0
$\begingroup$

Take $X_1 , X_2 , \cdots $ are $iid$ with zero mean. Take $$Z_n = \frac{X_1 + \cdots X_n}{\sqrt{n}} \stackrel{d}{\rightarrow} X$$ and $$Z_{2n} = \frac{X_1 + \cdots X_{2n}}{\sqrt{2n}} \stackrel{d}{\rightarrow} X$$ Name the characteristic function of $X$ to be $f(\xi)$

I managed to show the following: $$f(\xi) = f(\frac{\xi}{\sqrt{2}})^2$$ and that $f(\xi)$ is the characteristic function of a Gaussian when $f \in C^2(\mathbb{R})$

Now I want to show that, if we replace $\frac{1}{\sqrt{n}}$ with $\frac{1}{n}$, then $Z_n$ is distributed Cauchy-Lorentz when $f(\xi) = f(-\xi)$ or $f(\xi) = 1$

I started as so: $$f(\xi) = \mathbb{E}e^{i \xi X} = \mathbb{E}e^{i \xi \frac{X_1 + \cdots X_n}{n}} = \mathbb{E}\prod_{i=1}^{n}e^{i\xi \frac{X_i}{n}}=(\mathbb{E}e^{i\xi \frac{X_1}{n}})^n$$ Then by Taylor expantion $$(\mathbb{E}e^{i\xi \frac{X_1}{n}})^n = \mathbb{E}[1 + \frac{i \xi}{n}X + O(n^{-2})]^n \stackrel{n \rightarrow \infty}{\rightarrow} e^{i\xi (0)} = 1$$

This is however far from a cauchy Lorentz distribution... Where did I go wrong?

Thank you for your insight!

EDIT: I think the reason we cannot use the Strong Law of Large Numbers:

Suppose $\mathbb{E}|X| < \infty $ then $\bar{X_n}$ converges almost surely to $\mathbb{E}X$ In this case, we do not know that $E|X_i| < \infty$ only that $EX_i = 0$

$\endgroup$
  • $\begingroup$ I may be misunderstanding something. Why do you expect a Cauchy distribution to begin with? If you normalize by $n$, the Law of Large Numbers should kick in and give you a limit of zero. $\endgroup$ – Clement C. Oct 14 '18 at 14:39
  • $\begingroup$ @ClementC. I think the idea comes in when we add constraints on the Characteristic functions. Take the Gaussian case. We give that $f \in C^2(\mathbb{R})$. Which implies that the second derivative of the characteristic function exists and is continuous, hence the second moment, allowing us to get the CLT. In the second case, the constraint on $f$ is that $f(\xi) = f(- \xi)$ or $f(\xi) = 1$ $\endgroup$ – rannoudanames Oct 14 '18 at 14:42
  • $\begingroup$ But adding a constraint on the MGF does not change the fact that the rest of the constraints (iid, zero mean) imply convergence to zero almost surely. $\endgroup$ – Clement C. Oct 14 '18 at 14:43
  • $\begingroup$ @ClementC. That was my thought process also... unless this is a trick problem, I doubt it however. I see the contraint on the characteristic function to be key here. similarily to the CLT case $\endgroup$ – rannoudanames Oct 14 '18 at 14:47
0
$\begingroup$

If the iid $X_i$ have zero mean, the sequence $Z_n=(X_1+\dots+X_n)/n$ converges to 0, almost surely. Hence your random variable $X$ is equal to 0 with probability 1, and has the constant function $\phi_X(\xi)=1$ as characteristic function. As is well known, non degenerate Cauchy rvs do not have expected values, so there is something fishy with your problem statement.

Another symptom of hinkiness is your Taylor expansion. If all you know about the rv is that it has a first moment, you are not entitled to either equality in your last displayed equation. You can expand $\exp(i\xi X_1/n)$ in a power series in $\xi$ but you have no warrant for interchanging this summation with $\mathbb E$.

$\endgroup$
  • $\begingroup$ I think the reason we cannot use the Strong Law of Large Numbers: Suppose $\mathbb{E}|X| < \infty $ then $\bar{X_n}$ converges almost surely to $\mathbb{E}X$ In this case, we do not know that $E|X_i| < \infty$ only that $EX_i = 0$ $\endgroup$ – rannoudanames Oct 14 '18 at 15:16
  • $\begingroup$ Just what does your hypothesis that the $x_i$ have "zero mean" mean? $\endgroup$ – kimchi lover Oct 14 '18 at 15:17
  • $\begingroup$ $E[X_i] = 0$ , i wonder if we can conclude that $E|X| < \infty$ $\endgroup$ – rannoudanames Oct 14 '18 at 15:20
  • 1
    $\begingroup$ No. Why-ever would you think so? Look at para 2.5 of the wikipedia article en.wikipedia.org/wiki/Expected_value . Do you have a textbook on measure theory? $\endgroup$ – kimchi lover Oct 14 '18 at 15:31
  • 1
    $\begingroup$ @rannoudanames Having an expectation by definition means $\mathbb{E}[|X|]<\infty$. That does not mean that $X$ is square integrable, though -- there are many counter examples. $\endgroup$ – Clement C. Oct 14 '18 at 16:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.