# Analytic form for the quadrant probability of a (standard) bivariate t distribution?

#### spunky

##### Doesn't actually exist
hellows! i may need this further along my manuscript and quick google searches aren't really being helpful, so maybe someone around here knows.

for the case of the bivariate standard normal distribution, the quadrant probability is given by:

$$P(x_{1}\leq0,x_{2}\leq0)=P(x_{1}\geq0,x_{2}\geq0)=\int_{-\infty }^{0}\int_{-\infty }^{0}P(x_{1},x_{2})dx_{1}dx_{2}=\frac{1}{4}+\frac{sin^{-1}\rho}{2\pi}$$

(or at least the quadrants going in the same direction).

i may find myself needing a nice, close-formed thingy like that one but for the (standard) bivariate t distribution, but goolging stuff like that doesn't give me many results that i could use.

does anyone know if there is a nice closed formed expression for the (standard) bivariate t distribution like there is one for the bivariate standard normal distribution? or there isn't any?

thanks!

#### Jake

Re: Analytic for the quadrant probability of a (standard) bivariate t distribution?

It seems unlikely since, according to wiki, there is no analytic expression for the CDF of multivariate t.

#### rogojel

##### TS Contributor
Re: Analytic for the quadrant probability of a (standard) bivariate t distribution?

hi,
on an off chance - we used to use a book by the authors Abramovici and Stegun (something like a hanbook of series and integrals) way before google was implemented It used to be, that if a formula is not in the A/S it does not exist.

#### BGM

##### TS Contributor
Re: Analytic form for the quadrant probability of a (standard) bivariate t distributi

I think most of you already know this but let's present this once again before trying to generalize further.

Let $$Z_1, Z_2$$ be i.i.d. standard normal random variables. From Cholesky Decomposition, we know that

$$X_1 = Z_1, X_2 = \rho Z_1 + \sqrt{1 - \rho^2} Z_2$$

will jointly follows a bivariate standard normal with correlation $$\rho \in (-1, 1)$$. The joint pdf will be in terms of

$$\begin{bmatrix} x_1 & x_2 \end{bmatrix} \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}^{-1} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$$

(*)
Now for any $$(X_1, X_2)$$ with joint pdf solely depending the above quadratic form, we try to use the same back-trasform listed in above into $$(Z_1, Z_2)$$,
and see how it affect the joint pdf:

$$\frac {\partial x_1} {\partial z_1} = 1, \frac {\partial x_1} {\partial z_2} = 0, \frac {\partial x_2} {\partial z_1} = \rho, \frac {\partial x_2} {\partial z_2} = \sqrt{1 - \rho^2}$$

and thus the Jacobian is
$$|1 \times \sqrt{1 - \rho^2} - 0 \times \rho| = \sqrt{1 - \rho^2}$$

and independent of $$z_1, z_2$$. The resulting joint pdf will be in terms of

$$\begin{bmatrix} z_1 & \rho z_1 + \sqrt{1 - \rho^2}z_2 \end{bmatrix} \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}^{-1} \begin{bmatrix} z_1 \\ \rho z_1 + \sqrt{1 - \rho^2}z_2 \end{bmatrix}$$

$$= \frac {1} {1 - \rho^2}\begin{bmatrix} z_1 & z_2 \end{bmatrix} \begin{bmatrix} 1 & \rho \\ 0 & \sqrt{1 - \rho^2} \end{bmatrix} \begin{bmatrix} 1 & -\rho \\ -\rho & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \rho & \sqrt{1 - \rho^2} \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix}$$

$$= \frac {1} {1 - \rho^2}\begin{bmatrix} z_1 & z_2 \end{bmatrix} \begin{bmatrix} 1 - \rho^2 & 0 \\ 0 & 1 - \rho^2 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix}$$

$$= \begin{bmatrix} z_1 & z_2 \end{bmatrix} \begin{bmatrix} z_1 \\ z_2 \end{bmatrix}$$

$$= z_1^2 + z_2^2$$

So by the transformation, we see that the joint pdf of $$(Z_1, Z_2)$$ has a circular contour, and thus is rotational symmetric about the mean (i.e. the origin). Therefore for any pair of straight lines intersecting at the mean and dividing $$\mathbb{R}^2$$ into four regions, the probability that $$(Z_1, Z_2)$$ fall into a particular region will be equal to the proportion of the angle at centre from $$2\pi$$, similar to the case when we calculating the area of a sector from circle.

And that is why we have the very neat closed-form result:

$$\Pr\{X_1 > 0, X_2 > 0\}$$

$$= \Pr\{Z_1 > 0, \rho Z_1 + \sqrt{1 - \rho^2} Z_2 > 0\}$$

$$= \Pr\left\{Z_1 > 0, Z_2 > - \frac {\rho} {\sqrt{1 - \rho^2}} Z_1 \right\}$$

Now the region enclosed by

$$\left\{z_1 > 0, z_2 > - \frac {\rho} {\sqrt{1 - \rho^2}} z_1 \right\}$$

in the $$(z_1, z_2)$$-plane consist of two bounds, one is the vertical $$z_2$$-axis, and the other one is a straight line passing through origin with slope $$- \frac {\rho} {\sqrt{1 - \rho^2}}$$. The angle at centre will be

$$\frac {\pi} {2} - \tan^{-1}\left(- \frac {\rho} {\sqrt{1 - \rho^2}}\right) = \frac {\pi} {2} + \sin^{-1}\rho$$

Dividing by $$2\pi$$ yields Spunky posted result.

Now back to the question: In the above calculation starting from (*), we merely required that the joint pdf solely in terms of the quadratic form (we do not need to require it to be multivariate normal), so it should holds for any ellipitical distribution, including the standard bivariate t-distribution mentioned. The bivariate t-distribution mentioned was referring to
https://en.wikipedia.org/wiki/Multivariate_t-distribution#Derivation

So my guess will be the same result also applies to all elliptical distributions (note that in bivariate t, the joint pdf has a circular contour does not mean $$Z_1, Z_2$$ are independent). I hope this guess is correct Feel free to point out if the argument is wrong.