I computed mean, variance and covariance of the Dirichlet distribution. To do so, I computed \(E[x_k]\), \(E[x_k^2]\) and \(E[x_i x_j]\). This is the first time I've dealt with multivariate distributions. The mean should be the weighted sum of the vectors x in the simplex so I can consider one component at a time and compute \(E[x_k]\). But what is \(x^2\)? Is \(x^2 = [x_1^2, x_2^2, \ldots ,x_N^2]^T\) and so I can compute \(E[x_k^2]\) individually? And what about the covariance? Are variance and covariance about the components of the vectors x?
Dirichlet distribution
- Thread starter Kiuhnm
- Start date
- Tags dirichlet distribution
There do have component-wise operation, but in many occasion you will not consider \( \mathbf{X}^2 \) where \( \mathbf{X} \) is not a square matrix.
When you are dealing with the multivariate distribution, you will consider the variance-covariance entry, with the \( (i, j) \)-th entry being \( Cov[X_i, X_j] \).
When you are dealing with the multivariate distribution, you will consider the variance-covariance entry, with the \( (i, j) \)-th entry being \( Cov[X_i, X_j] \).
I'm not sure I understand your answer completely. Maybe I should just look up the definitions of mean, variance and covariance for multivariate distributions.
Edit: OK, I think I understand now. \(E[x] = [E[x_1], \ldots, E[x_n]]^T\) and then there is the covariance matrix. Bottom line, we can compute a component/entry at a time.
Edit: OK, I think I understand now. \(E[x] = [E[x_1], \ldots, E[x_n]]^T\) and then there is the covariance matrix. Bottom line, we can compute a component/entry at a time.
Last edited: