If $X$ is a random variable with absolutely continuous distribution function $F(x)$, then find the distribution of $Y= \log(1-F(x))$
Find the distribution of $Y=\log(1-F(x))$ if $X$ is a r.v with absolutely c.d.f. $F(x)$
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probability-distributions
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0Importantly, it is not really $\log(1-F(x))$ but $\log(1-F(X))$. Accordingly, you'll want to know what the distribution of $1-F(X)$ is when $X$ has continuous CDF $F$. Surprisingly this doesn't depend on what $F$ actually is, provided it is at least continuous. – 2017-01-03
1 Answers
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I guess you mean that $Y=\log(1-F(X))$. In this case let $F^-$ denote generalized inverse of $F$. Then we get applying the basic properties of $F^-$: \begin{align} F_Y(x) &= \mathbb P (Y \leq x) = \mathbb P \bigl(F(X) \geq 1- \exp(x) \bigr)= \mathbb P \Bigl(X \geq F^-\bigl(1- \exp(x) \bigr)\Bigr)\\ &= 1- F\bigl(F^-\bigl(1- \exp(x)\bigr)\bigr) = \exp(x), \quad x < 0. \end{align} Since $Y\leq 0$ we get $F_Y(x) = 1$ for $x\geq 0$.