Define in some probability space the random variables $X_2,X_3,\ldots$, where each $X_n$ r.v. is uniformly distributed in the $n-$th unitary sphere
$$ B_n(0,1)=\{ x\in \mathbb{R}^n: \vert \vert x \vert \vert ^2 \leq 1\}$$
where $\vert \vert x \vert \vert$ is the $\mathbb{R}^n$ Euclidean norm. By definition of uniform distribution, is $A$ is a measurable subset of $B_n(0,1)$ then $\mathbb{P}(X_n \in A) = \operatorname{vol}(A)/\operatorname{vol}(B(0,1))$, where $\operatorname{vol}$ is the $n-$dimensional Lebesgue measure.
Let $R_n=\vert \vert X_n \vert \vert$. Show that $R_n \to 1$ in probability when $n \to \infty$.
This is how I have tried to understand this problem:
- As you sample your variables you are increasing the dimension of the sphere and the previous samples belong to a subset $A$ such that $\mathbb{P}(A)=0$ in $B_n(0,1)$ since $A$ will have at least one ''flat'' component.
- To prove that $R_n \to 1$ in probability is equivalent as to show that with probability $1$ you will find the $X_n$ r.v. in the surface of the $n-$sphere. (or for every $\epsilon>0$, close enough to the surface of the sphere)
2.1) This is where it gets confusing, as if the samples are uniformly distributed in $B_n(0,1)$ it makes no sense to me to prove that the r.v should fall on the surface. In particular, for every $\epsilon>0$ you can define the subset $A_{\epsilon}$ as the ball $B_n(0, 1-\epsilon)$ and then $\mathbb{P}(X_n \in A_\epsilon) \neq 0$.
- Besides changing the dimension of the unit ball, I don't see how $n\to \infty$ affects the outcome of the r.v. $X_n$.
Any guidance is highly appreciated. Maybe (possibly) I am not understanding the problem correctly.
Best Answer
$$\begin{align}P(|R_n - 1| < \epsilon) &= P(|\|X_n\| - 1| < \epsilon) \\ &= P( 1 - \epsilon < \|X_n\| < 1+\epsilon) \\ &= P(\|X_n\| > 1-\epsilon) \\ &= 1 - \frac{\text{Vol}(B_n(0, 1-\epsilon))}{\text{Vol}(B_n(0, 1)}. \end{align}$$
Can you show that $\lim_{n \to \infty} \frac{\text{Vol}(B_n(0, 1-\epsilon))}{\text{Vol}(B_n(0, 1)} = 0$?