# the limit of a probability density function

I have one maths challenge which I could not resolve. The question is the following. Let f(x) denotes the probability density of a continuous RV., X. I want to know the limit f(x) as x approaches to either -∞ or +∞. It might seem silly, but I have do idea how to do that.

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You don't you know how to prove it or don't even have idea of the answer? – leonbloy Jul 6 '12 at 13:13
@leonbloy I have no idea of the answer. I am less technical guy. There maybe a proof for a specific type of density, but I have no idea how it can be done for a generalized PDF. – Dere Jul 6 '12 at 13:16
It may have no limit, for example $f(x)=\sum_{j\geq 1} 2^{-j}\chi_{(2n,2n+1)}$. – Davide Giraudo Jul 6 '12 at 13:18
The limit depends on the $f(x)$. Write down it, please. – Riccardo.Alestra Jul 6 '12 at 13:19
@Riccardo.Alestra; you may suggest me a few examples of some known f(x). – Dere Jul 6 '12 at 13:22

## 2 Answers

It is not necessary that a density have any limit as $x\to\infty$ or $x\to -\infty$. The sole criteria for a function $f$ to be a density are $f\ge 0$ a.e. and that $$\int_{-\infty}^\infty f(x)\mbox{d}x = 1.$$

Such a function could have high, narrow spikes at either or both ends. It is, in fact, possible to have $$\limsup_{n\to\infty} f(x) = \infty$$ and $$\limsup_{n\to-\infty} f(x) = \infty.$$ Try to construct such an example. It is not terribly hard.

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what about the continuity of $f$? – Seyhmus Güngören Apr 6 '14 at 12:55

Assuming that this question arose in an undergraduate (non-measure-theoretic) course in probability, the answer that the OP (or perhaps the OP's instructor) might have been looking for might be that $$\lim_{x\to\infty} f(x) = \lim_{x\to -\infty} f(x) = 0.$$ The rationale would be that if $f(x)$ were converging to a positive number $c$ as $x \to \infty$, then $\int_{-\infty}^{\infty}f(x)\,\mathrm dx$ would diverge. In more detail for the OP's benefit, there would exist an $x_0$ such that for all $x > x_0$, $|f(x) - c| < \frac{c}{2} \Rightarrow f(x) > \frac{c}{2}$, and so $$\int_{-\infty}^{\infty}f(x)\,\mathrm dx \geq \int_{x_0}^{\infty}f(x)\,\mathrm dx \geq \int_{x_0}^{\infty}\frac{c}{2}\,\mathrm dx.$$ Indeed, not only must $f(x)$ converge to $0$ as $x \to \pm\infty$ but the convergence must be faster than $|x|^{-1}$.

Of course, since $f(x)$ can be viewed as a representative member of a class of density functions that differ only on a set of measure zero, one can find other functions in this class for which $\limsup f(x) = \infty$, e.g. $$fx) = \begin{cases} \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right), &x ~\text{is irrational},\\ |x|, &x ~\text{is rational.} \end{cases}$$

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