1$ and in general can be made to be 0 for symmetrical distributions. $\endgroup$ – mpiktas Sep 26 '13 at 16:19
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    $\begingroup$ My comment above answers clarifies your doubt about the last step. The equality of distribution functions implies equality in distribution, but not almost sure equality. Sorry for writing the previous comment without reading the whole answer. $\endgroup$ – mpiktas Sep 26 '13 at 16:22
  • $\begingroup$ Thanks a lot for you comment. I edited my answer given my understanding of what you wrote. I hope this works now... $\endgroup$ – Martin Van der Linden Sep 26 '13 at 16:49
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    $\begingroup$ Final nitpicking, usually the notation is $X\sim F$ where X is a random variable, $F$ is the distribution function and $\sim$ means distributed as. To note that two random variables have the same distribution notation $X=^dZ$ is used. The notation though is a minor matter, the understanding is much more significant, keep up the good work! $\endgroup$ – mpiktas Sep 26 '13 at 17:33
  • $\begingroup$ Good to know the convention, I edit! Thanks for all you comment, really helpfull at clarifying the difference between equality of r.v. and equal distribution of r.v. $\endgroup$ – Martin Van der Linden Sep 26 '13 at 18:09

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