Consider the problem of finding a nice upper bound to the following.
$$
I:=\int_{\mathbb{R}^N} \pi^{-N/2} \frac{\prod_{i=1}^N \exp(-(x_i-a)^2)}{\sum_{i=1}^N \exp(-x_i^2)+1}dx_{1:N}
$$
I would like an upper bound that converges to $0$ as $N\rightarrow\infty$ (as I think $I$ does).
A tentative is as follows. Using some decompositions kindly suggested by Pavel Gubkin and Sangchul Lee here, we can write the above as $J+K$, where
$$
J = \int_{\mathbb{R}^N} \pi^{-N/2} \prod_{i=1}^N \exp(-(x_i-a)^2)dx_{1:N} = 1
$$
$$
K = -\int_{\mathbb{R}^N} \pi^{-N/2}\prod_{i=1}^N \exp(-(x_i-a)^2) \frac{\sum_{i=1}^N \exp(-x_i^2)}{\sum_{i=1}^N \exp(-x_i^2)+1}dx_{1:N}
$$
We could bound $K$ with
$$
K \leq -\int_{\mathbb{R}^N}\pi^{-N/2} \prod_{i=1}^N \exp(-(x_i-a)^2) \frac{\sum_{i=1}^N \exp(-x_i^2)}{N+1}dx_{1:N} = \frac{N\exp(-a^2/2)}{\sqrt{2}(N+1)}
$$
Now this bound converges to $-\frac{\exp(-a^2/2)}{\sqrt{2}}$, whereas I think our $K$ actually converges to $-1$ for any $a$, and (thus cancels asymptically with $J$). Is there a way to obtain a nicer upper bound on $K$ that actually converges to $-1$?
Any suggestion is appreciated.
Best Answer
We use a probabilistic approach for this problem.
First, we transform the integral $I$ by denoting $x_i = \frac{y_i}{\sqrt{2}}$ for $i = 1,...,n$ $$\begin{align} I &= \int_{\mathbb{R}^n} \pi^{-n/2} \frac{\prod_{i=1}^n \exp(-(x_i-a)^2)}{\sum_{i=1}^n \exp(-x_i^2)+1}dx_{1:n} \\ &= \int_{\mathbb{R}^n} (2\pi)^{-n/2} \frac{\prod_{i=1}^n \exp(-\frac{(y_i-\sqrt{2}a)^2}{2})}{\sum_{i=1}^n \exp(-\frac{1}{2}y_i^2)+1}dy_{1:n} \\ & = \color{red}{ \mathbb{E} \left(\frac{1}{\sum_{i =1}^n e^{-\frac{1}{2}Y_i^2} +1} \right)} \tag{1} \end{align}$$
where $Y_i$ are i.i.d and follow the normal distribution $\mathcal{N}(\sqrt{2}a, 1)$ for all $i$
We notice that the function $f(x): = e^{-\frac{1}{2}x}$ is convex (as $f''(x) = \frac{1}{4}e^{-x}>0$), applying the Jensen's inequality for $Y_1^2,...,Y_N^2$ we have $$\frac{1}{n}\sum_{i =1}^n e^{-\frac{1}{2}Y_i^2} \ge \exp \left(-\frac{1}{2}\cdot\frac{1}{n}\sum_{i =1}^n Y_i^2 \right) \tag{2}$$
From $(1),(2)$, we deduce $$I \le \mathbb{E} \left(\frac{1}{n\exp \left(-\frac{1}{2n}\sum_{i =1}^n Y_i^2 \right) +1} \right) \tag{3} $$
Let us denote $S_n := \frac{1}{2}\sum_{i =1}^nY_i^2$, then $$(3) \implies \color{red}{I \le \mathbb{E} \left(\frac{1}{ne^{-\frac{1}{n}S_n } +1} \right) =\underbrace{\mathbb{E} \left(\frac{ \mathbf{1}_{\{S_n \le z_n \}}}{ne^{-\frac{1}{n}S_n } +1} \right)}_{\text{Term }1} + \underbrace{\mathbb{E} \left(\frac{\mathbf{1}_{\{S_n \ge z_n \}}}{ne^{-\frac{1}{n}S_n } +1} \right)}_{\text{Term }2} } \tag{4}$$
with $z_n \in \mathbb{R}$ that will be determined later.
We will use the Chernoff bound for $S_n$. The moment-generating function $M(t)$ of $S_n$ is $$M(t) = \mathbb{E} (e^{tS_n}) = \mathbb{E}^n (e^{\frac{t}{2}Y_1^2}) = \left(\frac{e^{a^2 \frac{t}{1-t}}}{\sqrt{1-t}} \right)^n$$
and then for all $t > 0$ and for all $z_n >0$ $$\begin{align} \mathbb{P}(S_n \ge z_n) \le M(t)\cdot e^{-tz_n} &= \frac{e^{n\cdot a^2 \frac{t}{1-t}-tz_n}}{(1-t)^{n/2}} \\ &= \exp\left( n\cdot a^2 \frac{t}{1-t}-tz_n - n \cdot \frac{\ln(1-t)}{2} \right) \\ &= \exp\left( n\cdot \left( a^2 \frac{t}{1-t} - \frac{\ln(1-t)}{2} \right)-tz_n \right) \tag{5} \end{align}$$
In particular, $(5)$ holds true for $t = \frac{1}{2}$: for all $z_n > 0$ $$\color{red}{\mathbb{P}(S_n \ge z_n) \le \exp\left( n\cdot \left( a^2 + \frac{\ln(2)}{2} \right)-\frac{1}{2} z_n \right) } \tag{6}$$
Return now to the $(4)$, we have
$$ \begin{align} &\text{Term }1 \le \mathbb{E} \left(\frac{ \mathbf{1}_{\{S_n \le z_n \}}}{ne^{-\frac{1}{n}z_n } +1} \right) < \frac{ 1}{ne^{-\frac{1}{n}z_n } +1} \\ &\text{Term }2 < \mathbb{E} \left( \mathbf{1}_{\{S_n \ge z_n \}}\right) = \mathbb{P} \left( S_n \ge z_n \right) \tag{7} \end{align}$$
We choose $\color{red}{z_n = 2 \left(a^2 +\frac{\ln(2)}{2} \right)n + n} $ in the inequality $(6)$, from $(7)$ we have
$$ \begin{align} &\text{Term }1 < \frac{ 1}{ne^{-2 \left(a^2 +\frac{\ln(2)}{2} \right)1 - 1 } +1} \xrightarrow{n \to +\infty} 0\\ &\text{Term }2 < \mathbb{P} \left( S_n \ge z_n \right) \le \exp\left( -\frac{n}{2} \right) \xrightarrow{n \to +\infty} 0 \tag{8} \end{align}$$
From $(4)$ and $(8)$ we obtain the upper bound that converges to $0$ when $n \to +\infty$:
$$\color{red}{I \le \frac{ 1}{ne^{-2 \left(a^2 +\frac{\ln(2)}{2} \right)1 - 1 } +1} +\exp\left( -\frac{n}{2} \right) } $$
Q.E.D