Upper Bound on $E[\frac{1}{1-X}]$ where $E[X]=a$ and $0

Question

$X$ is a discrete random variable that can take values from $(0,1)$. Since $\varphi(x)=1/x$ is a convex function, we can use Jensen's inequality to derive a lower bound: $$ E\left[\frac{1}{1-X}\right]\ge \frac{1}{1-E[X]}=\frac{1}{1-a} $$ Is it possible to derive an upper bound?

Answer 1

There is no upper bound.

Intuitively, if $X$ has substantial support along a sequence approaching $1$, then $1/(1-X)$ could have a divergent (arbitrarily large) expectation. To show there is no upper bound, all we have to do is find a combination of support and probabilities that achieves the desired expectation of $a$. The following explicitly constructs such an $X$.

---

Assume $0 \lt \lambda \lt 1$ (to be chosen later) and $s \gt 1$ (also to be chosen later). Let $X$ take on the values $$a_n = 1 - \lambda n^{-s}$$ with probabilities $$p_n = \frac{n^{-s}}{\zeta(s)},$$ $n = 1, 2, \ldots $. Then

$$a = \mathbb{E}(X) = \sum_{n=1}^\infty p_n a_n = \frac{1}{\zeta(s)}\sum_{n=1}^\infty n^{-s}\left(1 - \lambda n^{-s}\right) = 1 - \lambda \frac{\zeta(2s)}{\zeta(s)}.$$

The range of $f(s) = \zeta(2s)/\zeta(s)$ is the interval $(0,1)$, as this partial graph indicates:

Selecting $\lambda$ such that $1-a \lt \lambda \lt 1$, pick $s \gt 1$ for which $f(s) = (1-a)/\lambda$; that is, $a = 1 - \lambda f(s)$. This constructs an $X$ with all the stated properties.

Consider

$$\mathbb{E}\left(\frac{1}{1-X}\right) = \sum_{n=1}^\infty p_n \frac{n^s}{\lambda} = \frac{1}{\lambda\zeta(s)}\sum_{n=1}^\infty 1.$$

The sum diverges. Consequently no upper bound is consistent with the stated conditions.