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Question:

Let $(X_1,X_2,X_3)\sim N_3\left[\mathbf0, \begin{pmatrix}1&\rho_{12}&\rho_{13}\\\rho_{12}&1&\rho_{23}\\\rho_{13}&\rho_{23}&1\end{pmatrix} \right]$

Show that

(i) $$P(X_1>0,X_2>0,X_3>0)=\frac{1}{8}+\frac{\sin^{-1}\rho_{12}+\sin^{-1}\rho_{13}+\sin^{-1}\rho_{23}}{4\pi}$$

(ii) $$1+2\rho_{12}\rho_{13}\rho_{23}\ge \rho_{12}^2+\rho_{13}^2+\rho_{23}^2$$

My problem: I was solving test papers when I found this..

I had this sum done for a Bivariate Normal Distribution.

(X1, X2)~ BN( 0, 0,sigma_1^2, sigma_2^2, rho )

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I know the formula of a Trivariate Normal Distribution.

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How can I prove the given relation (i) in the question, in a method like the Bivariate one was solved ?

And how do I solve part (ii) ? Please help. Thank you.

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    $\begingroup$ (ii) is solved at stats.stackexchange.com/questions/72790, which relates it to the determinant of the correlation matrix. For (i), consider transforming the variables into a standard Normal distribution. (The formula is recognizable as a solid angle subtended by the transformed first octant, divided by the total angle $4\pi.$) Working directly from your formula for the PDF is likely to be a long and painful process. $\endgroup$ – whuber Nov 28 '18 at 18:44
  • $\begingroup$ @whuber The marginals are already standard normal, aren't they? $\endgroup$ – StubbornAtom Nov 29 '18 at 18:11
  • $\begingroup$ @Stubborn They are, but the trivariate distribution is not. By "standard Normal" I meant standard Trivariate Normal: that is, the mean is $(0,0,0)$ and the variance is the unit matrix. $\endgroup$ – whuber Nov 29 '18 at 20:20
  • $\begingroup$ @Stubborn Let me try again: the desired transformation will re-express the event $X_1\gt0,X_2\gt 0,X_3\gt 0$ in an eigenbasis of the covariance matrix. In that representation the distribution is spherical, showing that the answer is simply the fractional solid angle subtended by the transformed first octant. (Nice derivation in your answer, BTW, and +1 for that.) $\endgroup$ – whuber Nov 29 '18 at 20:30
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For the first question, indeed there is no need to work directly with the pdf of $(X_1,X_2,X_3)$.

Let the desired probability be $$p=P(X_1>0,X_2>0,X_3>0)$$

Analogous to the two-variable case, we have $$(X_1,X_2,X_3)\stackrel{d}{=}(-X_1,-X_2,-X_3)$$

So due to symmetry we must have

$$p=P(-X_1>0,-X_2>0,-X_3>0)=P(X_1<0,X_2<0,X_3<0)\tag{1}$$

Continuing from $(1)$,

\begin{align} 1-p&=P\left[\{X_1>0\}\cup\{X_2>0\}\cup\{X_3>0\}\right] \\\\&=P(X_1>0)+P(X_2>0)+P(X_3>0)-P(X_1>0,X_2>0) \\\\&\quad-P(X_2>0,X_3>0)-P(X_1>0,X_3>0)+p \\\\&=\frac{3}{2}-\left[\frac{3}{4}+\frac{1}{2\pi}(\sin^{-1}\rho_{12}+\sin^{-1}\rho_{23}+\sin^{-1}\rho_{13})\right]+p \end{align}

Or,

$$\\1-\left[\frac{3}{4}-\frac{1}{2\pi}(\sin^{-1}\rho_{12}+\sin^{-1}\rho_{23}+\sin^{-1}\rho_{13})\right]=2p$$

Finally,

$$p=\frac{1}{8}+\frac{1}{4\pi}(\sin^{-1}\rho_{12}+\sin^{-1}\rho_{23}+\sin^{-1}\rho_{13})$$


As a side note, I think that for the derivation of the expression for $P(X_1>0,X_2>0)$, it suffices to evaluate the corresponding double integral using a polar transformation. The details are not very messy. What you have done is perfectly fine by the way.

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