Let $X$ denote a real-valued random variable with distribution function $F$ and characteristic function $\phi$. Suppose that $\phi$ satisfies the following condition:

$$\lim_{T\to\infty}\int_{-T}^{T}\phi(t) ~dt = 2.$$

What can be said about the distribution?


The distribution is symmetric across the $x = 0$ axis (?)

The distribution is absolutely continuous.

  • $\begingroup$ Without some kind of reasoning or working, your 'attempt' looks like a guess. $\endgroup$
    – Glen_b
    Jun 28, 2014 at 4:48

1 Answer 1


Remember that $$ F_X(a)-\lim_{x\uparrow a}F_X(x) = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T e^{-ita}\varphi_X(t)\,dt \, . \qquad (*) $$

If you make $a=0$ in this formula, what happens? The product rule for limits will be useful. Can you use this to say something about the distribution function at the origin?

  • 1
    $\begingroup$ What I can see is that if a = 0, then the $e^{-ita}$ becomes 1. As T approaches infinity, the $\frac{1}{2T}$ term becomes zero. I must be missing something. $\endgroup$
    – statsguyz
    Jun 28, 2014 at 7:54
  • $\begingroup$ If $\lim_{x\to\infty}g(x)=0$ and $\lim_{x\to\infty}h(x)=2$, what can you say about $\lim_{x\to\infty}g(x)h(x)$? $\endgroup$
    – Zen
    Jun 28, 2014 at 12:35
  • $\begingroup$ Also, since $F_X$ is right continuous (check this: stats.stackexchange.com/questions/25238/…), the LHS of $(*)$ is the size of the jump of $F_X$ at $a$. $\endgroup$
    – Zen
    Jun 28, 2014 at 12:40
  • 1
    $\begingroup$ Suppose that $$\lim_{x\to c} f(x) =L$$ and $$\lim_{x\to c} g(x) =M$$. Then $$ \lim_{x\to c} [f(x) g(x)] = \lim_{x\to c} f(x) \lim_{x\to c} g(x) = L M $$ This is what I have for the product rule of limits. So then the equation should equal zero if we simply have L = 2 and M = 0. $\endgroup$
    – statsguyz
    Jun 29, 2014 at 2:06
  • 1
    $\begingroup$ Ah ok I think I understand. So basically we know that this function (like all distribution functions) is right continuous, but we are able to ascertain that $Pr(X=0)=0$ from the given information. $\endgroup$
    – statsguyz
    Jun 29, 2014 at 2:47

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