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Let $X_1,X_2,\dots$ be independent, symmetric random variables with characteristic functions $\varphi_{1},\varphi_{2},\dots$

Prove:

If $\varphi_{1},\varphi_{2},\dots$ is differentiable at zero, then $\frac{X_1+\dots +X_n}{n}$ converges in probability to zero.


My attempt:

Use the Weak Law of Large Numbers for not identical distributed variables. For this I need

  • the mean of $X_i$ is zero,
  • independence of the $X_i$,
  • finite variance.

The mean is zero, because the variables are symmetric. The independence also is given. I conclude the finite variance should be shown by the differentiability of $\varphi_n$. For the derivatives it holds that $$ \varphi_n^{(k)}(t)=i^k\mathbb E\left(X_n^ke^{itX_n}\right) $$ so I know that $$ \varphi_n^{(1)}(0)=i\mathbb E(X_n) $$ exists for any $n = 1,2,\dots$. But I don't get the information about the variance or the second moment.

Hints? Thoughts?

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What is $\varphi$? –  Byron Schmuland Feb 21 '12 at 13:17
    
$\varphi$ is the characteristic function. So the precondition is that all characteristic functions of $X_1$,... are differentiable at $0$. –  user25070 Feb 21 '12 at 13:45
    
The characteristic function of what? I mean, $\varphi_{1}$ is the characteristic function of $X_1$, $\varphi_{2}$ is the characteristic function of $X_2$, etc. But $\varphi$ is the characteristic function of nothing. –  Byron Schmuland Feb 21 '12 at 13:48
2  
Be careful. Having a symmetric distribution implies that if the mean exists it must be 0. There are symmetric distributions with no mean, such as the standard Cauchy distribution. And you are going to have to do without the condition of finite variance; it won't follow from the conditions you have. –  Nate Eldredge Feb 21 '12 at 14:09
    
I hope I fixed the notation with $\varphi_n$ - please take a look if it is what you wanted to say. –  Ilya Feb 21 '12 at 14:11
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1 Answer

If the only assumptions are independence and symmetry, the result is not true. For a counterexample, assume $X_n=a_nY_n$ for your favorite sequence $(a_n)$ of real numbers such that $a_n\to\infty$ quickly, and your favorite sequence $(Y_n)$ of i.i.d. well behaved random variables, say centered Bernoulli or standard normal.

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To sum up, the Edit history of the question shows that a perfectly valid question (every i.i.d. sequence with a hypothesis on their common characteristic function satisfies a weak law of large numbers) was, later on, transformed into an incorrect statement: Edit #2 cancels the equidistribution hypothesis, without which the result does not hold. The OP might have been interested in restoring the original (correct) version of the question but this never happened. –  Did Apr 28 '12 at 8:04
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