Assume X$_1$, ... , X$_n$ are i.i.d distributed random variables. And each them of has the same probability density function as f(x) = $\frac{1}{\pi}$$\frac{1}{1+ x^2}$. How can we compute the probability density function for X = $\frac{1}{n}$ $\sum_{i=1}^{n}$X$_i$?

Some idea is to write the cumulative distribution function:

F$_{X_1+X_2+...+X_n}$(a) = P(X$_1$ + X$_2$ + ... + X$_n$ $\leq$ a) = $\idotsint$ f$_{X_1}$(x$_1$)f$_{X_2}$(x$_2$) $\cdots$ $\cdots$ f$_{X_n}$(x$_n$)d$_{x_1}$d$_{x_2}$$\cdots$d$_{x_n}$

But feel hard to continue. Any suggestions?

  • 3
    $\begingroup$ Try using the characteristic function (if you know about this). Note that $X_i$ come from a standard Cauchy distribution (see en.wikipedia.org/wiki/Cauchy_distribution; $x_0=0$ and $\gamma=1$ there for the standard Cauchy distribution). See stats.stackexchange.com/questions/238246/… for a proof with the characteristic function. $\endgroup$ – Minus One-Twelfth Mar 28 at 8:31
  • $\begingroup$ thanks, is there any way to calculate it? btw, why can we say they have same pdfs if their characteristic functions are the same? $\endgroup$ – Jonny Mar 28 at 21:44

It also has density $\frac 1 {\pi} \frac 1 {1+x^{2}}$. The best way to prove this is to use characteristic functions: since $Ee^{itX_1}=e^{-|t|}$ and $(e^{-|\frac t n|})^{n}=e^{-|t|}$ we get the conclusion easily.

  • $\begingroup$ Thanks, btw, how can we prove the characteristic function for standard Cauchy distribution? $\endgroup$ – Jonny Mar 28 at 19:57
  • $\begingroup$ ignore previous comment. any suggestion on traditional way to calculate it rather than using characteristic functions? $\endgroup$ – Jonny Mar 28 at 22:29

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