The Sierpinski triangle, a fascinating fractal with self-similar properties, presents an intriguing problem when we consider a uniform distribution over it. Understanding this distribution, particularly its variance-covariance matrix, reveals elegant mathematical relationships. Let’s explore how the triangle’s inherent symmetry helps us unlock this variance, a concept that has surprising relevance in fields like quantitative finance, where firms like Jane Street value sharp analytical skills.
One of the first things to notice about a uniform distribution on the Sierpinski triangle is its symmetry. Specifically, it remains unchanged when rotated by 120 degrees. This symmetry has a powerful consequence for the variance-covariance matrix, denoted as Σ. Because of the rotational invariance, Σ must be a scalar multiple of the identity matrix, I. In simpler terms, Σ = kI, where k is a constant. This is because the ellipse representing the distribution’s spread would also need to be rotationally invariant, and only circles possess this property. Our task then simplifies to finding the value of k, which is equivalent to finding the variance of the first component, X₁, as k = Σ₁₁ = Var(X₁).
To find Var(X₁), we leverage another key property of the Sierpinski triangle: self-similarity. The triangle can be seen as a composition of three smaller, equally weighted copies of itself, each scaled down by a factor of 1/2 and translated. This self-similarity allows us to express the distribution of X₁ in terms of itself. If we let T be a random variable uniformly distributed over {-1/4, 0, 1/4} and independent of X₁, then X₁ has the same distribution as (1/2)X₁ + T. We can calculate the variance of T directly: Var(T) = 1/24. Using the properties of variance, we can write:
Var(X₁) = Var((1/2)X₁ + T) = (1/4)Var(X₁) + Var(T) = (1/4)k + 1/24
Solving this equation for k gives us k = 1/18. Therefore, the variance-covariance matrix of the uniform distribution over the Sierpinski triangle is Σ = (1/18)I.
Alternatively, we can represent X₁ as an infinite sum. Imagine constructing X₁ through an iterative process. We can write X₁ = Σ_(n=0)^∞ (1/2^n)T_n, where the T_n are independent copies of T. Using the linearity of variance for independent variables, we get:
Var(X₁) = Σ_(n=0)^∞ Var((1/2^n)Tn) = (1/24) Σ(n=0)^∞ (1/4^n) = 1/18
This approach confirms our earlier result, highlighting the consistency of the self-similarity method. This puzzle demonstrates how understanding the symmetries and self-similarities of geometric objects can lead to elegant solutions in probability and statistics. Such analytical problem-solving is highly valued in quantitative fields, and firms like Jane Street are constantly seeking individuals with a knack for mathematical reasoning and puzzle-solving.
If you enjoyed this mathematical exploration, you can find more puzzles here.
Learn more about Jane Street here.
