log likelihood function of a cauchy distribution

Question

What is the log likelihood function of a random varible x with cauchy distribution (0,1)? I've tried to work it out. I think its $\log (1+x)^2$. Is that correct?

Answer 1

Assuming you have a random sample $X_1, X_2, \ldots X_n$ from a population with a (0,1) cauchy distribution, i.e. the $X_i$'s are i.i.d. with pdf $$f(x)=\frac{1}{\pi (1+x^2)}$$ for $x \in \mathbb R$. Then the likelihood function is $$\mathcal L(x;0,1)=\prod_{i=1}^{n}\frac{1}{\pi (1+x_i^2)}=\frac{1}{\pi^n}\cdot\frac{1}{\prod_{i=1}^{n}(1+x_i^2)}=\pi^{-n}\cdot\prod_{i=1}^{n}(1+x_i^2)^{-1}$$ and the log-likelihood function is $$\mathcal l(x;0,1)=\ln(\mathcal L (x;0,1))=-n\ln\pi-\sum_{i=1}^{n}\ln(1+x_i^2)$$