21
$\begingroup$

Let $X$ be a random variable with a continuous and strictly increasing c.d.f. function $F$ (so that the quantile function $F^{−1}$ is well-defined). Define a new random variable $Y$ by $Y = F(X)$. Show that $Y$ has 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?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Michael Hardy Jul 15 '14 at 21:41
  • 3
    $\begingroup$ see the text of the question. X is continuous! $\endgroup$ – user162381 Jul 15 '14 at 21:44
  • 7
    $\begingroup$ 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 '14 at 14:38
  • $\begingroup$ Thanks @binkyhorse - that reference is really good. $\endgroup$ – Math1000 Apr 5 '15 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 at 20:10
23
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Are all CDF's of continuous densities invertible? $\endgroup$ – tintinthong Aug 22 '17 at 7:43
  • 1
    $\begingroup$ @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 '17 at 22:12
  • $\begingroup$ Strictly increasing and continuous CDF is needed. $\endgroup$ – Dani Feb 20 at 15:36
6
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $\int f(x) dx$.

$\endgroup$
-1
$\begingroup$

Let x be a RV with a PDF f(x) and CDF F(x), then: dF= f(x)dx

Or g(F)dF=f(x)dx

Comparing the two equations we see that g(F) must be uniform.

$\endgroup$
-2
$\begingroup$

Since it is shown by JimmyK4542 that the cdf is equal to $y$, differentiating this with respect to $y$ will yield 1, which will be the pdf of $Y$. And a random variable that is uniformly distributed on $[0,1]$ should have a pdf equal to 1 by definition. Therefore, $Y$ is uniformly distributed over $[0,1]$.

$\endgroup$

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.