47
$\begingroup$

Let $X$ be a random variable with a continuous and strictly increasing c.d.f. $F$ (so that the quantile function $F^{−1}$ is well-defined). Define a new random variable $Y$ by $Y = F(X)$. Show that $Y$ follows a uniform distribution on the interval $[0, 1]$.

My initial thought is that $Y$ is distributed on the interval $[0,1]$ because this is the range of $F$. But how do you show that it is uniform?

$\endgroup$
7
  • 3
    $\begingroup$ This is not true in cases where there's a discrete component. For example, suppose $X=\left.\begin{cases} 1/2 & \text{with probability }1/2, \\ W & \text{with probability }1/2,\end{cases}\right\}$ and $W$ is uniformly distributed on $[0,1]$, and that the choice between whether $X=1/2$ or not is independent of $W$. Then the cdf of $X$ has no values between $1/4$ and $3/4$, so it cannot be uniformly distributed on $[0,1]$. It is, however, true of continuous distributions. $\endgroup$ Jul 15, 2014 at 21:41
  • 4
    $\begingroup$ see the text of the question. X is continuous! $\endgroup$
    – user162381
    Jul 15, 2014 at 21:44
  • 9
    $\begingroup$ By the way, it is not necessary that $F$ is a strictly increasing CDF, continuity is sufficient. Just define the quantile function the usual way as a generalized inverse via $F^-(y)=inf\{x\in\mathbb{R}: F(x)\geq y\}$. See the proof of Proposition 3.1 in Embrechts, P., Hofert, M.: A note on generalized inverses. Mathematical Methods of Operations Research 77(3), 423-432 link for a very careful and detailed explanation. $\endgroup$
    – binkyhorse
    Jul 16, 2014 at 14:38
  • 1
    $\begingroup$ Thanks @binkyhorse - that reference is really good. $\endgroup$
    – Math1000
    Apr 5, 2015 at 22:33
  • 1
    $\begingroup$ @s0ulr3aper07 By Proposition 3.1 in the paper I linked above, yes. Prop. 3.1: Let F be a distribution function and X ~ F. (a) If F is continuous, F(X)∼U[0,1]. The paper includes a detailed proof. $\endgroup$
    – binkyhorse
    Mar 27, 2019 at 20:10

6 Answers 6

52
$\begingroup$

Let $F_Y(y)$ be the CDF of $Y = F(X)$. Then, for any $y \in [0,1]$ we have:

$F_Y(y) = \Pr[Y \le y] = \Pr[F(X) \le y] = \Pr[X \le F^{-1}(y)] = F(F^{-1}(y)) = y$.

What distribution has this CDF?

$\endgroup$
3
  • 1
    $\begingroup$ Are all CDF's of continuous densities invertible? $\endgroup$ Aug 22, 2017 at 7:43
  • 2
    $\begingroup$ @tintinthong: not always completely, but enough. If you define $G(y)= \inf\{x:F_X(x) \ge y\}$ then $F_X(G(y))=y$ when $y \in (0,1)$ $\endgroup$
    – Henry
    Aug 22, 2017 at 22:12
  • 3
    $\begingroup$ Strictly increasing and continuous CDF is needed. $\endgroup$ Feb 20, 2019 at 15:36
10
$\begingroup$

$$ Prob(Y\leq x)=P(F(X)\leq x)=P(X\leq F^{-1}(x))=x \\ $$ The last equality is from the definition of the quantile function.

$\endgroup$
0
3
$\begingroup$

Let $y=g(x)$ be a mapping of the random variable $x$ distributed according to $f(x)$. In the mapping $y=g(x)$ you preserve the condition of probability density (namely you counts the same number of events in the respective bins)

$$ h(y)dy=f(x)dx $$

where h(y) is the probability distribution of $y$

if $h(y)=1$ (uniform distribution) we have

$$ dy=g'(x)dx=f(x)dx $$

This means that $$ g(x)=\int f(x)dx $$

namely the function $g(x)$ that maps the random variable $x$ distributed according $f(x)$, into a random variable $y$ distributed uniformly is his own cumulative distribution function $\int f(x) dx$.

