Exercise: kernel smoothing

Kernel smoothing #

Suppose that we have a bi-infinite sequence \( \{f(n)\}_{n\in\mathbb{Z}}\) of real numbers, that is, a real number for each integer \( n \in \mathbb{Z}\). In fact, consider the following one: \[ \ldots, 9,7,1,3,9,7,1,3,9,7,1,3,9,7,1,3,9,7,1,3,9,7,1,3, \ldots\] I would like to propose the following computation. We will form another bi-infinite sequence \( \{h_1(n)\}_{n\in\mathbb{Z}}\) via the formula: \[h_1(n) = \tfrac{1}{3} f(n-1) + \tfrac{1}{3} f(n) + \tfrac{1}{3} f(n+1)\] Repeating the computation, this time replacing \(f(n)\) with \(h_1(n)\), we get another bi-infinite sequence \( \{h_2(n)\}_{n\in\mathbb{Z}}\). Since we are not doing anything else, we can continue and get a sequence of bi-infinite sequences \(\{h_k\}_{k \in \mathbb{N}}\).

Question: What happens to the sequence \(\{h_k\}_{k \in \mathbb{N}}\) as \( k \rightarrow \infty\)? I am looking for an intuitive answer, no formal proof is required.

Solution

For simplicity, suppose that the sequence \(f(n)\) is periodic like above. Then one can prove that the sequences \(h_k\) tend to the constant sequence whose value is the average value of \(f\). On the other hand, if \(f(n)\) is unbounded, then the sequences \(h_k\) may not tend to a limit.

Question: Describe what happens if instead we use a slightly different process, this time letting: \[h_1(n) = \tfrac{1}{5} f(n-2)+\tfrac{1}{5} f(n-1) + \tfrac{1}{5} f(n) + \tfrac{1}{5} f(n+1) + \tfrac{1}{5} f(n+2)\] Or: \[h_1(n) = \tfrac{1}{4} f(n-1) + \tfrac{1}{2} f(n) + \tfrac{1}{4} f(n+1)\]

Solution

The limiting answer remains the same, although the rate of convergence may be different.

The above process is an example of a surprisingly useful construction. We will define and study it formally in class.