I'm doing this problem and I have some doubts:
Let $X$ be a continuous uniform random variable on $(0,1]$. Find the pdf and the density function of $Y=-\frac{1}{\lambda} \ln(X)$, with $\lambda >0$.
Here's what I've done:
We know that
$$F_{X}(t)=\cases{0\ \ \ \text{if}\ \ \ t\leq 0\\t\ \ \ \text{if}\ \ \ t\in (0,1]\\1\ \ \ \text{if}\ \ \ t> 1}.$$
From here, we want $F_{Y}(t)$. This is:
$$F_{Y}(t)=P\{Y\leq t\}=P\{-\frac{1}{\lambda} \ln(X)\leq t\}=P\{X\geq e^{-\lambda t}\}.$$
How do I see from here who is $F_{Y}(t)?$ Can I see it using the density function of $X$?
Best Answer
We have \begin{align} F_Y(t) &= \mathbb P(X\geqslant e^{-\lambda t})\\ &= 1- \mathbb P(X\leqslant e^{-\lambda t})\\ &= 1 -( e^{-\lambda t}\mathsf 1_{(0,1]}(e^{-\lambda t}) + \mathsf 1_{[1,\infty)}(e^{-\lambda t}))\\ &= (1 - e^{-\lambda t})\mathsf 1_{[0,\infty)}(t), \end{align} so that $Y\sim\operatorname{Exp}(\lambda)$. (Note that $0<e^{-\lambda t}\leqslant 1$ for $0\leqslant t<1$.)