# surface unit sphere

Suppose $\sum_i c_i = 0$ and $Z$ is random draw from the collection of unit vectors, i.e. it has Euclidean norm $\lVert Z \rVert = 1$. Of course $E\left[\sum_i c_iZ_i^2 \right]=0$. What can we say about $P\left(\sum_i c_i Z_i^2 > 0\right)$? I don't think this probability is always one half. Can anybody help to find the probability?

$\it{Example}:$ Consider $c_1 = 1$, $c_2 = 1$ and $c_3 = -2$. Then, we are interested in $P\left(Z_1^2 + Z_2^2 - 2 Z_3^2 > 0\right)$. It is not difficult to see that $\left\{z:z_1^2 + z_2^2 - 2 z_3^2 = 0, z_1^2 + z_2^2 + z_3^2 =1\right\} = \left\{z: z_1^2 + z_2^2 = \frac{2}{3}, z_3 =±\frac{1}{3}\sqrt{3}\right\}$

such that the set $\left\{z_1^2 + z_2^2 - 2 z_3^2 = 0, \lVert z\rVert=1\right\}$ consists of two circles with centres $(0,0,±\frac{1}{3}\sqrt{3})$, radius $\sqrt{\frac{2}{3}}$, and normal vector pointing in the $z_3$ direction. For this example, the question is then to find the surface of the unit sphere enclosed within the two circles, i.e. $P\left(Z_1^2 + Z_2^2 - 2 Z_3^2 > 0\right)=P\left(-\frac{1}{3}\sqrt{3}< Z_3<\frac{1}{3}\sqrt{3}\right)$.

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What do you mean by $Z^2$? From the context it appears that you intend $Z\cdot Z$, but that would always yield a value of 1. –  Tpofofn Nov 26 '12 at 11:39
@Tpofofn: Where do you see $Z^2$? –  joriki Nov 26 '12 at 11:41
@Sander: You can get norm bars with proper spacing using \lVert and \rVert. –  joriki Nov 26 '12 at 11:41
@joriki: The norm bars are improved now. –  Sander Nov 26 '12 at 11:48
You seem to know the way to calculate the probability in terms of the surface area, or equivalently, the solid angle. Then what is your questioning point? –  Sangchul Lee Nov 26 '12 at 12:11

The general case only requires slight modification, as your example contains all the key ingredient for the calculation. Indeed, let $c=-\min\{c_1,c_2,c_3\}$ and consider the following equivalent inequality:

$$(c+c_1)Z_1^2 + (c+c_2)Z_2^2 + (c+c_3)Z_3^2 > c.$$

Assuming without loss of generality that $c_1 > c_2 > c_3$, we have $c=-c_3$ and the inequality above yields the region

$$(c_1-c_3)Z_1^2 + (c_2-c_3)Z_2^2 > c,$$

The outside of an elliptical cylinder. Thus the problem reduces to find the area of the corresponding surface region on the sphere. Of course it may extremely hard (or even impossible) to find a closed form for the surface integral, the problem becomes simple when $c_1 = c_2$.

Honestly, I haven't tried evaluating the integral for the general case yet as it's midnight here and I have to go to bed. When I'm awake I will try it again and supplement this answer if there is something new.

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Just some wild thoughts: As $\sum_i c_i Z_i^2>0 \Leftrightarrow \sum_i c_i (rZ_i)^2>0$ for any $r>0$, your problem is equivalent to the evaluation of $P\left(Q:=\sum_i c_i X_i^2>0\right)$, where the $X_i$s are i.i.d. standard Normal. So it can be further reduced to $P\left(\sum_{c_j>0} c_j X_j^2>\sum_{c_k<0} c_k X_k^2\right)$. In other words, you are comparing two weighted sum of chi-square distributions. There are a few papers on numerical computation of the CDF of weighted sum of chi-square distributions (this paper, in particular, seems to spell out the CDF of the previously mentioned $Q$ directly). I hope they can help.

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