Assume a pdf $f(x)$ is continuous along $-\infty$ to $+\infty$. Does this assumption guarantee that $f(+\infty)=f(-\infty)=0$? How to prove? Thanks in advance.
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2No. Think of the addition of spikes with vanishing bases centered at points going to infinity. – Did Jun 25 '14 at 16:00
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I think it is important to clarify that by $f(\pm \infty)$ we are referring to $$\lim_{x \to \pm \infty} f(x).$$ In such a case, the answer is no even for a continuous density with support on $\mathbb R$. – heropup Jun 25 '14 at 16:22
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Here is a setting worth keeping in mind when looking for examples. Consider some sequences $(\mu_n)$, $(\sigma^2_n)$ and $(p_n)$ with $\sigma^2_n\gt0$, $p_n\gt0$, and $\sum\limits_np_n=1$, and the function $$ f=\sum_np_n\,g_{\mu_n,\sigma^2_n}, $$ where, for every $(\mu,\sigma^2)$, $g_{\mu,\sigma^2}$ is the gaussian density with mean $\mu$ and variance $\sigma^2$.
Then $f$ is a PDF since $\sum\limits_np_n=1$. However, if $\mu_n\to\infty$ and $\sigma_n\ll p_n$ when $n\to\infty$ (say, $\mu_n=n$, $p_n=1/2^n$ and $\sigma_n^2=1/8^n$), then $$ f(\mu_n)\geqslant p_ng_{\mu_n,\sigma^2_n}(\mu_n)=\frac{p_n}{\sqrt{2\pi\sigma_n^2}}, $$ hence the density $f$, far from converging to zero at infinity, is actually unbounded at infinity.
Did
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1@DanielLittlewood I believe so, since, on every bounded interval, $f$ is the sum of a regular function (the sum of the $n$ first terms of the series) and a uniformly small one (the rest). Crucial estimate: the series $\sum\limits_np_n\sigma_n^{-1}\mathrm e^{-\mu_n^2/2\sigma_n^2}$ is summable. – Did Jun 25 '14 at 16:23
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@Did: Thank you for this useful answer. I think that if the limit of f at infinity exists, it must be zero. In the example you show the limit at the boundary doesn't exist because of the spikes you are mentioning, otherwise how come it is infinite, and f is still continuous and integrable? – Avocaddo Jun 22 '19 at 02:22
