# average of random variable

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?

• 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. – Minus One-Twelfth Mar 28 at 8:31
• thanks, is there any way to calculate it? btw, why can we say they have same pdfs if their characteristic functions are the same? – 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.