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Let $X \sim \operatorname{Exp}(1)$, and show $Y = F_X(X)$ has uniform distribution on $[0,1]$.

I calculated $F_Y$, since the cumulative distribution function identifies a distribution. We have: \begin{align*} F_Y(y) &= P(Y \leq y) \\ &= P(1 - e^{-X} \leq y) \\ &= P(X \leq -\ln(1 - y)) \tag{for y < 1}\\ &= F_X(-\ln(1 - y)) \\ &= 1 - \exp(\ln(1 - y)) \\ &= y. \end{align*}

So I got $F_Y(y) = \boldsymbol{1}_{(-\infty, 1)}(y)$, but I should have gotten instead $F_Y(y) = \boldsymbol{1}_{[0, 1]}(y) + H(y - 1)$, which is completey different! So where did I go wrong?

Note: $H(x)$ is the Heaviside step function.

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    $\begingroup$ For proving that $Y$ has uniform distribution on $[0,1]$ it is enough to show that $F_Y(y)=y$ for $y\in(0,1)$. Isn't that exactly what you did? Your remark "So I got $F_Y(y)=1_{(-\infty,1)}(y)$" makes no sense. $\endgroup$ – drhab Jul 19 '16 at 8:21
  • $\begingroup$ @drhab Initially $y \in \mathbb R$, but in order for the logarithm to make sense it has to be $y < 1$. And for the uniform distribution it's not simply $F_Y(y) = y$, because it also has to be exactly $F_Y(y) = 0$ for $y < 0$ and $F_Y(y) = 1$ for $y > 1$, doesn't it? $\endgroup$ – rubik Jul 19 '16 at 8:43
  • $\begingroup$ It is indeed not simply $F_Y(y)=y$ but if we have $F_Y(y)=y$ on $(0,1)$ then automatically we also have $F_Y(y)=1$ if $y\geq1$ and $F_Y(y)=0$ if $y\leq0$. This because we know that $F_Y$ is a CDF. $\endgroup$ – drhab Jul 19 '16 at 10:05
  • $\begingroup$ In general, the CDF of the uniform distribution is on an interval $[a,b]$ is $$\begin{cases} 0, & x < a, \\ \frac{x - a}{b - a}, & a \le x \le b, \\ 1, & x \ge b \end{cases}.$$ For the interval $[0,1]$ this implies the CDF should be $$\begin{cases} 0, & x < 0, \\ x, & 0 \le x \le 1, \\ 1, & x \ge 1 \end{cases}.$$ The function you gave is $$1_{[0,1]}(y) + H(y - 1) = \begin{cases} 0, & x < 0 \\ 1, & x \ge 0 \end{cases} = 1_{[1, \infty)}.$$Maybe you wanted to write $F_Y(y) = y \cdot 1_{[0,1]}(y) + H(y - 1)$, which gives the CDF I described above. $\endgroup$ – Viktor Glombik Jun 25 at 19:19
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If $F$ is a CDF and it must be shown that $F$ is the CDF corresponding with the uniform distribution on $[0,1]$ then it is enough to prove that $F(y)=y$ for every $y\in(0,1)$.

This because the values of $F$ on elements not in $(0,1)$ are determined by the following rules for a CDF:

If $y\geq1$ then $1\geq F(y)\geq F(z)=z$ for every $z\in(0,1)$ and consequently $F(y)=1$.

If $y\leq0$ then $0\leq F(y)\leq F(z)=z$ for every $z\in(0,1)$ and consequently $F(y)=0$.


We make use of the facts that $0\leq F(y)\leq1$ must be true for any $y$, and $F$ is monotonically increasing.

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  • $\begingroup$ Ok, I get it. Thanks for clearing up my misunderstanding! $\endgroup$ – rubik Jul 19 '16 at 10:08

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