Smallest constant $C$ to bound $\mathbb{E}[\tfrac{1}{\overline{X}_n + 1}] \leq C~\tfrac{1}{\mathbb{E}[X] + 1}$?

Question

Let $X_1, \dots, X_n$ be random real-valued random variables in the interval $[0, a]$. Assume they are independently and identically distributed. Let $\mu = \mathbb{E}[X_i]$ denote their common mean.

Define their average $\overline{X}_{n} = n^{-1} S_n$ where $S_n = \sum_{i=1}^n X_i$.

Let $f(t) := (t + 1)^{-1}$.

Question: What is the smallest constant $C = C(a, n) \geq 1$ such that we have $$ \mathbb{E}[f(\overline{X}_n)] \leq C~f(\mu)? $$

It should be emphasized that the constant $C$ is universal: it is valid for any law of $X_i$, supported on $[0, a]$. It should be dependent only on $a, n$.

Comments:

• Necessarily $C \geq 1$. Note that by Jensen's inequality, we have the following inequality, $ \mathbb{E}[f(\overline{X}_n)] \geq f(\mu), $ since $f$ is a convex function on the nonnegative reals.

• Additionally, I claim that $C \leq 1 + \tfrac{a}{n}$. I have a proof of this below. (I would also be interested if there is a simpler way to establish this. I tried, thinking I could possibly improve the constant, to obtain it by Taylor expansion of the function $f$ around $t = \mu$, but failed to recover it.)

Let $X_{n+1}$ be independent of $\{X_i\}_{i=1}^n$, but identically distributed. Then \begin{multline*} \mathbb{E} [f(\overline{X}_n)] = \frac{n}{\mu} \mathbb{E} \Big[\frac{X_{n+1}}{S_n + n}\Big] = \frac{n}{\mu} \mathbb{E} \Big[\frac{X_{n+1}}{S_{n+1} + n} \frac{S_{n} + n + X_{n+1}}{S_{n} + n}\Big] \\\leq \Big(1 + \frac{a}{n}\Big) \frac{n}{\mu} \mathbb{E} \Big[\frac{X_{n+1}}{S_{n+1} + n} \Big].\qquad \mbox{(1)} \end{multline*} Since $X_{j}$, $j \leq n + 1$ are exchangeable, we also have $$ \mathbb{E} \Big[\frac{X_{n+1}}{S_{n+1} + n} \Big] = \frac{1}{n+1} \mathbb{E} \Big[\frac{\overline{X}_{n+1}}{\overline{X}_{n+1} + n/(n+1)} \Big] \stackrel{{\rm (*)}}{\leq} \frac{\mu}{(n+1)\mu + n} \leq \frac{\mu}{n} \frac{1}{\mu + 1}. \quad \mbox{(2)} $$ Above, we have used Jensen's inequality in (*) with mapping $z \mapsto z/(z + n/(n+1))$ which is concave on the nonnegative reals. Combining bounds (1) and (2), we get a bound $C(a, n) \leq 1 + a/n$.

Answer 1

Here is the solution to the problem. It turns out that the sharpest constant is in fact $$ C(a, n) = \sup_{\lambda \in [0, 1]} \sum_{j=0}^n \frac{\lambda n a + n }{ja + n} \lambda^j (1-\lambda)^{n-j} \binom{n}{j}. $$

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We use $\mathsf{Ber}(p)$ and $\mathsf{Bin}(n ,p)$ to denote the Bernoulli and Binomial distributions with parameter $p \in [0, 1]$ and positive integer $n$.

Let $Y_i = a~\mathsf{Ber}(X_i/a)$. Define the sums $$ S_n = X_1 + \dots + X_n, \quad \mbox{and} \quad T_n = Y_1 + \dots + Y_n. $$ Note that $\mathbb{E}[T_n \mid X_1, \dots, X_n] = S_n$. Additionally, note that $T_n = a~\mathsf{Bin}(n, \lambda)$ in distribution, where $a\lambda = \mathbb{E}[X]$.

Therefore, using the convexity of the map $t \mapsto (n + t)^{-1}$, we have

$$ \frac{\mathbb{E}((1 + \tfrac{1}{n}S_n)^{-1})}{(1 + \mathbb{E}[X])^{-1}} = \frac{n~\mathbb{E}((n + S_n)^{-1})}{(1 + \mathbb{E}[X])^{-1}} \leq \frac{n~\mathbb{E}((n + T_n)^{-1})}{(1 + \mathbb{E}[X])^{-1}} = \frac{\mathbb{E}((1 + \tfrac{1}{n}T_n)^{-1})}{(1 + \mathbb{E}[Y])^{-1}} $$ This implies that with $f(z) = 1/(1+ z)$, $$ \sup_{X:X \in [0, a]~a.s.} \frac{\mathbb{E}(f(\overline{X}_n))}{f(\mathbb{E} X)} = \sup_{\lambda \in [0, 1]} \frac{\mathbb{E}[f(\tfrac{a}{n}~\mathsf{Bin}(n, \lambda))]}{f(a \lambda)}. $$ The right hand side is equal to $$ \sup_{\lambda \in [0, 1]} n(1 + a\lambda) \sum_{j=0}^n \frac{1}{ja + n} \lambda^j (1-\lambda)^{n-j} \binom{n}{j}. $$ This proves the claim.

Addendum: The original post claimed that the optimum in the variational expression above is attained at $\lambda = 1/2$. This is false. It is not actually possible to compute the optimum analytically (to my knowledge).

Answer 2

I think a smoothing argument shows that the maximum is attained for the random variable $X$, where $X = a$ with probability $1/2$ and $X=0$ with probability $1/2$. This yields$$C(a,n) = \frac{n(2+a)}{2^{n+1}}\sum_{k=0}^n \frac{1}{ka+n}{n \choose k}.$$

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Let's start with $n=1$ (and arbitrary $a > 0$).

Fix $\lambda \in [0,1]$. We claim the maximum of $\mathbb{E}\left[\frac{1}{1+X}\right]$ over all rv's $X$ on $[0,a]$ with $\mathbb{E}[X] = a\lambda$, is attained when $X = a$ with probability $\lambda$ and $X = 0$ with probability $1-\lambda$.

We quickly note that the above claim resolves the $n=1$ case, since $(1+a\lambda)\left(\lambda\frac{1}{1+a}+(1-\lambda)\right)$ is maximized when $\lambda = 1/2$, no matter what $a$ is.

Anyways, the claim follows from a smoothing argument. Indeed, the inequality $$\frac{1}{1+b} < \frac{1}{2}\cdot\frac{1}{1+(b-\Delta)}+\frac{1}{2}\cdot\frac{1}{1+(b+\Delta)}$$ holds for any $0 < \Delta < b$. Therefore, any "mass" of $X$ strictly between $0$ and $a$ can be broken up by moving half of it down a bit and half of it up a bit, preserving $\mathbb{E}[X]$ while increasing $\mathbb{E}\left[\frac{1}{1+X}\right]$.

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Hopefully the above is rigorous for you enough. It's not too hard to make it (more) rigorous.

The same argument should work for any $n \ge 1$; I know that the maximum still occurs at $\lambda = 1/2$, though the analogue of that centered inequality (to reduce to the optimization over $\lambda$ problem) becomes a bit more complicated, though shouldn't be too bad (I'll do it later).

Answer 3

This is not a complete answer, but builds on the argument of mathworker21.

The goal of this extended comment is to show a "simple" proof of the claim that $$ C(a, 1) = \Big(1 + \frac{a}{2}\Big) \Big(\frac{1}{2} \frac{1}{1+a} + \frac{1}{2}\Big). $$

To show this claim, it suffices to show that the constant above can be placed in the following inequality for $C$, $$ \mathbb{E}[f(X)] \leq C~f(\mathbb{E} X), \quad \mbox{where}~f(t) = (1+t)^{-1},~\qquad\mbox{(*)} $$ where $X$ has any distribution supported on $[0, a]$. (This is because we know that $C(a, 1)$ is attained for $X$ taking values $\{0, a\}$, equiprobably.)

Then (*) follows from Grüss-inequality [see Cor. 4, 1], which says that for any random variable $X$ and measurable functions $\phi, \psi$, we have $$ \mathbb{E} \phi(X) \mathbb{E}[\psi(X)] \leq \mathbb{E}[\phi(X) \psi(X)] + \frac{(M_\phi - m_\phi) (M_\psi - m_\psi)}{4}. $$ Above, $m_\phi \leq \phi \leq M_\phi$, and $m_\psi \leq \psi \leq M_\psi$.

To see how (*) follows, take $\phi = f, \psi = 1/f$. Then we have $$ M_\phi = 1, \quad m_\phi = (a+1)^{-1}, \quad M_\psi = a + 1, \quad m_\psi = 1. $$ Therefore, we have $$ \mathbb{E}(f(X)) \leq \Big(1 + \frac{a^2/(a+1)}{4} \Big) f(\mathbb{E}(X)) = C(a, 1) ~f(\mathbb{E}(X)). $$ This proves the claim.

[1] Xin Li, R. N. Mohapatra, and R. S. Rodriguez. Gruss-Type Inequalities. Journal of Mathematical Analysis and Applications 267, 434–443 (2002).