creating vectors with normal distribution of lengths

Question

First consider vectors of unit length, say on the unit sphere. Now I want to give some magnitude to these vectors and I want the magnitude to be chosen from the normal distribution. In one dimension that is easy:

RandomReal[NormalDistribution[0,x]]

However, if my vector is arbitrarily pointed I cannot take from this distribution since I cannot have a negative magnitude.

The only way I can think of giving the vector a normal distribution of lengths is by using

RandomReal[HalfNormalDistribution[y]]

which forces the randomly chosen quantity to be > 0. Does this seem like the proper procedure?

Answer 1

Technically, the normal distribution is defined on the real line, $\mathbb{R}$, but vector lengths are nonnegative numbers, elements of $\mathbb{R}_{0+}$. So you can't really have the lengths of your vectors satisfy a normal distribution. You have to choose a distribution for the lengths that has the correct domain, $\mathbb{R}_{0+}$.

One thing you can do, and my best guess at what you really want, is to use the restriction of the normal (or half-normal) distribution to the nonnegative numbers. That would be accomplished exactly like you said, by choosing the length from HalfNormalDistribution.

If the angular distribution of the vectors (the distribution of their directions) has reflection symmetry, in the sense that a vector is just as likely to point in any particular direction $\hat{n}$ as it is to point in the opposite direction $-\hat{n}$, then you actually can multiply the unit vectors you have by random numbers chosen from the full normal distribution. If the random number is negative, it will switch the direction of your vector, but since $P(\hat{n}) = P(-\hat{n})$, the probability of switching any one direction to the opposite direction is the same as the probability of switching the opposite direction to the original direction. Those two effects cancel out. If the angular distribution is not symmetric, however, then using the full normal distribution will have the effect of symmetrizing it, i.e. you'll wind up with a new distribution of directions $P'$ such that $P'(\hat{n}) = \frac{1}{2}\bigl(P(\hat{n}) + P(-\hat{n})\bigr)$.

Answer 2

Theorem. This works pretty well:

randomVector := RandomVariate[NormalDistribution[], 3];

Proof. Create a large data set and display it for the fun of it.

points = Table[randomVector, {100000}];
Graphics3D[{PointSize[0], Point[points]}]

Statistics time! Histogram plus Maxwell distribution etc,

norms = Norm /@ points;
hlist = HistogramList[norms];
Show[
    Histogram[norms],
    Plot[\[Pi] Max[hlist[[2]]] r^2 PDF[NormalDistribution[], r], {r, 0, Max[hlist[[1]]]}, PlotStyle -> ColorData[1][2]]
]

Looks pretty good to me. $\square$

(David Zaslavsky's comment is of course still valid, thete's no normal distribution on $\mathbb R^+$.)