= Gaussian smoothing of an integer-valued distribution
{c}
{title2=$g_\sigma(z)=\sum_{n\in\mathbb Z}p_n\varphi_\sigma(z-n)$}
Let $X$ take integer values with probabilities $p_n$, and let $Y_\sigma$ be an independent centered <normal distribution> with positive <standard deviation> $\sigma$. Conditioning on $X$ gives the <probability density function> of their sum as $g_\sigma(z)=\sum_np_n\varphi_\sigma(z-n)$, where $\varphi_\sigma$ is the normal density. Nonnegativity and <Tonelli's theorem> show that this density integrates to one. For small $\sigma$, isolated atoms become narrow smooth peaks with their original probability masses.
For each fixed real $t$, <dominated convergence> gives
$$
\sigma g_\sigma(\sigma t)=\mathbb E\varphi_1(t-X/\sigma)\longrightarrow\varphi_1(t)\qquad(\sigma\to\infty),
$$
since $X/\sigma\to0$ almost surely and $\varphi_1$ is bounded. Thus every fixed integer-valued distribution acquires a Gaussian central profile on this expanding scale, without any moment assumption on $X$. This is not a claim about uniform relative error in the tails or about finite moments. For example, if $p_n=3/(\pi^2n^2)$ for $n\ne0$ and $p_0=0$, the <expectation> of $|X|$ is infinite. Since $\mathbb E|Y_\sigma|<\infty$, the triangle inequality forces $\mathbb E|X+Y_\sigma|=\infty$ too. Symmetry then gives equal infinite positive and negative first moments, so there is no ordinary mean. The second moment is also infinite. A Gaussian-looking center can therefore coexist with a <heavy-tailed distribution>.
Back to article page
