1
$\begingroup$

I have a multivariate student distribution fitted on some data on 4 dimensions (so I know the parameters). I am trying to calculate the $P(X_4\le x_4| X_1=x_1, X_2=x_2, X_3=x_3)$ but falling short.

I found a paper stating that conditional distribution of multivariate students are student distributions but that's about it. Is there a close form formula that I should use or do I have to run a Monte-Carlo kind of process?

$\endgroup$

1 Answer 1

2
$\begingroup$

The $p$-dimensional $t$ distribution has its density given by $$f_p(\mathbf x;\nu,\boldsymbol\mu,\boldsymbol\Sigma) =\frac{\Gamma\left[(\nu+p)/2\right]}{\Gamma(\nu/2)\nu^{p/2}\pi^{p/2}\left|{\boldsymbol\Sigma}\right|^{1/2}}\left[1+\frac{1}{\nu}({\mathbf x}-{\boldsymbol\mu})^{\rm T}{\boldsymbol\Sigma}^{-1}({\mathbf x}-{\boldsymbol\mu})\right]^{-(\nu+p)/2}$$ Hence \begin{align}f(x_4|x_1,x_2,x_3)&\propto f_4(\mathbf x;\nu,\boldsymbol\mu,\boldsymbol\Sigma)\\ &\propto \left[1+\frac{1}{\nu}({\mathbf x}-{\boldsymbol\mu})^{\rm T}{\boldsymbol\Sigma}^{-1}({\mathbf x}-{\boldsymbol\mu})\right]^{-(\nu+4)/2}\\ &\propto \left[1+\frac{1}{\nu}(a(x_4-\mu_4)^2+b(x_4-\mu_4)+c)\right]^{-(\nu+4)/2} \end{align} the last term being obtained by expanding$$({\mathbf x}-{\boldsymbol\mu})^{\rm T}{\boldsymbol\Sigma}^{-1}({\mathbf x}-{\boldsymbol\mu})$$as a second degree polynomial in terms of $x_4-\mu_4$. With $a,b,c$ depending on $(x_1-\mu_1,x_2-\mu_2,x_3-\mu_3)$ as well as $\boldsymbol\Sigma$. Since \begin{align}\frac{1}{\nu}(a(x_4-\mu_4)^2+b(x_4-\mu_4)+c) &=\frac{1}{\nu}\{a(x_4-\mu_4+{b}/{2a})^2+c-b^2/4a\} \end{align} the conclusion is that $$f(x_4|x_1,x_2,x_3)\propto \left[1+\frac{1}{\nu+3} \frac{(x_4-\eta_4)^2}{\sigma_4^2}\right]^{-(\nu+4)/2}$$ where $$\eta_4=\mu_4-\frac{b}{2a}$$ and $$\sigma_4^2=\frac{1}{\nu+3}\frac{\nu+c-\frac{b^2}{4a}}{a}$$ is indeed the density of a $t$ distribution with $\nu_4=\nu+3$ degrees of freedom.

$\endgroup$
4
  • $\begingroup$ Thanks for your reply. I don't understand what are referring $x$ , $\eta$ and $\nu$ . I know $\nu$ is the degree of freedom but in that case specifically is it $\nu_4$ and $\sigma_4$ $\endgroup$
    – Mayeul sgc
    May 31 at 9:19
  • $\begingroup$ I had forgotten an index, see the corrected version. The other symbols are arbitrary, you need to do by yourself the algebra for the quadratic form expansion. Except for the degree of freedom, which is $\nu+3$. $\endgroup$
    – Xi'an
    May 31 at 9:46
  • $\begingroup$ thank you! do you have any book or paper to suggest for this kind of topics ? It took me some time to get your answer and I feel I might need some theory background on those. $\endgroup$
    – Mayeul sgc
    Jun 1 at 7:16
  • $\begingroup$ This is standard calculus, so I cannot suggest a particular book for this... $\endgroup$
    – Xi'an
    Jun 1 at 7:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.