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

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?

Attempt:

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

The distribution is absolutely continuous.

• Without some kind of reasoning or working, your 'attempt' looks like a guess. Commented Jun 28, 2014 at 4:48

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?

• 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. Commented Jun 28, 2014 at 7:54
• If $\lim_{x\to\infty}g(x)=0$ and $\lim_{x\to\infty}h(x)=2$, what can you say about $\lim_{x\to\infty}g(x)h(x)$?
– Zen
Commented Jun 28, 2014 at 12:35
• 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 $a$.
– Zen
Commented Jun 28, 2014 at 12:40
• 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. Commented Jun 29, 2014 at 2:06
• 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. Commented Jun 29, 2014 at 2:47