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I am studying the (asymptotic) behavior of the p.d.f of the random variable $Y = X + Z$, where $X$ is an r.v. with any distribution function $F(x)$ such that $\int_{-\infty}^{\infty} x^2 \mathrm{d}F(x) \leq P $, and $Z$ is independent of $X$, with a Gaussian density

$p_Z(z) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{z^2}{2\sigma^2}}$.

So the density of $Y$ is the convolution $p_Y(y) = \int_{-\infty}^{\infty}\ p_Z(y-x)\ \mathrm{d}F(x)$.

Applying Jensen's inequality, we know the tail of $p_Y$ decays no faster than a (scaled) Gaussian function with some mean and variance $\sigma^2$. Also by bounding $y^2p_Z(y-x)$, we see that $p_Y$ decays no slower than $cy^{-2}$. Of course the tail of $p_Y$ depends very much on $F(x)$. I am wondering in which area people study such relations, and what literature I should look for. I know in image processing, people use "Gaussian kernel smoothing/filtering" methods, but they are more interested in the algorithms etc.



P.S. (in response to Yuri) Currently I am able to analyze the tail of $p_Y$ in some specific cases. For instance, if both ends of $F(x)$ decay faster than any Gaussian, then the tail of $p_Y$ is roughly the same Gaussian as $p_Z$. On the other hand, if $X$ has a density that decays slower than any Gaussian and is not changing too fast (e.g. reciprocals of polynomials, or exponential), then $p_Y$ has the same tail as the density of $X$. However, these regular examples are not general enough to represent all distributions $F(x)$. For example, if the tail of $X$ is some times Gaussian, sometimes polynomial/exponential, then what can we say about the tail of $p_Y$?

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Fourier transformation gives $\hat{p_Y}(\xi) = \hat{p_Z}(\xi) \hat{dF}(\xi)$, where $\hat{p_Z}$ is a Gaussian, so in the frequency domain you have exact control of the influence of $\hat{dF}$. Perhaps you can then use Fourier analysis to draw conclusions for the asymptotics of $p_Y$. – Patrik Wahlberg Nov 20 '11 at 11:24

1 Answer 1

If $X$ has polynomial tails, then so does $Y$.

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Thanks. Here the length of response is too limited, please see the P.S. part of my question. – Wei Mao Nov 20 '11 at 18:22

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