$\frac{X_1+X_2}{2}$ with $X_1\perp X_2\sim C(0,1)$

Question

Let $X\sim C(0,1)$ with density $f_X(x):=\frac{1}{\pi(1+x^2)}$.

1. Find the probability of event $B=(-1<X<1)$.

$\rightarrow \mathbb{P}(-1<X<1)=\int_{-1}^{1}\frac{1}{\pi(1+x^2)}dx=\frac{1}{2}$

2. Find the law of $Y=\frac{1}{X}$.

$\rightarrow Y\sim C(0,1)$

3. Find the law of $Z=\sigma X+\mu$, with $\mu \in \mathbb{R}$ and $\sigma >0$.

$\rightarrow F_z(z)=\mathbb{P}(Z\leq z)=\mathbb{P}(X\leq \frac{z-\mu}{\sigma})=… $

Now, I guess, I have to conclude that $Z\sim N(0,1)$. Right?

4. If $X_1 \perp X_2 \sim C(0,1)$, find the law of $W=\frac{X_1+X_2}{2}$.

$\rightarrow $ Here I've read that the sum of $n$ Cauchy distributed random variables is $C(0,1),\forall n$, but I didn't really understand why. How can I prove it?

Thanks in advance for any clarification.

Answer 1

1 and 2 look good to me.

Regarding 3, your answer isn't correct. Note that $X \sim C(0, 1)$ and NOT $\mathcal{N}(0, 1)$.

You have clearly shown that $$F_{Z}(z) = F_{X}\left(\dfrac{z - \mu}{\sigma}\right)$$ so taking derivatives, $$F^{\prime}_{Z}(z) = F^{\prime}_{X}\left(\dfrac{z - \mu}{\sigma}\right)\cdot \dfrac{1}{\sigma} = \dfrac{1}{\sigma}f_{X}\left(\dfrac{z - \mu}{\sigma}\right)$$ and I will leave it to you to see what this is.

Regarding 4, I haven't tried to do this myself, but you could probably do this by way of characteristic functions. The characteristic function of $X \sim C(0, 1)$ is $$\varphi_{X}(t) = \exp(-|t|)$$ (setting $x_0 = 0$ and $\gamma = 1$ in the linked equation) and so by basic properties of characteristic functions, if $X_1, \dots, X_n \overset{\text{iid}}{\sim} C(0, 1)$, $$\varphi_{X_1 + X_2 + \cdots + X_n}(t) = \varphi_{X_1}(t)\varphi_{X_2}(t)\cdots \varphi_{X_n}(t) = [\exp(-|t|)]^n = \exp(-n|t|)$$ so $S_n = \sum_{i=1}^{n}X_i \sim C(0, n)$ and thus using linear transformations, $$\varphi_{S_n/n}(t) = \varphi_{S_n}(t/n)=\exp\left(-n\left|\dfrac{t}{n}\right|\right) = \exp(-|t|)$$ so $S_n/n \sim C(0, 1)$.