If $X \sim \exp(\lambda)$, then what is the exact distribution of $Y=e^{-\lambda X}$?

Question

If $X \sim \exp(\lambda)$, then what is the exact distribution of $Y=e^{-\lambda X}$?

After my calculation of the probability density function of $f(y)=\dfrac{1}{y\cdot(e^{\ln y})}$ that should be $1$.

And that is wired. I don't know how to convert into a reasonable distribution function.

Answer 1

1. For $y < 0$, $$F_Y(y) = P(Y \leq y) = P(e^{-\lambda X} \leq y) = 0 $$ because $e^{-\lambda x} \geq 0$ for all $x \in \mathbb{R}$.

2. For $y > 1$, $$F_Y(y) = P(Y \leq y) = P(e^{-\lambda X} \leq y) = 1 $$ because $e^{-\lambda x} \leq 1$ for all $x \geq 0$.

3. For $0 \leq y \leq 1$ \begin{align*} F_Y(y) = P(Y \leq y) &= P(e^{-\lambda X} \leq y)\\ &= P(\ln(e^{-\lambda X}) \leq \ln(y))\\ &= P(-\lambda X \leq \ln(y))\\ &= P\left (\lambda X \geq \ln\left (\frac{1}{y}\right )\right )\\ &= P\left (X \geq \frac{1}{\lambda} \ln \left(\frac{1}{y} \right) \right )\\ &= \int_{\frac{1}{\lambda} \ln \left(\frac{1}{y} \right)}^{\infty} \lambda e^{-\lambda x} dx\\ &= - \left [0 - e^{-\lambda \frac{1}{\lambda} \ln \left(\frac{1}{y} \right)} \right ] \\ &= e^{ -\ln \left(\frac{1}{y} \right)}\\ &= e^{\ln(y)}\\ &= y \end{align*}

Answer 2

Observe that the event $0 < Y \leq 1$ has probability $1$, so, if $y\leq0$, then $F_Y(y) = 0$ and if $y>1$, then $F_Y(y) = 1$. So, let $0<y\leq1$ and now note that \begin{align*} F_Y(y) = \textsf{P}(Y \leq y) &= \textsf{P}(e^{-\lambda X} \leq y) \\ &= \textsf{P}(-\lambda X \leq \log y) \\ &= \textsf{P} \Big(X \geq -\frac{1}{\lambda} \log y \Big) \quad [\textrm{because } \lambda>0] \\ &= 1-F_X\Big(-\frac{1}{\lambda} \log y \Big). \end{align*} Now, the distribution function of $X$ is given by $$F_X(x) = \begin{cases} 0 & \textrm{if } x<0 \\ 1-e^{-\lambda x} & \textrm{if } x \geq 0 \end{cases} $$ and since $\displaystyle -\frac{1}{\lambda} \log y \geq 0$, it follows that $\displaystyle F_X\Big(-\frac{1}{\lambda} \log y \Big) = 1-y$. In conclusion, $$F_Y(y) = \begin{cases} 0 & \textrm{if } y \leq 0 \\ y & \textrm{if } 0 < y \leq 1 \\ 1 & \textrm{if } y > 1 \end{cases} $$