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?
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Hello! Can you please add a link to the notes and the specific page that contains this statement?("Essential implication" is also a bit vague, so it will help to clarify this). – Sarvesh Ravichandran Iyer May 19 '21 at 12:02
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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
tommik
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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 $0$.
Mostafa Ayaz
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