In this exercise, we will generate a point that is uniformly distributed on the *surface* of a three dimensional sphere.

Let $X$, $Y$, and $Z$ be independent standard normal variables. Show that for any rotation $\theta$, we have that $\theta(X,Y,Z)$ has the same distribution as $(X,Y,Z)$. Hint: use the power of linear algebra.

It is true that there is only one distribution on the sphere that is rotationally-invariant. Using this fact, and the previous exercise, show that the random variable

is uniformly distributed on the sphere of radius $1$.

Code it and code a Poisson point process on the sphere of intensity $4$. Hint: recall that a sphere has surface area $4\pi r^2$.

We will also plot the simulation.

- The rotation invariance of $(X, Y, Z)$ follows from the fact that every rotation can be represented as a linear transformation $A$, where $A A^t = I$. Also recall the following fact, which we state as an exercise:

Suppose $X \sim N(\mu, V)$, where $\mu \in \mathbb{R}^n$ and $V \in \mathbb{R}^{n \times n}$ is a positive definite symmetric covariance matrix, so that $X$ has a non-degenerate multivariate normal distribution. Let $A \in \mathbb{R}^{n \times n}$ be a matrix with $\det(A) \not = 0$. Think of $X \in \mathbb{R}^{n \times 1}$ as a column vector. What is the distribution of $Y=AX$?

Note that: The pdf for $X$ is given by $$f(x) = \frac{1}{\sqrt{(2\pi)^ n |\det (V)| }} \exp[ -\tfrac{1}{2} (x-\mu)^t V^{-1} (x - \mu) ],$$ where $x \in \mathbb{R}^{n \times 1}$, and $t$ denotes matrix transposition.

Clearly, the transformation $x \to Ax$ is a bijection. Set $$W = A V A^t,$$ so that $$W^{-1} = (A^{-1})^tV^{-1} A^{-1},$$ and $$\det(W) = \det(A) \det(V) \det(A).$$

We will show that $Y \sim N(A\mu, W)$. The change of variables formula gives that the pdf for $Y$ is given by $$ \begin{eqnarray*} y &\mapsto& f(A^{-1}y ) |\det(A) ^{-1}| \\ &=& \frac{1}{\sqrt{(2\pi)^ n |\det (V)| }} \exp[-\tfrac{1}{2} (A^{-1}y -\mu)^t V^{-1} (A^{-1}y - \mu) ]|\det(A) ^{-1}| \\ &=& \frac{1}{\sqrt{(2\pi)^ n |\det (V)|\det(A)|^2 }} \exp[-\tfrac{1}{2} (y -A\mu)^t (A^{-1})^tV^{-1} A^{-1}(y - A\mu) ] \\ &=& \frac{1}{\sqrt{(2\pi)^ n |\det (W)| }} \exp[-\tfrac{1}{2} (y -A\mu)^t W^{-1}(y - A\mu) ], \\ % \end{eqnarray*} $$ as required.

With this fact in hand, note that you start with the covariance matrix given by the identity, since $X,Y$, and $Z$ are independent. Observe that for any rotation will simple

*cancel out*in the algebra and leave you with the same joint pdf.It easily follows that since the distribution of $(X,Y,Z)$ is invariant under rotations, so too is $U$, from which it follows that $U$ is uniformly distributed on the sphere.

In [1]:

```
import numpy as np
def point():
x=np.random.normal()
y=np.random.normal()
z=np.random.normal()
norm= np.sqrt(x**2 + y**2 + z**2)
p = (1/norm)*np.array([x,y,z])
return p
def psphere(L):
L = L * 4* np.pi
N = np.random.poisson(L)
if (N==0):
return None
else:
p=point()
x=np.array(p[0])
y=np.array(p[1])
z=np.array(p[2])
for i in range(N-1):
pp = point()
x=np.append(x,pp[0])
y=np.append(y,pp[1])
z=np.append(z,pp[2])
points = [x,y,z] # we store the coordinates in this way for easy plotting
return points
print(psphere(0.1))
print(psphere(0.3))
real = psphere(4)
more = psphere(100)
```

In [2]:

```
#allows for interacting graph, which we can play with using mouse
%matplotlib notebook
from mpl_toolkits import mplot3d
import matplotlib.pyplot as plt
#plt.style.use('seaborn-poster')
fig = plt.figure(figsize = (10,10))
ax = plt.axes(projection='3d')
ax.grid()
ax.scatter(more[0], more[1], more[2], c = 'r', s = 50)
ax.set_title('Poisson points on a sphere')
# Set axes label
#ax.set_xlabel('x', labelpad=20)
#ax.set_ylabel('y', labelpad=20)
#ax.set_zlabel('z', labelpad=20)
plt.show()
```