Why is this exponentially distributed

Question

At some point in my notes they essentially imply that when $u\sim \mathcal{U}(0, 1)$ then $-\log(u)\sim \text{ExponentialDistribution}(\lambda=1)$. Clearly this isn't true since by the integral transform we should use the inverse CDF?

Answer 1

Consider that the CDF of you uniform is

$$F_U(u)=u$$

and setting

$$Y=-\log U$$

the CDF of $Y$ is, by definition

$$\mathbb{P}[Y\leq y]=\mathbb{P}[-\log U\leq y]=\mathbb{P}[U>e^{-y}]=1-e^{-y}$$

which is exactly the CDF of a negative exponential with mean 1

Answer 2

This is true. Since the CDF of exponential distribution is exponential, its inverse becomes in logarithmic form. Another reason for the correctness of the result, is that $u$ varies within $[0,1]$, so its logarithm must change within $[-\infty,0]$. Since $-\log 0^+=\infty$, the exponential random variable $-\log u$ will be expanded to infinite when $u$ is close to $0$.