Suppose I have random variable $X \sim \exp(1)$, then we have a random variable Y such thath
$$Y = 1+ e^{-sX}$$
When I find the PDF for $Y$, it came out to be
$$f_Y(y) = -e^{x(s-1)}$$
Steps to find the PDF Let $F_Y(y)$ be the CDF of new variable $Y$, $$F_Y(y) = P[Y \leq y] = P[1+ e^{-sX} \leq y] = P\Large[\small X \geq \frac{-\ln(y-1)}{s}\Large]$$
$ P\Large[\small X \geq \frac{\ln(y-1)}{s}\Large]$ is the Complimentary CDF of $X \sim \exp(1)$ and equals to $\Large e^{-\frac{\ln(y-1)}{s}}$
$$F_Y(y) = \Large e^{-\frac{\ln(y-1)}{s}}.$$
Now, if I take derivate of $F_Y(y)$, I get the PDF as above to be a negative value.
If now I replace $y = 1 + e^{-sx}$, then I get the above result.
$P[X \geq \frac {-\ln (y-1)} s] =e^{\ln (y-1) /s}$ and not $e^{-\ln (y-1) /s}$; the derivative of this is $\frac 1 {s(y-1)} e^{\ln (y-1) /s}$ for $y >1$.
[$P{(X>t)}=e^{-t}$. Put $t=\frac {-\ln (y-1)} s$ in this].