Is this true: $P(X>x) = P(\log(X)>\log(x))$ I can't find any theorem regarding this. I know it works for normal / lognormal distributions and as the logarithm is an affine transformation and the cdf is increasing it seems plausible but i can't find any reason/proof in favour or against.
Many thanks!
Alex
 A: The simple/heuristic explanation is for why this is true is that $[X > x]$ occurs if-and-only-if $[\log X > \log x]$ provided that both $X$ and $x$ are non-negative. Hence $[X > x]$ and $[\log X > \log x]$ are just two names for the same event, so the probability is the same. 
I think the above description gets at the heart of what is going on, but since you asked for a "proof" I will also give something formal. Consider a probability space $(\Omega, \mathcal F, P)$ and let $X: \Omega \to \mathcal X$ be a random element and suppose $g: \mathcal X \to \mathcal Y$ is a bijection between the sets $\mathcal X$ and $\mathcal Y$.
Let $A$ be such that $[X \in A] = \{\omega : X(\omega) \in A\}$ is measurable, and recall that we define $P(X \in A)$ and $P(g(X) \in g(A))$ to be $$P(\{\omega \in \Omega: X(\omega) \in A\}) \quad \text{and} \quad P(\{\omega \in \Omega: g(X(\omega)) \in g(A)\})$$
respectively. However, because $g$ is a bijection, it is true that $x \in A$ if-and-only-if $g(x) \in g(A)$, i.e., 
$$
\{\omega \in \Omega: X(\omega) \in A\} =
\{\omega \in \Omega: g(X(\omega)) \in g(A)\}.
$$
Hence, $P(X \in A) = P(g(X) \in g(A))$. 
To apply this to your problem, let $\mathcal X = [0, \infty]$ and $\mathcal Y = [-\infty, \infty]$ and $g(x) = \log x$. The set $A$ is $(x, \infty]$ and $g(A) = (\log x, \infty]$. 
A: You can make your transformation more general. It just needs to be measurable. 
If you define $\mathbb{P}_Y(Y \in A) = \mathbb{P}_Y(g(X) \in A) = \mathbb{P}_X(X \in g^{-1}(A))$ for a measureable $g$, where pre-image is defined as $g^{-1}(A) = \{ x : g(x) \in A\}$, then I think reasons why $\mathbb{P}_Y$ is still a valid probability measure is because the pre-image satisfies some properties, namely:


*

*$g^{-1}\left(\bigcup_i A_i\right) = \bigcup_i g^{-1}(A_i)$, and

*$g^{-1}(A^c) = \left[g^{-1}(A)\right]^c$.


This will mean $\mathbb{P}_Y$ is a valid probability measure because it satisfies all the axioms still, and so it will work for any set in the sigma field you throw into it, in particular the one you're looking at.
