Hey I have some questions about this two exercises:
Consider $(X_n){n∈\mathbb{N}}$, a sequence of independent and identically distributed real-valued random variables with an absolutely continuous distribution function and density $f{X_n} (x) = \frac{1}{π(1+x^2)}$.
a) Show that for the characteristic function $ϕ_{X_1}$, it holds: $ϕ_{X_1} (u) = e^{−|u|}$ , $∀u ∈ \mathbb{R}$.
b) Show that $\frac{X_1+···+X_n}{n}$ has the same distribution as $X_1$.
For part a), I wanted to use the following theorem:
"If the characteristic function φX of a random variable X is integrable, then FX is absolutely continuous, and therefore X has a probability density function. In the univariate case (i.e., when X is scalar-valued), the density function is given by $f_{X}(x)=F_{X}'(x)={\frac {1}{2\pi }}\int _{\mathbf {R} }e^{-itx}\varphi _{X}(t),dt.$"
So, I have ${\frac {1}{2\pi }}\int _{-\infty}^{\infty}e^{-itx}e^{-|t|},dt={\frac {1}{2\pi }}\int {-\infty}^{\infty}e^{-t(ix+1)},dt=\frac {1}{2\pi (ix+1)}[e^{-t(ix+1)}]{-\infty}^{\infty}$
And now I am unsure about how to proceed. I suspect that I may have made an error by not paying attention to the absolute value. Can someone help me with this?
Moving on to part b),I initially thought of solving it simply by:
$F^{\frac{X_1+···+X_n}{n}}=P\left(\frac{X_1+···+X_n}{n}\leq x\right)\overset{\text{since iid}}=P\left(\frac{X_1+···+X_1}{n}\leq x\right)=P\left(\frac{nX_1}{n}\leq x\right)=P(X_1\leq x)=F^{(X_1)}$
However, I felt that it seemed too simple, and I found a different solution online that used the characteristic function:
$ϕ_{\frac{X_1+···+X_n}{n}} (u)=E[e^{iu(\frac{X_1+···+X_n}{n})}]=\prod_{i=0}^{n}E[e^(iuX_1/n)]=e^{-|u|n/n}=e^{-|u|}$
Now, I am unsure if my solution is correct or if I made any mistakes. Can someone help I with this?
