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Suppose $X$ and $Y$ are independent, standard normal random variables. I'm trying to compute the probability of the event $$ \{X \leq x, Y \leq kX\} $$ where $k$ is a positive constant. The probability is given by the integral

$$ \int_{-\infty}^x \int_{-\infty}^{ks} \phi(t)\,\phi(s)\, dt \, ds $$

where

$$ \phi(s) = \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}s^2} $$

I'm only able to find a closed form solution when $k = 1$. Is it possible to obtain a closed form solution for arbitrary $k > 0$? I've tried the usual integration tools but seem to be coming up empty.

Thanks!

David
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  • Many people do not consider expressions in $\Phi$ to be truly 'closed form'. – BruceET Jan 01 '16 at 19:03
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    Actually, when $k=1$ the result is $\frac12\Phi(x)^2$, not $2\varphi(x)\Phi(x)$. Other particular values are $\min{\Phi(x),\frac12}$ when $k\to-\infty$, $\frac12\Phi(x)$ when $k=0$, and $(\Phi(x)-\frac12)^+$ when $k\to+\infty$. Additionally, for every $(k,x)$, $$\frac{\partial}{\partial k}P(X\leqslant x,Y\leqslant kX)=-\frac1{\sqrt{2\pi}}\frac1{1+k^2}\varphi(x\sqrt{1+k^2}).$$ – Did Jan 02 '16 at 14:43

1 Answers1

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Just as we must rely on printed tables (or numerical integration) to get most univariate normal probabilities, it is difficult to get closed forms for many multivatiate normal probabilities. There may be a trick I don't know to handle the particular case of this Problem, but my impression is that such problems are often handled via 'zonal polynomials'.

For an uncorrelated bivariate normal distribution, simulation is especially easy. Here is an approximation based on a million simulated points with $x = 1$ and $k = 2$. The answer should be correct to 2 or 3 decimal places.

 m = 10^6;  x = 1;  k = 2
 z1 = rnorm(m);  z2 = rnorm(m)
 mean(z1 < x & z2 < 2*z1)
 ## 0.342797

The dark points in the plot below satisfy the desired condition.

enter image description here

BruceET
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  • Yeah it's probably easier to obtain a numerical approximation using the Gauss-Hermite quadrature: https://en.wikipedia.org/wiki/Gauss%E2%80%93Hermite_quadrature – David Jan 01 '16 at 20:07
  • For a case where the normals are not independent see 'http://math.stackexchange.com/questions/1598940'. – BruceET Jan 03 '16 at 23:20