Find the distribution of $Y = 1-e^{-x}$ $X$is the standard exponential

Question

$X$ is the standard exponential find the distribution of $Y = 1-e^{-x}$

$P(Y\le y) = P(1-e^{-x} \le y) = P(e^{-x} \ge 1-y) = P(-x \ge ln(1-y))= P(x \le -ln(1-y)) = 1-e^{ln(1-y)} = 1-(1-y) = y$ since for exp(1) $P(X\le x) = 1-e^{-x}$

$P(Y=y) = dP(Y\le y)/dy = 1$

What is wrong in the logic/calculations?

Answer 1

Be carefull

Y is a continuos variable, so $P(Y = y) = 0$

You mean $f_{Y}(y) = dP(Y\leq y)/dy = 1$

Thats make sense if and only if $y\in [0,1]$

Because there no exist $ln(1-y)$ if $y>1$.

Then we have $f_{y} = 1$ in $[0,1]$ and $f_{y} = 0$ in $[0,1]^{c}$

Then $Y$~$U([0,1])$