CDF of distance to a point in 3D-sphere

probability distributionsprobability theory

Let's say we have a unit sphere and want to generate uniformly distributed points inside it. We use rejection to do so, and as such we place the sphere inside a $2$x$2$x$2$ bounding box and generate points uniformly inside of there, rejecting any points that don't end up inside the sphere.

If I'm correct then there is a .23 rejection rate.

But let's say we introduce a random variable $L$ that is the distance from the center of the sphere to a random point inside the sphere. What would the CDF of $L$ be?

Best Answer

Hint: the volume of a sphere of radius $L$ is $\frac{4}{3} \pi L^3$.

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[Math] Random number generation inside an interval based on cdf (Zipf and Exponential)

The quantile function $x = Q(q)$ maps unit interval into the domain of your distribution, which is $\mathbb{R}^+$ in your case. One can perform the truncation of this distribution to any interval measurable with your distribution. Let it be $a<=x<b$. Then the c.d.f. of the truncated distribution will be $F_{[a,b)}(x) = \chi_{[a,b)}(x) (F(x)-F(a) )/(F(b)-F(a))$, where the $\chi$ is the indicator function for the interval. It is also easy to find quantile of this truncated distribution. $Q_{[a,b)}(q) = Q(F(a) + q*(F(b)-F(a))$. This last statement gives your algorithm to generate random numbers.

Here is the simulation experiment for standard exponential truncated to interval $[0,2)$.

histogram of random sample generated by the described method

EDIT

Quantile of Zipf distribution is not available in closed form, but can be numerically computed by finding the $$Q(q) = \mathop{sup} \{ x \in \mathcal{D} : cdf(x) \leq q \}$$

Once this is implemented, you can generate random variates from the truncated Zipf distribution (albeit the method is not very efficient) as so:

lb=1; ub=6;
di = ZipfDistribution[3/2];
fa = CDF[di, lb-1]; fb = CDF[di, ub];
Q[q_] := Quantile[di, fa + q*(fb - fa)]

Histogram of truncated zipf random sample

[Math] Average distance between random points inside a cube

I take it that the $U_i$ are independent. As $\binom{n}{2}=n(n-1)/2$ I assume that you mean to investigate $$ R=\frac{2}{n(n-1)}\sum_{j<i}|U_i-U_j|. $$ The distribution of $R$ is not any of the well-known distributions. By linearity, $E[R]=E[|U_i-U_j|]$ which you can find by integration over $\text{box}\times\text{box}$, i.e. $$ E[R]=\frac{1}{\operatorname{vol}(\text{box})^2}\int_{\text{box}\times\text{box}}|x-y|\,d^3x\,d^3y. $$

Best Answer

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[Math] Random number generation inside an interval based on cdf (Zipf and Exponential)

[Math] Average distance between random points inside a cube

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