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$X\in Un(0,1)$, form $U=-ln(1-x)$.

I want to calculate $f_U$. My attempt:

$F_U(t)=P(U\leq t)=P(-ln(1-X)\leq t)=P(1-X \geq e^{-t})=P(X\leq 1-e^{-t})=F_X(1-e^{-t})$.

Since $X\in Un(0,1)$. It has distribution $F_X(t)=\begin{cases} 0 &t<0 \\t & 0\leq t \leq 1 \\ 1 &t>1 \end{cases}$

Then $F_X(1-e^{-t})=\begin{cases} 0 &1-e^{-t}<0 \\1-e^{-t} & 0\leq 1-e^{t} \leq 1 \\ 1 &1-e^{-t}>1 \end{cases}$

Am I on the right track, should I just simplify the inequalities above for the different cases. And then just differentiate $F_X(1-e^{t})$ with respect to $t$?

How do I solve the second and third inequality for $t$?

$0\leq 1-e^{-t} \leq 1$ corresponds to $0\leq e^{-t} \leq 1$, Since I cant take the natural logirthm of 0 what do I do?

fejz1234
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  • You are generally on the right track. However, in the third equality in your attempt, you made an error which propagated through the rest of your attempt. – Ian Oct 19 '17 at 17:43
  • Ah yep it should be $e^{-t}$. So do I just solve those inequalities in $F_X(1-e^{-t})$ for $t$? And then differentiate each interval with respect to $t$ in order to obtain $f_U$ ? – fejz1234 Oct 19 '17 at 17:47
  • That's correct! – Mike Earnest Oct 19 '17 at 18:08
  • One suggestion - $1 - X$ has the same distribution as $X$. So you can replace $1 - X$ by $X$. – Abhiram Natarajan Oct 19 '17 at 18:09
  • How do I solve $0\leq 1-e^{-t} \leq 1$? It corresponds to $0 \leq e^{-t} \leq 1$. But I cant take the natural-logarithm of 0. – fejz1234 Oct 19 '17 at 20:27
  • @AbhiramNatarajan I dont understand how you deduce that $1-X$ has same distribution as $X$. How do I see that? – fejz1234 Oct 20 '17 at 17:07
  • @fejz1234. Since it is a uniform distribution on (0, 1), intuitively, you can see that the probability density of $\alpha$ is the same as the probability density of $1 - \alpha$. If you want a rigorous proof, you can say $F_{1 - X}(\alpha) = f_X(1 - \alpha) = f_X(\alpha)$, where $f_X(\cdot)$ is the probability density at $\cdot$, and the second equality is due to uniformity of distribution.

    In other words, the uniform distribution is symmetric around it's mean.

     – Abhiram Natarajan Oct 20 '17 at 17:10
  • Ok, but could you help me with my previous comment regarding that inequality. How do I solve for $t$? – fejz1234 Oct 20 '17 at 17:24
  • For $0 \le e^{-t} \le 1$, you get $0 \le t < \infty$. – Abhiram Natarajan Oct 21 '17 at 18:42
  • @AbhiramNatarajan
    Would you mind explaining how? I couldn't take the natural logartihm on both sides since its not defined for 0, how do you obtain $0\leq t\leq \infty$?     – fejz1234 Oct 21 '17 at 20:52

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