1

The problem is as stated in the title. I am looking for an example or a disproof, whether there exists a continuous density function on the whole real line with infinite expected value.

Once again:

The density function $f$ must be

$1.$ Positive on $\mathbb{R}$

$2.$ Continuous on $\mathbb{R}$

$3.$ $E[X]=\infty$ with $X\sim f$

Seyhmus Güngören
  • 7,848
  • 1
    Would something like a Cauchy density on $\mathbb R^+$ and a zero-mean normal density on $\mathbb R^-$ work? The variance of the normal density can be adjusted to make both functions have the same value at $0$ and both densities have slope $0$ at $0$. – Dilip Sarwate Jan 26 '15 at 12:43
  • @DilipSarwate I don't mind. It can be constructed as how we want. – Seyhmus Güngören Jan 26 '15 at 12:52

2 Answers2

2

Expanding on my comment, consider the density $$f(x) = \begin{cases}\displaystyle\frac{1}{\pi(1+x^2)}, & x > 0,\\ \displaystyle \frac{1}{\pi}\exp\left(-\frac{x^2}{\pi}\right),&x < 0,\\ \displaystyle\frac{1}{\pi}, & x = 0\end{cases}$$ which can be recognized as a standard Cauchy density on $\mathbb R^+$ and a zero-mean normal density with variance $\frac{\pi}{2}$ on $\mathbb R^-$. Note that $f(x)$ is positive and continuous on $\mathbb R$ as desired, and more strongly, it also differentiable on $\mathbb R$ (at $x=0$, both the left derivative and the right derivative have value $0$). If $f(x)$ is not required to be differentiable everywhere (that is, continuity is sufficient), we could, for example, set $f(x)$ to $\frac 1\pi\exp(2x/\pi)$ for $x<0$ instead of the normal density.

Thus, $\displaystyle\int_0^\infty xf(x)\,\mathrm dx$ is unbounded while $\displaystyle \int_{-\infty}^0 xf(x)\,\mathrm dx =- \frac 12$, and so $\displaystyle\int_\infty^\infty xf(x)\,\mathrm dx = E[X] = \infty$.

Dilip Sarwate
  • 25,197
  • I think it is not differentiable at $0$. I started to wonder if the condition "continuous" would be differentiable? – Seyhmus Güngören Jan 26 '15 at 13:37
  • @SeyhmusGüngören: Both functions are symmetric about $x=0$, so they both have zero derivative there. – TonyK Jan 26 '15 at 13:54
  • @TonyK I just used mathematica and got the result "indeterminate" for point $0$. Used simply the piecewise function. – Seyhmus Güngören Jan 26 '15 at 13:56
  • @SeyhmusGüngören: Well, Mathematica is just confused then. The function is clearly differentiable at $x=0$. – TonyK Jan 26 '15 at 14:11
  • @SeyhmusGüngören Thanks for fixing the typo in the Cauchy density. The $f(x)$ I specified is continuous at $0$ and is also differentiable at $0$ since, as TonyK points out, both the Cauchy and the normal density have derivative $0$ at $0$. (I had said so in my comment but hadn't included it above; I will fix my answer when I have some time). – Dilip Sarwate Jan 26 '15 at 14:15
  • @DilipSarwate that is okay. everything is clear. Thanks. – Seyhmus Güngören Jan 26 '15 at 14:16
  • @Dilip: In your last sentence, I think both instances of $\int f(x)dx$ should be $\int xf(x)dx$. – TonyK Jan 26 '15 at 19:26
  • @TonyK Thanks, I am fixing the whole answer. – Dilip Sarwate Jan 26 '15 at 19:35
1

Start with the function $f(x) = \dfrac{1}{1+x^2}$. This is not quite right, because

  • it's symmetric, so we don't get $E[X]=+\infty$;
  • $\int_{\mathbb R}f \ne 1$.

But $\int_0^\infty xf(x)\mathrm{dx} = +\infty$, so you can modify it to meet your requirements:

  1. for negative arguments, change it to a function that tends to $0$ fast enough as $x \to -\infty$, e.g. $f(x)=e^x$ for $x < 0$;
  2. multiply it by a suitable constant so that $\int_{\mathbb R}f = 1$.
TonyK
  • 64,559