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For a real-valued random variable $X \geq 0$,

We have $1 - \frac{1}{1+E \left[x\right]} \geq E \left[ 1 - \frac{1}{1+x} \right]$ (Jensen's inequality).

We want to get a tight constant gap between $1 - \frac{1}{1+E \left[x\right]}$ and $E \left[ 1 - \frac{1}{1+x} \right]$, i.e., $ \lvert 1 - \frac{1}{1+E \left[x\right]} - E \left[ 1 - \frac{1}{1+x} \right] \rvert \leq \epsilon_0$.

Any hints for this inequality?

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Taylor expanding $\frac{1}{1+x}$ around $x = E[x]$ gives us \begin{align*} \text{Gap} &= \left|1 - \frac{1}{1+E[x]} - E\left[1 - \frac{1}{1+x}\right] \right| \\&= \left|E\left[\frac{1}{1+x}\right] - \frac{1}{1+E[x]} \right| \\ &= \left|\frac{1}{1+E[x]} - E\left[\frac{x - E[x]}{(1 + E[x])^2}\right] + E\left[\frac{(x - E[x])^2}{(1 + E[x])^3}\right] - \cdots - \frac{1}{1+E[x]}\right| \\ &\le \frac{\text{Var}(x)}{(1 + E[x])^3} \end{align*}

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  • $\begingroup$ How can we guarantee $ | - E \left[ \frac{x - E[x]}{(1+E[x])^2} \right] + \cdots | \leq \frac{\text{Var}(x)}{(1+E[x])^3}$? $\endgroup$ – Inkyu Bang Jun 5 at 12:30
  • $\begingroup$ The first term $E[x-E[x]] = 0$ and the second term is $\text{Var}(x)/(1+E[x])^3$ Since the series is alternating and your random variable is always positive, the remaining terms only serve to decrease your sum, so we can truncate and leave with that inequality. $\endgroup$ – Tom Chen Jun 9 at 22:27

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