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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?

Bob
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    A characteristic function proof of this is trivial. You need some calculus to get a formula for $E\exp(itX)$, but the rest is then easy. – kimchi lover Nov 19 '17 at 22:40
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    You are trying to solve the problem bass ackwards. Find the characteristic function of a Cauchy random variable (hint: it is not what you have written), and then claim that the characteristic function of $Z$ is just the square of that. You need to work just a tad more to show that the distribution of $Z/2$ is the same as the distribution of $X$. Don't forget the arithmetic average part of the question. – Dilip Sarwate Nov 20 '17 at 00:18
  • @DilipSarwate The characteristic function of a Cauchy variable was given in the book so I am free to use that. I agree that I need to show that the distribution of $\frac{Z}{2}$ is the same as the distribution of $X$ but I have no idea how to do that. – Bob Nov 20 '17 at 00:30

1 Answers1

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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 $a$.

Joaquin San
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    You assume that $t\ge0$, I guess. – zhoraster Nov 20 '17 at 10:00
  • I do not understand why the following is true:$$\mathbf{E}[e^{i\frac{t}{2}X_2}]=e^{a \frac{t}{2}}$$ Does it come from the fact that we already know the characteristic function of a Cauchy variable and we are doing a substitution on the Cauchy variable? – Bob Nov 20 '17 at 15:24
  • You claimed that the characteristic function of $X$, was $ \phi_{X} (\omega) = e^{a \omega} $, I am just evaluating at $ \omega = \frac{t}{2} $. But if you are concerned with doing everything in the complex sense, then the characteristic function is not exactly that, it needs a modulus. The modulus wouldn't affect the proof I did, just replace $t$ for $|t|$. – Joaquin San Nov 20 '17 at 23:26
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    Let me just edit everything. – Joaquin San Nov 20 '17 at 23:27
  • If you still have doubts, maybe remember de definition of the characteristic function: $$ \phi_X (t) = \mathbf{E}[e^{itX}]$$ – Joaquin San Nov 21 '17 at 00:02