$\endgroup$
2
$\begingroup$

Here is an approach that does not use the quantile function whatsoever - the only property used is that independent copies of $X$ have zero probability of being equal. (The main ingredient in my argument is conditional expectation.)

Consider the cumulative distribution function of $X$, namely $$ F(t)=\mathbb P(X\leq t). $$ Your random variable - which I will suggestively call $U$ instead of $Y$ - can be described by starting with two independent and identically distributed random variables $X,Z$ and considering the conditional probability $$ U=\mathbb P(X\leq Z\mid Z). $$ Then, for all integers $n\geq 1$, we can represent $U^n$ as follows. Let $X_1,X_2,\ldots,X_n,Z$ be independent and identically distributed. By independence, $$ \mathbb P\bigl(X_1\leq Z,X_2\leq Z,\ldots, X_n\leq Z\bigm\vert Z\bigr)=U^n, $$ and thus by the tower property $$ \mathbb EU^n=\mathbb P(X_1\leq Z,X_2\leq Z,\ldots, X_n\leq Z)=\mathbb P\bigl(Z=\max(X_1,X_2,\ldots,X_n,Z)\bigr). $$ Since $X_1,\ldots,X_n,Z$ are iid, each of them is equally likely to be the maximum and therefore $$ \mathbb EU^n=\frac{1}{n+1}. $$ Thus $U$ has the same moments as a uniformly distributed random variable on $[0,1]$. Since $U$ is supported in $[0,1]$ as well, it follows (by the uniqueness of the Hausdorff moment problem) that $U$ is uniformly distributed, as desired.

$\endgroup$
2
$\begingroup$

Let $y\in(0,1)$. Since $F$ is continuous, there exists $x\in\mathbb{R}$ s.t. $F(x)=y$. Thus, $$ \mathsf{P}(Y\le y)=\mathsf{P}(F(X)\le F(x))=F(x)=y, $$ i.e., $Y\sim\text{U}[0,1]$. In order to see the first equality we don't need continuity. Specifically, since any cdf is right-continuous, \begin{align} \{F(X)\le F(x)\}&=\{\{F(X)\le F(x)\}\cap\{X\le x\}\}\cup \{\{F(X)\le F(x)\}\cap\{X>x\}\} \\ &=\{X\le x\}\cup \{\{F(X)=F(x)\}\cap\{X>x\}\}, \end{align} and $\mathsf{P}(\{F(X)=F(x)\}\cap\{X>x\})=0$.

$\endgroup$
0
$\begingroup$

For a proof of this problem when $F_X(x)$ is strictly increasing, refer to JimmyK4542's answer. Let's assume $F_X(x)$ is just non-decreasing (there are intervals such as $[a,b]$ where $F_X(x') = c$ for $x'\in[a,b]$). We define $G(y)$ similar to what Henry's comment suggests: $$ G(y)=\inf\{x:F_X(x)\gt y\}$$ Now substituting this expression in what Jimmy has written will give us: $$ F_Y(y) = \Pr[Y \le y] = \Pr[F_X(X) \le y] = \Pr[X \le G(y)] = F_X(G(y))= y \label{eq:I}\tag{I}$$

We need to show that:

  1. $F_X(x)\le y \rightarrow x \le G(y)$
  2. $F_X(G(y))=y$

The second argument is easier to prove. We have the following expression almost according to definitions: $$ F_X(G(y))= F_X(\inf\{x:F_X(x)\gt y\})= y$$ Now for the first argument, we can still use what $G(y)=\inf\{\cdots\}$ implies; if $F_X(x)\le y$, then $x\le \inf\{x:F_X(x)\gt y\};$ hence $x\le G(y)$.

With the two arguments proved and a substitution in \ref{eq:I}, we have proved the main argument.

$\endgroup$
1
  • $\begingroup$ I personally believe this problem is a severe case for abuse of notation, and a bad professor's problem. $\endgroup$
    – mahyar
    Feb 19 at 1:46

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.