Here is a problem and my attempt to solve it. Am I right so far?
Note: I did look at the following URL: Proving the sum of two independent Cauchy Random Variables is Cauchy . However, I did not understand it. I am thinking that there is some theorem that uses the convolution that helps in this case but if there is, I do not know what it is.
Thanks
Bob
Problem:
Prove that if $X_1$ and $X_2$ are independent and have the same Cauchy distribution, then their arithmetic mean also has this distribution.
Answer:
Let $Z = X_1+X_2$. I want to show that $Z$ has the Cauchy distribution. Let $f_x(u)$ and $f_z(u)$ be the density functions of
$X_1$ and Z. To find this density function, I use the idea of a characteristic functions. Let $\phi_z(\omega)$ be the characteristic function for Z.
\begin{eqnarray*}
f_x(u) &=& \frac{a}{\pi(u^2+a^2)} \\
\phi_x(\omega) &=& e^{a\omega} \\
\phi_z(\omega) &=& \phi_x(\omega) \phi_x(\omega) = e^{a\omega} e^{a\omega} \\
\phi_z(\omega) &=& e^{2a\omega} \\
\end{eqnarray*}
Now, how do I complete the proof?
Best Answer
You are almost there. Let $W=\frac{1}{2}(X_1+X_2)$. The definition you have for the characteristic function is not precise:
$$ \mathbf{E}[e^{i \omega X_1}] = \mathbf{E}[e^{i \omega X_1}] = e^{a |\omega|} $$
Then compute the characteristic function of $W$.
$$\mathbf{E}[e^{itW}]=\mathbf{E}[e^{i\frac{t}{2}X_1}]\mathbf{E}[e^{i\frac{t}{2}X_2}]=e^{a \frac{|t|}{2}}e^{a \frac{|t|}{2}}=e^{a|t|}$$ Which corresponds to the characteristic function of a Cauchy with parameter $a$, by unicity of the characteristic function you conclude that $W$ is a Cauchy with parameter $a$.