What property of exponential distribution was used to infer this in the solution? And where did I mess up?

Question

I'm looking over this example of transformations of random variables:

Let $X,Y$ be an independent random variables, such that $X\sim exp(1), Y\sim exp(2)$.

We define $Z=e^{-X}+e^{-2Y}$.
Find $f_z(z)$.

My attempt (the idea):

And the solution was let $U=e^{-X}$, $V=e^{-2Y}$.
It was easy to find that $U\sim U(0,1)$, with $f_U(u)=f_X(-\ln(u))\frac{1}{u}=1$. Considering $g(x)=e^{-x}$, $g^{-1}(u)=-\ln(u)$.
Using this formula : $f_U(u)=f_X(g^{-1}(u))*|[g^{-1}(u)]'|$.

And for $V=e^{-2Y}$, I tried to do the same thing, defining $g(y)=e^{-2y}\Longrightarrow g^{-1}(v)=-\frac{\ln(v)}{2}$. $(g^{-1}(v))'=-\frac{1}{2v}$.
So $f_V(v)=e^{\ln(v)}\frac{1}{2v}=\frac{1}{2}, v\in(0,1)$.

And now using convulotion: $Z=V+U$.
$$\int_{-\infty}^{\infty}f_U(u)f_V(z-u)du=\int_{max\{z-1,0\}}^{min\{1,z\}}1*\frac{1}{2}du=$$ Whenever $z\in(0,1): f_Z(z)=\frac{z}{2}$.
Whenever $z\in(1,2): f_Z(z)=\frac{2-z}{2}.$ and otherwise $0$.

What the solution of the example used:
For $U$, it was the same calculations as me, but for $V$:
$V=e^{-2Y}$ and then $2Y\sim exp(0,1)$ so $V\sim (0,1)$.
And they reached $f_Z(z) = z$ whenever $z\in (0,1)$.
$f_Z(z)=1-z$ whenever $z\in (1,2)$. and otherwise $0$.

I don't understand how they easily got that $V$ is uniform, and why they reached another PDF of $Z$, or where I messed up calculating $V$, would appreciate any help!

Answer 1

Your mistake is in the calculation of $f_V(v)$.

$Y \sim exp(2) \implies f_Y(y) = 2 e^{-2y}$

Now $V = e^{-2Y} \implies g^{-1}(v) = - \frac{ \ln v}{2}$

So,
$f_V(v) = f_Y(g^{-1}(v))*|[g^{-1}(v)]'| = 2 e^{\ln v} \cdot \frac{1}{2v} = 1$

As the support of $V$ is $(0, 1)$, we have $V \sim U(0,1)$

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The solution simply uses the fact that if $Y \sim exp(2)$ then $2Y\sim exp(1)$ which you can show as below -

If $Y \sim exp (1)$ and $Z = 2Y$

$f_Z(z) = f_Y(\frac{z}{2}) \cdot \frac{1}{2} = e^{-z} \implies Z \sim exp(1)$.