# Upper Bound on a en exponential function

I am trying to upper bound the following function and find its growth rate:

$$\begin{equation} \psi(y) \stackrel{\triangle}{=} \int_{0}^{\infty}\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right)f(x)\,dx,~y>0, \end{equation}$$ where $$f(x)>0$$ satisfies $$\int_{0}^{+\infty}f(x)\,dx = 1$$ (it is a pdf of the random variable $$X$$). I would like to find the growth rate of $$\psi(y)$$ for large values of $$y$$, i.e., would like to know if $$\psi(y)$$ grows like $$O(y^{-k})$$, for some $$k>1$$.

Here are my attempts so far: It is straightforward to show that the function $$\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right)\in[0,1]$$ is increasing in $$x$$ for $$y>x>0$$ and is decreasing in $$x$$ for $$0.

Then one can break the bounds of the integral to $$[0,y/2], \, [y/2,3y/2], \, [3y/2, y^{3/2}],\,[y^{3/2},\infty)$$ and then evaluate each integral. Doing so will give the following:

\begin{align} \int_{0}^{y/2}\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right)f(x)\,dx &= O(y^{-k}), \,k>1\\ \int_{y/2}^{3y/2}\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right)f(x)\,dx &\stackrel{?}{=}O(y^{-k}),\, k>1\\ \int_{3y/2}^{y^{3/2}}\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right)f(x)\,dx &= O(y^{-k}), \,k>1\\ \int_{y^{3/2}}^{\infty}\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right)f(x)\,dx &= O(y^{-k}), \,k>1 \end{align}

Therefore, if we can prove that the second integral also grows like $$O(y^{-k}),\,k>1$$, then we are done. However, I have not been able to show this and am stuck. Any help would be highly appreciated!

• Looks like some assumption on $f(x)$ is also needed. With these assumptions, one can only say that the second integral converges to $0$, but no convergence rate can be further proven. Mar 20, 2020 at 0:35
• @Stefano The only assumption of $f(x)$ is that it is the pdf of the random variable $X$. Thus it is nonnegative and its integral over all possible values of $x$ is equal to 1. How did you prove that the second integral converges to 0? Mar 20, 2020 at 1:32

The statement is false in general. For a counterexample consider a pdf $$f$$ defined as follows

$$f(x) = \sum_{n=1}^{\infty} \frac{1}{2^{n+1}} I_{(y_{n}-1;y_{n}+1)}(x)$$

where $$(y_{n})_{n \in \mathbb{N}}$$ is an increasing sequence that will be determined later and where $$I_{A}(x)$$ is the indicator function of the set $$A \subset \mathbb{R}$$. To fix ideas, let's further assume $$y_{n+1}-y_{n}> 2$$ so that the different addenda in the sum have disjoint support.

Then we can provide the following estimate from below to the second integral in your statement \begin{align} \mathfrak{I}_2 (y) &\dot= \int_{y/2}^{3y/2} \exp \left( -\frac{(y-x)^2}{2(\sigma_0^2+\sigma_1^2x)} \right) f(x) \, \mathrm{d}x\\ &\geq \int_{y-1}^{y+1} \exp \left( -\frac{(y-x)^2}{2(\sigma_0^2+\sigma_1^2x)} \right) f(x) \, \mathrm{d}x\\ &\geq K \int_{y-1}^{y+1} f(x) \, \mathrm{d}x \end{align}

with $$K = \exp\left(-\frac{1}{4\sigma_0^2}\right)$$ for $$y$$ sufficiently large. Therefore

$$\mathfrak{I}_{2}(y_{n}) \geq \frac{K}{2^n}.$$

Since the right hand side is independent of $$y$$, a suitable choice of $$(y_{n})_{n \in \mathbb{N}}$$ proves $$\mathfrak{I}_{2}(y) \notin O(y^{-k})$$ for any $$k$$.

It is however possible to prove that $$\mathfrak{I}_{2}(y) \to 0$$ for $$y \to \infty$$. Choose a sequence $$(b_{n})_{n \in \mathbb{N}}$$ such that

$$\int_{0}^{b_n} f(x) \, \mathrm{d}x \geq 1-\frac{1}{n}.$$ Then $$\forall y \geq 2b_{n}$$ one has

\begin{align} \mathfrak{I}_2 (y) &= \int_{y/2}^{3y/2} \exp \left( -\frac{(y-x)^2}{2(\sigma_0^2+\sigma_1^2x)} \right) f(x) \, \mathrm{d}x\\ &\leq \int_{b_n}^{\infty} \exp \left( -\frac{(y-x)^2}{2(\sigma_0^2+\sigma_1^2x)} \right) f(x) \, \mathrm{d}x\\ &\leq \int_{b_n}^{\infty} f(x) \, \mathrm{d}x \leq \frac{1}{n} \end{align} from which the result easily follwos.

First, let us see an example in which $$\lim_{y\to \infty} y\psi(y) = \infty$$.

Let (log-Cauchy distribution) $$f(x) = \frac{1}{\pi x (1 + (\ln x)^2)}, \ x > 0.$$ We have, for sufficiently large $$y$$, \begin{align} \psi(y) &= \int_{0}^{\infty}\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right) \frac{1}{\pi x (1 + (\ln x)^2)}\,\mathrm{d} x\\ &\ge \int_{y - \sqrt{y}}^{y + \sqrt{y}}\exp\left(-\frac{(y-x)^2}{2(\sigma_0^2 + \sigma_1^2 x)}\right) \frac{1}{\pi x (1 + (\ln x)^2)}\,\mathrm{d} x \\ &\ge \int_{y - \sqrt{y}}^{y + \sqrt{y}}\exp\left(-\frac{\sqrt{y}}{2(\sigma_0^2 + \sigma_1^2 y/2)}\right) \frac{1}{\pi x (1 + (\ln x)^2)}\,\mathrm{d} x \\ &\ge \int_{y - \sqrt{y}}^{y + \sqrt{y}} \frac{1}{2}\cdot \frac{1}{\pi x (1 + (\ln x)^2)}\,\mathrm{d} x \\ &= \frac{\arctan(\ln(y + \sqrt{y})) - \arctan(\ln(y - \sqrt{y}))}{2\pi}\\ &= \frac{1}{2\pi} \arctan \frac{\ln(y + \sqrt{y}) - \ln(y - \sqrt{y})}{1 + \ln(y + \sqrt{y})\ln(y - \sqrt{y})} \end{align} where we have used $$\tan (x-y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}$$. Thus, we have \begin{align} \lim_{y\to \infty} y\psi(y) &= \lim_{y\to \infty} \frac{1}{2\pi} y \arctan \frac{\ln(y + \sqrt{y}) - \ln(y - \sqrt{y})}{1 + \ln(y + \sqrt{y})\ln(y - \sqrt{y})}\\ &= \lim_{y\to \infty} \frac{1}{2\pi} y \frac{\ln(y + \sqrt{y}) - \ln(y - \sqrt{y})}{1 + \ln(y + \sqrt{y})\ln(y - \sqrt{y})}\\ &= \infty \end{align} where we have used $$\lim_{z \to 0}\frac{\arctan z}{z} = 1$$.

Second, if $$\mathbb{E}[X]$$ and $$\mathbb{E}[X^2]$$ are both finite, one can prove that $$\psi(y) = O(y^{-2})$$.