I've come across the following claim regarding equivalence of functions of random variables, and I need help verifying whether this claim is true or not:
We are given the following function of random variable $X$:
$$G(X) = X - ln(X)$$
where $E(X) > 1$. Observe that $G(X)$ is also a monotonically increasing function for $X >= 1$.
Now the claim is that due to the aforementioned properties, $G(X)$ is equivalent to $F(X) = X$.
I fail to see how this claim is possible, unless perhaps $G(X)$ and $F(X)$ are equivalent in distribution, but that too is difficult to see. I will note that in practical applications, we are concerned only in the tails of distributions of $G(X)$ and $F(X)$.
Thanks in advance.
The only case where $X$ and $Y=G(X)$ have the same distribution is the trivial one $X = 1$ almost surely. This is ruled out by your condition $\mathbb E[X] > 1$.
Let $G(x) = x - \ln(x)$. Suppose $X$ is a random variable with $X > 0$ a.s., and $Y = G(X)$ has the same distribution as $X$.
Note that $G$ is a strictly convex function with $G(1) = 0$ and $G'(1) = 0$, so $G$ has a strict minimum at $1$, i.e. $G(x) \ge 1$ for all $x > 0$, with equality only at $x=1$. Since $Y \ge 1$ a.s., if its distribution is the same as that of $X$, we also have $X \ge 1$ a.s. But then $Y \le X$ a.s., with $Y < X$ whenever $X > 1$, so the only way $X$ and $Y$ can have the same distribution is that $X = 1$ a.s.