# Bivariate Student t-distribution (density representation)

It is well known that bivariate normal pdf can be written in terms of univariate pdfs: $$\frac{1}{2\pi\sigma_x\sigma_y\sqrt{1-\rho^2}}\exp\left(-\frac{(\xi_1^2+\xi_2^2-2\rho \xi_1\xi_2)}{2(1-\rho^2)}\right)=\frac{1}{\sigma_x\sigma_y\sqrt{1-\rho^2}}\phi\left(\frac{\xi_1-\rho\xi_2}{\sqrt{1-\rho^2}}\right)\phi\left(\xi_2\right)$$ Is there a similar result for bivariate Student t-distribution, that is, can a bivariate Student t-distribution be written in terms of univariate Student t-densities?

• where did you see the last equality derived, and what is $\phi$? Dec 7, 2020 at 1:33

There is a mistake in the expressions of the Normal bivariate pdf decomposition. As a general result, the conditional distribution of the first component of a Normal bivariate vector given the second component is $$X_1\mid X_2=\xi_2 \ \sim\ \mathcal{N}\left(\mu_1+\frac{\sigma_1}{\sigma_2}\rho(\xi_2 - \mu_2),\, (1-\rho^2)\sigma_1^2\right).$$ Hence, assuming $$\mu_1=\mu_2=0$$, the conditional density of $$X_1$$ given $$X_2=\xi_2$$ is $$\frac{1}{\sigma_1\sqrt{1-\rho^2}}\,\phi\left(\frac{(\xi_1-\frac{\sigma_1}{\sigma_2}\rho\xi_2)^2}{\sigma_1\sqrt{1-\rho^2}}\right)$$ and the equality should be \begin{align}\frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\exp\left(-\frac{\sigma_1^{-2}\xi_1^2+\sigma_2^{-2}\xi_2^2-2\rho \xi_1\xi_2/\sigma_1\sigma_2}{2}\right)\\=\frac{1}{\sigma_1\sigma_2\sqrt{1-\rho^2}}\phi\left(\frac{(\xi_1-\frac{\sigma_1}{\sigma_2}\rho\xi_2)^2}{\sigma_1\sqrt{1-\rho^2}}\right)\phi\left(\xi_2/\sigma_2\right)\end{align} with $$\phi(\cdot)$$ denoting the standard Normal pdf.
Reading through this paper, if $$(\boldsymbol X_1,\boldsymbol X_2)$$ is distributed from a $$p$$ dimensional multivariate Student's $$\mathfrak{t}$$ distribution $$\mathfrak{t}_{p}(\nu,\boldsymbol\mu,\boldsymbol\Sigma)$$ [LaTeX copied from Wikipedia] $$\frac{\Gamma\left(\frac{\nu+p}{2}\right)}{(\nu\pi)^{\frac{p}{2}}\Gamma(\frac{\nu}{2})\left|{\boldsymbol\Sigma}\right|^{1/2}}\left[1+\frac{1}{\nu}({\mathbf x}-{\boldsymbol\mu})^{\rm T}{\boldsymbol\Sigma}^{-1}({\mathbf x}-{\boldsymbol\mu})\right]^{-\frac{\nu+p}{2}}$$ then both the marginal and the conditional distributions of $$\boldsymbol X_1$$ given $$\boldsymbol X_2$$ are also $$p_1$$ dimensional multivariate Student's $$\mathfrak{t}$$ distributions: $$\boldsymbol X_1 \sim \mathfrak{t}_{p_1}(\nu,\boldsymbol\mu_1,\boldsymbol\Sigma_{11})$$ and \begin{align}\boldsymbol X_1|\boldsymbol X_2 \sim \mathfrak{t}_{p_1}\big(&\nu+p_2,\boldsymbol\mu_1+\boldsymbol\Sigma_{12}\boldsymbol\Sigma^{-1}_{22}(\boldsymbol X_2−\boldsymbol \mu_2),\\&\dfrac{\nu+(\boldsymbol X_2-\boldsymbol\mu_2)^\text{T}\boldsymbol \Sigma^{−1}_{22}(\boldsymbol X_2-\boldsymbol\mu_2)}{\nu+p_2}\boldsymbol \Sigma_{11|2}\Big) \end{align} where$$\boldsymbol \Sigma_{11|2}=\boldsymbol \Sigma_{11}−\boldsymbol \Sigma_{21}\boldsymbol \Sigma^{−1}_{22}\boldsymbol \Sigma_{12}$$ This is easily proved by using the demarginalisation of the Student's $$\mathfrak{t}$$ as a mixture of a Normal variate with a chi-squared variate: $$\boldsymbol X|q\sim\mathcal N_p(\boldsymbol\mu,\boldsymbol\Sigma/q),\qquad q\sim\chi^2_\nu/\nu$$
• Thank you! I derived it for the bivariate case: $t_{\nu,2}(x,y;0,C) = \frac{1}{\sqrt{(1-\rho^2)\frac{x^2+\nu}{\nu+1}}} \times t_{\nu,1}(x)\times t_{\nu+1,1}\left(\frac{y-\rho x}{\sqrt{(1-\rho^2)\frac{x^2+\nu}{\nu+1}}}\right).$
• Wikipedia also says $\frac{\Gamma\left(\frac{\nu+2}{2}\right)}{(\nu\pi)^{\frac{2}{2}}\Gamma(\frac{\nu}{2})}=\frac{1}{2\pi}$, which means that the bivariate t-distribution is $$\frac{1}{2\pi \left|{\boldsymbol\Sigma}\right|^{1/2}}\left[1+\frac{1}{\nu}({\mathbf x}-{\boldsymbol\mu})^{\top}{\boldsymbol\Sigma}^{-1}({\mathbf x}-{\boldsymbol\mu})\right]^{-\frac{\nu+p}{2}}$$ do any sources derive/confirm this to be the bivariate t-distribution? Dec 7, 2020 at 1:29
• @Dev Yes, the Wikipedia article on the multivariate t distribution gives this formula with $p=2,$ of course.