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I am working with the Cauchy random variable. I have a sequence of iid Cauchy r.v. $X_i$ with parameters $0,\lambda>0$, i.e. $\phi_{X_1}(t)=e^{-\lambda|t|}$.

I think I have shown that $\frac{S_n}{n} \to X_1$ in distribution. Here's the proof, using the charateristic functions: $\phi_{\frac{S_n}{n}}(t)=\mathbb{E}(e^{it\frac{S_n}{n}}) = \mathbb{E}(e^{\frac{it}{n}(X_1+\ldots+X_n)}) = \mathbb{E}(\prod_{j=1}^n e^{\frac{it}{n}X_j}) = \prod_{j=1}^n(\mathbb{E}(e^{\frac{it}{n}X_j})) = \mathbb{E}(e^{\frac{it}{n}X_1})^n = \exp(-\lambda|\frac{t}{n}|)^n= \exp(-\lambda|t|)=\phi_{X_1}(t)$

I am trying to prove now that $S_n/n$ does not converge to $X_1$ in probability. I am trying to do so by contradiction, supposing it does and then proving $S_n/n - S_{2n}/2n $ does not converge in distribution to $0$ (thus contradicting the first statement). But I did not manage to do so because every time I extended the computation it did go to the characteristic function of $0$ (if you have any suggestion in this direction they will be very well accepted). Since my problem was about the factorization of the $2$ and $n$ I started trying with $S_{2n}/n - 2S_n/n$ which should go to $0$ in probability and I seemed to reach a contradiction, but I am not sure it works formally.

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    $\begingroup$ The error is replacing the second term in the limit with $X_1$, the convergence is only in distribution. To see that it gives inconsistent result you may replace it with $X_2$ and note that $X_2-X_1$ is Cauchy distributed with scale parameter $2\lambda$ and the probability isn’t 0 for any $\epsilon$. $\endgroup$
    – dioid
    Commented Apr 30, 2018 at 11:52

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I think you are expected to show that $S_n/n$ does not converge to any random variable in probability. Here is a proof: if $S_n/n$ converges in probability then $S_n/n-S_{2n}/\{2n\} \to 0$ in probability. Since convergence in probability implies convergence in distribution we must have $Ee^{it (S_n/n-S_{2n}/\{2n\})} \to 1$ for each $t$. The characteristic function is $e^{-\lambda |t|}$ for each $n$, so we have a contradiction.

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  • $\begingroup$ Thanks for the answer, but the problem asks if, since $S_n/n$ converges to $X_1$ in distribution, it converges to it also in probability. Also, in your proof, why did you use $2n$? $\endgroup$
    – user1868607
    Commented Apr 30, 2018 at 8:34
  • $\begingroup$ Also, can you explain to me why the characteristic function is $e^{|t|}$ and not $e^{-\lambda|t|}$ ? @Kavi $\endgroup$
    – user1868607
    Commented Apr 30, 2018 at 9:25
  • $\begingroup$ I have corrected the characteristic function. In the comment you say something just opposite to what is in the original question. Convergence in distribution does not imply convergence in probability. To prove that $S_n /n$ does not converge in probability I am proving that it is not Cauchy. The choice of $S_n/n -S_{2n}/{2n}$ is only a convenience and it makes the proof easy. $\endgroup$
    – Kavi Rama Murthy
    Commented Apr 30, 2018 at 10:16
  • $\begingroup$ sorry to insist, but isn't $E(\exp(it(S_n/n-S_{2n}/2n))=E(\exp(it(S_n/n))/E(\exp(it(S_{2n}/2n)) \to e^{-\lambda|t|}/e^{-\lambda|t|} = 1$? $\endgroup$
    – user1868607
    Commented Apr 30, 2018 at 10:25
  • $\begingroup$ No, it isn’t. $S_n$ and $S_{2n}$ are not independent. Simplify to sum of independent variables first. $\endgroup$
    – dioid
    Commented Apr 30, 2018 at 11:55

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