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 $0<y<x$.

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!

  • $\begingroup$ 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. $\endgroup$
    – Stefano
    Mar 20, 2020 at 0:35
  • $\begingroup$ @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? $\endgroup$ Mar 20, 2020 at 1:32

2 Answers 2


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[4]{y}}^{y + \sqrt[4]{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[4]{y}}^{y + \sqrt[4]{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[4]{y}}^{y + \sqrt[4]{y}} \frac{1}{2}\cdot \frac{1}{\pi x (1 + (\ln x)^2)}\,\mathrm{d} x \\ &= \frac{\arctan(\ln(y + \sqrt[4]{y})) - \arctan(\ln(y - \sqrt[4]{y}))}{2\pi}\\ &= \frac{1}{2\pi} \arctan \frac{\ln(y + \sqrt[4]{y}) - \ln(y - \sqrt[4]{y})}{1 + \ln(y + \sqrt[4]{y})\ln(y - \sqrt[4]{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[4]{y}) - \ln(y - \sqrt[4]{y})}{1 + \ln(y + \sqrt[4]{y})\ln(y - \sqrt[4]{y})}\\ &= \lim_{y\to \infty} \frac{1}{2\pi} y \frac{\ln(y + \sqrt[4]{y}) - \ln(y - \sqrt[4]{y})}{1 + \ln(y + \sqrt[4]{y})\ln(y - \sqrt[4]{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 $\psi(y) = O(y^{-2})$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .