PDF of $Y = \log X$ when $ X \sim\text{Exp}(1)$

Question

$X$ is a exponential R.V with $\lambda=1$.

Im asked to find the probability density function of a r.v, $Y = \log X$

I compute it as follows:

$P( Y < y1) = P (\log X < y_1) = P ( X < e^{y_1} ) = F_X(e^{y_1})$ [ $F_X$ is the c.d.f]

$f_Y(y) = d/dy [ F_X (e^y)] = (e^y)(f_X(e^y)) = e^y [ e^{-(\exp y)} ]$

Question is whats the domain of the p.d.f of $Y$, $f_Y(y)$?

Since $X$ is exponential its defined from $x \ge 0$ , but $\log X$ is undefined at $X = 0$

so is $Y$ still a valid r.v if it's missing one point in the sample space of $X$?

Answer 1

If $X$ takes values in $(0,\infty)$ then $\log X$ takes values in $(-\infty,\infty)$. Let $$F_X(t)=\left(1-e^{-\lambda t}\right)\cdot\mathsf 1_{(0,\infty)}(t)$$ be the CDF of $X$, then for $y\in\mathbb R$ we have \begin{align} F_Y(y) &= \mathbb P(\log X\leqslant y)\\ &= \mathbb P(X\leqslant e^y)\\ &= F_X(e^y)\\ &= \left(1-e^{-\lambda e^y}\right)\cdot\mathsf 1_{(0,\infty)}(e^y)\\ &= 1 -e^{-\lambda e^y}. \end{align} The density is obtained by differentiating: $$ f_Y(y) = \frac{\mathsf d}{\mathsf dy} F_Y(y) = \lambda e^{y-\lambda e^y}. $$ When $\lambda=1$, this reduces to $e^{y-e^y}$.