Dirichlet distribution - moments

  • Context: Graduate 
  • Thread starter Thread starter Boot20
  • Start date Start date
  • Tags Tags
    Distribution Moments
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 3K views
Boot20
Messages
8
Reaction score
0
For a Dirichlet variable, I know the means and covariances, that is,

[itex]E[X_i] = \alpha_i/\alpha_0[/itex]
[itex]Cov[X_i,X_j] = \frac{ \alpha_i (\alpha_0I[i=j] - \alpha_j)}{\alpha_0^2(\alpha_0 + 1)}[/itex]

But how can I prove these facts?
 
Last edited:
Physics news on Phys.org
[itex]p(x_1,...,x_{k-1}) = \frac{1}{B(\alpha) } \left[ \prod^{K}_{i=1} x_i^{\alpha_i - 1} \right][/itex]

[itex]E[X_1] = \frac{1}{B(\alpha) } \int^1_0 ...\int^{1 - \sum^{K}_{i=2}x_i}_0 x_1 \left[\prod^{K}_{i=1} x_i^{\alpha_i - 1} \right] d x_1 ... d x_{k}[/itex]
 
Last edited:
Boot20 said:
[itex]p(x_1,...,x_{k-1}) = \frac{1}{B(\alpha) } \left[ \prod^{K}_{i=1} x_i^{\alpha_i - 1} \right][/itex]

[itex]E[X_1] = \frac{1}{B(\alpha) } \int^1_0 ...\int^{1 - \sum^{K}_{i=2}x_i}_0 x_1 \left[\prod^{K}_{i=1} x_i^{\alpha_i - 1} \right] d x_1 ... d x_{k}[/itex]

Hint: the expectation integrals closely resemble the definition of the normalizing constant B (which itself can be expressed in terms of gamma functions).