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

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 $$a, n$$.

• 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 $$C(a, n) \leq 1 + a/n$$.

• As you argued, the constant is already asymptotically optimal (for $a=o(n)$), nice proof! The only way I see to identify the smallest constant is to identify distributions for which the inequality holds with equality (or sequences of distributions that enforce it, but I think it is attained). This is fairly demanding, but I think it might actually be possible. What you want to do is to completely explain the Jensen gap, over all possible distributions. Commented Oct 3, 2022 at 23:16
• Indeed, one way to interpret my question is to completely characterize the Jensen gap, which as you say is quite a strong requirement. For instance one could wonder if it is possible to replace $C$ by $1 + a/n^2$. This would also be asymptotically optimal in the sense that for fixed $a$, it is $1 + o(1)$ in the limit $n \to \infty$. Commented Oct 4, 2022 at 1:49
• Secondly, I am wondering if there is a simpler proof of my claim $1 + a/n$. One could try to do this by Taylor expanding $\mathbb{E}[1/(\overline{X}_n + 1)] = 1 /(\mu+ 1) + \mathbb{E}[\xi^{-3} (\overline{X}_n - \mu)^2]$. However, the difficulty is that $\xi$ is random in the sense that it lies between $\overline{X}_n$ (random) and $\mu$. Commented Oct 4, 2022 at 1:51
• The constant $C$ has to be valid for any distribution. It cannot be that you select $C$ based on the distribution. Commented Oct 7, 2022 at 16:26
• I have added a few clarifying comments to emphasize this point. Commented Oct 7, 2022 at 16:28

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

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).

• Nice! This is a cleaner solution. I haven't seen this trick (of defining and using the $Y_i$'s) before. Let me explain what I had in mind and what the difference between our two answers is. What I had in mind was a "smoothing" argument, where you show in a "smooth" fashion that $X$ being a Bernoulli random variable is best (you could also assume you have are working with a maximizer $X$ and do a proof by contradiction). As a toy example, say we wanted to minimize $x_1^2+\dots+x_n^2$ subject to $x_1+\dots+x_n = 1$ and $0 \le x_i \le 1$ for each $i$. We claim the minimum is $\frac{1}{n}$, Commented Oct 10, 2022 at 9:49
• obtained by having each $x_i = \frac{1}{n}$ (of course this can be proved immediately from Cauchy-Schwarz but never mind that for now). We can prove this by taking any $(x_1,\dots,x_n)$ satisfying $x_1+\dots+x_n = 1$ and $0 \le x_i \le 1$ for each $i$ and supposing that some $x_i \not = x_j$; say $x_1 > x_2$ for concreteness. We may then define $(x_1',\dots,x_n')$ by $x_1'=x_1-\epsilon, x_2' = x_2+\epsilon$, and $x_i' = x_i$ for $3 \le i \le n$, where $\epsilon \in (0,\frac{x_1-x_2}{2})$. Then $x_1'+\dots+x_n' = 1$, $x_i' \ge 0$ for each $i$, and a brief computation shows Commented Oct 10, 2022 at 9:53
• $(x_1')^2+\dots+(x_n')^2 < x_1^2+\dots+x_n^2$. Therefore, we can "smoothly" transform any $(x_1,\dots,x_n)$ into (the minimizer) $(\frac{1}{n},\dots,\frac{1}{n})$ (or we can assume the initial $(x_1,\dots,x_n)$ is a minimizer and get a contradiction from $(x_1',\dots,x_n')$). [As a remark, this kind of proof is used to establish the characterization of eigenvalues as solutions to the optimization problem $\sup_{||v||_2 = 1} v^T M v$.] By analogy, what your proof does is jump immediately from $(x_1,\dots,x_n)$ to $(\frac{1}{n},\dots,\frac{1}{n})$; i.e., you go from $X_i$ to $Y_i$ in one step. Commented Oct 10, 2022 at 10:00
• I think the denominator in your first inequality is $1/f$ instead of $f$, but that's not a problem for the proof since it's just a constant.
– p.s.
Commented Oct 13, 2022 at 0:11
• @p.s. fixed, thanks!! Commented Oct 15, 2022 at 5:20

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$$C(a,n) = \frac{n(2+a)}{2^{n+1}}\sum_{k=0}^n \frac{1}{ka+n}{n \choose k}.$$

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 $$\mathbb{E}\left[\frac{1}{1+X}\right]$$.

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).

• Interesting. I am not sure I completely follow your smoothing argument. However, if correct it is interesting to note that your $C(a,n)$ is able to be written as $\tfrac{n (2 + a)}{2} \cdot \mathbb{E}[\frac{1}{aX + n}]$, where $X \sim \mathsf{Bin}(n, 1/2)$. Using the heuristic $\mathbb{E}[\frac{1}{aX + n}] \approx \frac{2}{(a+2)n}$ would give $C(a,n) \approx 1$. In other words, if your claim is correct, it seems the order of approximation is much better than $a/n$ as my previous bound indicates. Commented Oct 8, 2022 at 15:04
• @DrewBrady Do you understand it for $n=1$? If not, I can try to add some details. Commented Oct 8, 2022 at 15:20
• Does your argument work by considering a random variable $Y = a$ when $X \geq x_0$ and $Y = 0$ otherwise, for some $x_0$? Commented Oct 8, 2022 at 21:07
• $\lambda \in [0,a]$ should be $\lambda \in [0,1]$, no? Commented Oct 9, 2022 at 20:25
• @leonbloy yea, thx Commented Oct 9, 2022 at 21:16

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).