show/hide this revision's text 3 continuous not necessary.

We have $\mu_n \uparrow \infty$. Proof: let $$G(y) = \frac{\int_y^\infty x f(x) dx}{\int_y^\infty f(x) dx}$$ so that $\mu_{n+1} = G(\mu_n)$. Clearly $G$ is a continuous function and $G(y) > y$ for all $y$. But if $\mu_n \to \mu$ for some finite $\mu$ we must have $G(\mu) = \mu$, a contradiction.

More generally, this should show that if $X$ is a continuous random variable with essential supremum $M$, and we define $G(y) = E[X | X \ge y]$ for $y < M$, then the iterates $G^n(y) \to M$.
show/hide this revision's text 2 Punctuation

We have $\mu_n \uparrow \infty$. Proof: let $$G(y) = \frac{\int_y^\infty x f(x) dx}{\int_y^\infty f(x) dx}$$ so that $\mu_{n+1} = G(\mu_n)$. Clearly $G$ is a continuous function and $G(y) > y$ for all $y$. But if $\mu_n \to \mu$ for some finite $\mu$ we must have $G(\mu) = \mu$, a contradiction.

More generally, this should show that if $X$ is a continuous random variable with essential supremum $M$, and we define $G(y) = E[X | X \ge y]$ for $y < M$, then the iterates $G^n(y) \to M$.
show/hide this revision's text 1

We have $\mu_n \uparrow \infty$. Proof: let $$G(y) = \frac{\int_y^\infty x f(x) dx}{\int_y^\infty f(x) dx}$$ so that $\mu_{n+1} = G(\mu_n)$ Clearly $G$ is a continuous function and $G(y) > y$ for all $y$. But if $\mu_n \to \mu$ for some finite $\mu$ we must have $G(\mu) = \mu$, a contradiction.

More generally, this should show that if $X$ is a continuous random variable with essential supremum $M$, and we define $G(y) = E[X | X \ge y]$ for $y < M$, then the iterates $G^n(y) \to M$.