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I'm given (X,Y) ~ standard bivariate normal (p) -(assume this is the greek letter ro).

I'm asked to find the P(XY>0) as functions of p and other values as indicated.

I know by definition, two random variables X and Y are said to be bivariate normal if and only if aX+bY has a normal distribution. Though, I'm not certain I'm able to satisfy this axiom. Is the product of two normal distributions univariate normal? How should I approach this question?

Tommyixi
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  • You can probably start out by splitting it into two cases:

    $P(XY>0) = P(XY>0, X>0) + P(XY>0, X<0) = P(Y > 0, X >0) + P(Y < 0, X < 0)$.

     –  Sep 29 '16 at 23:49
  • Consider the density of $(X,Y)$, which can be shown to have the form (rewrite the expression from here: https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Density_function)

    $f(x,y)=\frac{1}{2\pi\sqrt{1-\rho^2}}\exp\left(-\frac{(x-\rho y)^2}{2(1-\rho)^2}\right)\exp\left(-\frac{y^2}{2}\right)$.

     –  Sep 30 '16 at 00:29
  • With $g(y)=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{y^2}{2}\right)$ denoting the density of $Y$, we see that

    $f(x,y) = g(y) \cdot \frac{1}{\sqrt{2\pi(1-\rho^2)}}\exp\left(-\frac{(x-\rho y)^2}{2(1-\rho)^2}\right)$,

    thus the conditional density of $X$ given $Y=y$ is $h(x,y)=\frac{1}{\sqrt{2\pi(1-\rho^2)}}\exp\left(-\frac{(x-\rho y)^2}{2(1-\rho)^2}\right)$.

    This means that conditional on $Y=y$, $X\sim\mathcal{N}(-\rho y, 1-\rho^2)$.

     –  Sep 30 '16 at 00:29
  • Also due to symmetry, $P(Y>0,X>0)=P(Y<0,X<0)$. –  Sep 30 '16 at 00:29
  • Interesting. Though, I was hoping not to compute this probability via integrating against the joint density. I thought perhaps I could do this by treating both as if they were independent variables. – Tommyixi Sep 30 '16 at 02:55
  • I have not been able to follow trough with my line of thought presented in the above comments. I end up with

    $ P(X>0,Y>0) = \int_0^\infty \Phi(\rho y / \sqrt{1-\rho^2}) g(y) , \mathrm{d}y$,

    where $\Phi$ is the distribution function of a standard normal distribution and $g$ is the density.

    If you assume independence, then $P(X>0,Y>0)=P(X>0)P(Y>0)=1/4$ and you get that $P(XY>0)=1/2$.

     –  Sep 30 '16 at 12:50
  • Yes! Thank you so much! They key was to notice I could breakup the probability into a joint probability as long as I made sure both variables were independent of one another. – Tommyixi Oct 01 '16 at 04:18
  • refer to this other post – G Cab Oct 15 '17 at 13:44

1 Answers1

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Assuming the Bivariate Normal has mean $0$. We can do the following:

First, let $Z:=\frac{X}{Y}$ denote the ratio between the two components of the bivariate normal then $Z \sim Cauchy (\rho\frac{\sigma_x}{\sigma_y},\frac{\sigma_x}{\sigma_y}\sqrt{1-\rho^2})$. Therefore for any number $z$ the distribution function is: $F_z=\frac{1}{2}+\frac{1}{\pi}\tan^{-1}\left(\frac{\sigma_yx-\rho\sigma_x}{\sigma_x\sqrt{1-\rho^2}}\right)$

You should note that $P(XY>0)=1-P(XY<0)$ and further $P(XY<0)=P(Z<0)$, now by taking $F_z(0)$ you get $\frac{1}{2}+\frac{1}{\pi}\tan^{-1}\left(-\frac{\rho}{\sqrt{1-\rho^2}}\right)$, now, $\tan^{-1}$ is and odd function and the trigonometric identity $\tan(\sin^{-1}(x))=\frac{x}{\sqrt{1-x^2}}$ imply that $P(Z<0)=\frac{1}{2}-\frac{1}{\pi}\sin^{-1}(\rho)$ and in your case $P(XY>0)=\frac{1}{2}+\frac{1}{\pi}\sin^{-1}(\rho)$

Daniel Ordoñez
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