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 $0<y<x$.
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!
 A: 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.
A: 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 $\psi(y) = O(y^{-2})$. 
