When given $X = −\frac{\ln(1 − U)}{\lambda}$, why is $X \sim \mathrm{Exp}(\lambda)$ and not $\mathrm{Exp}(-\lambda)?$

Question

When given $X = \frac{−\ln(1 − U)}{\lambda}$, why is the distribution of $X \sim \mathrm{Exp}(\lambda)$ and not $\mathrm{Exp}(-\lambda)?$ I solved for $X$ to get:

$-X=\cfrac{\ln(1-U)}{\lambda}$

$-\lambda X=\ln(1-U)$

Considering $1-U$ is equivalent to $U$, then $-\lambda X=\ln(U)$

$\exp(-\lambda X)=U$

I'm not exactly sure of why I can make the leap, but I've trained it into my brain that the above is the equivalent of $X \sim \mathrm{Exp}(-\lambda)$

So, why is the answer considered $X \sim \mathrm{Exp}(\lambda)?$

Answer 1

Using the fact that $\lambda>0$, the exponential function is strictly increasing and that $U$ is uniform one gets \begin{aligned} P[X\leq x] &=P[-\ln(1-U)/\lambda \leq x]=P[\ln\big((1-U)^{-1}\big)\leq\lambda x]\\ &=P[\frac{1}{1-U}\leq e^{\lambda x}]\\ & = P[1-U\geq e^{-\lambda x}]\leq P[U\leq 1-e^{-\lambda x}]\\ &=1-e^{-\lambda x} \end{aligned}

For that, one sees that $X$ has the distribution corresponding to the exponential distribution with parameter $\lambda$.

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One can also use the observation that $1-U$ and $U$ are equal in law. Then

\begin{aligned} P[-\ln(1-U)\leq x]&=P[-\ln(U)\leq x]=P[U^{-1}\leq e^{\lambda x}]\\ &=P[U\geq e^{-\lambda x}] = 1-P[U\leq e^{-\lambda x}]\\ &=1-e^{-\lambda x} \end{aligned}