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If $X\sim\text{Exp}(1)$, find a function $g$ so that $Y=g(X)$ is uniformly distributed over $(1,3)$.

I first tried to map $X$ onto $U\sim \text{Unif}(0,1)$, so that $X = F^{-1}(U)$ and $U=F(X).$

So I set $u=1- e^{-x}$ and found that $x=-\ln(1-u)$ for the inverse CDF.

I then tried to shift and scale $\text{Unif}(0,1)$ to $\text{Unif}(1,3)$ by stating that $½(Y-1)=U$ and plugged that into $u$ in the inverse function $x=-\ln(1-u).$

My answer turned out to be $Y=3 - 2e^{-X}$ but the correct solution is $Y = 2e^{-X} + 1$.

Can anyone please explain what I did wrong?

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$U\sim \text{Uniform}(1,3) \iff 4-U\sim \text{Uniform}(1,3)$

So both your answer and the answer in the book are correct.

What probably happened with the answer in the book is that instead of $F(X)=1-e^{-X}\sim \text{Uniform}(0,1)$, they took $1-F(X)=e^{-X}\sim\text{Uniform}(0,1)$ as well.

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    Thanks! A sidenote in the book says that we should assume x = -ln(u), whereas I found that x = -ln(1-u) by taking the inverse function of u=1-e^(-x). Why can we assume that x = -ln(u)? – 2017-01-07
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    Because if $U$ is Uniform$(0,1)$ then $1-U$ is Uniform$(0,1)$ as well. – 2017-01-07
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    ... and when an Exponential random variable is generated, $-\ln(U)$ is simpler than $-\ln(1-U)$ (by one subtraction), so it is the preferred method. – 2017-01-07