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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?

Inkyu Bang
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2 Answers2

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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*}

Tom Chen
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  • 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}$? – Inkyu Bang Jun 05 '19 at 12:30
  • 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. – Tom Chen Jun 09 '19 at 22:27
  • Notice that $\frac{1}{1-x}=1+x+x^2+\frac{x^3}{1-x}$. This gives $\frac{1}{1+X}=\frac{1}{1+\mathbb E[X]}\frac{1}{1-\frac{\mathbb E[X]-X}{1+\mathbb E[X]}}=\frac{1}{1+\mathbb E[X]}(1+\frac{\mathbb E[X]-X}{1+\mathbb E[X]}+(\frac{\mathbb E[X]-X}{1+\mathbb E[X]})^2+(\frac{\mathbb E[X]-X}{1+\mathbb E[X]})^3\frac{1}{1-\frac{\mathbb E[X]-X}{1+\mathbb E[X]}})$. This yields $\mathbb E[\frac{1}{1+X}]-\frac{1}{1+\mathbb E[X]}=\frac{\mathrm{Var}(X)}{(1+\mathbb E[X])^3}+\mathbb E[\frac{(\mathbb E[X]-X)^3}{(1+\mathbb E[X])^3(1+X)}]$. Why is the second part negative? – Matija Oct 19 '22 at 13:29
  • Also, notice that this is closely related to this question. – Matija Oct 19 '22 at 13:32
  • The argument given above is incorrect; you are essentially claiming that $$\Big|\frac{1}{1+x} - \frac{1}{1 + y}\Big| \leq \frac{(x-y)^2}{(1 + y)^3} \qquad \mbox{for any}~x, y > 0.$$ The above inequality is only true if $y \leq x$. In your application you have $y = \mathbb{E}[X]$ and $x = X$. Your inequality is only true if $\mathbb{E}[X] \leq X$. This can be seen from Matija's comment above. It is exactly in this setting that $(\mathbb{E}[X] - X)^3 \leq 0$. – Drew Brady Oct 19 '22 at 14:08
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Let's rephrase the inequality as $|\mathbb E[\frac{1}{1+X}]-\frac{1}{1+\mathbb E[X]}|\le\varepsilon$, i.e. we look at $\varphi(x)=\frac{1}{1+x}$. Let $X_n\in\{0,n\}$ with $\mathbb P(X_n=n)=1/\sqrt{n}$, then we have $\mathbb E[\frac{1}{1+X}]-\frac{1}{1+\mathbb E[X]}=1-\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\frac{1}{1+n}-\frac{1}{1+\sqrt{n}}$. Taking $n\rightarrow\infty$ shows that the gap tends to $1$. On the other hand, since $\varphi(x)$ is decreasing we have $0<\varphi\le 1$ and thereby $\mathbb E[\frac{1}{1+X}]-\frac{1}{1+\mathbb E[X]}<1$, whereas $\mathbb E[\varphi(X)]-\varphi(\mathbb E[X])\ge 0$ follows from Jensen's inequality, so the tight constant is $\varepsilon=1$. The above implies that the answer for a given expectation $\mathbb E[X]$ will still be trivial, given by $1-\varphi(\mathbb E[X])$. If the variance is additionally fixed, then an argument similar to the discussion of Hoeffding's inequality or this question can probably provide a non-trivial tight constant.

Matija
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