# Maximum likelihood estimation for Cauchy noise

What is the maximum likelihood estimator of the covariance matrix for a given vector in the presence of Cauchy noise?

How can we calculate it given that the Cauchy distribution has infinite variance?

• How do you define a multivariate Cauchy distribution? Do you mean a special multivariate t? In which case the scale matrix is well-defined if not necessarily a covariance matrix. Jul 20, 2015 at 8:43
• How does your question differ from that one? Jul 20, 2015 at 8:45
• I am not using multivariate Cauchy distribution. I am using standard Cauchy Distribution. Jul 20, 2015 at 8:55
• The multivariate Cauchy is a special case of the multivariate t with one degree of freedom. Jul 20, 2015 at 9:42

As explained in Nadarajah and Kotz (2007), given the log-likelihood function of the multivariate t distribution with parameters $(μ,R,ν)$, $$L(μ,R,ν)=−\frac{n}{2}\log|R|−\frac{ν+p}{2}\sum_{i=1}^n\log(ν+s_i)\,,$$ the maximum likelihood estimator can be found by an EM algorithm exploiting the latent Gaussian representation of the t.

The EM iteration is of the form $$μ^{(m+1)}=\text{average}(w^{(m)}_ix_i)\big/\text{average}(w^{(m)}_i)$$ and $$R^{(m+1)}=\text{average}(w^{(m)}_i\{x_i-μ^{(m+1)}\}\{x_i-μ^{(m+1)}\}^\text{T})\big/\text{average}(w^{(m)}_i)$$ where $$w^{(m)}_i=(ν+p)\big/\{\nu+(x_i−μ^{(m)})(^\text{T}R^{(m)})^{-1}(x_i−μ^{(m)})\}$$

• Thanks but as this is for multivariate t distribution as I need for cauchy me degree of freedom will be 1. We will get MLE of multivariate Cauchy. Jul 20, 2015 at 9:23
• Okay what will be our first assumption to start iteration. Jul 20, 2015 at 10:07
• Have you checked the basics of the EM algorithm? The starting point only matters in determining the limiting value of the algorithm and should thus be modified for multiple runs of EM. Jul 20, 2015 at 12:01
• Can you tell me what will be the distribution of the ML covariance matrix for a Cauchy noise? Aug 3, 2015 at 5:16
• Sorry sir, your answers are really helping me to learn, I am new to use stack exchange and a new learner may be due to this I missed but I found most of your answers useful. Aug 3, 2015 at 8:50