### Solution

Overview: the self-symmetry of the Sierpinski triangle leads to a linear equation that the covariance matrix \(\Sigma\) must satisfy; specifically, \(\Sigma = \tfrac{1}{4}\Sigma + \tfrac{1}{24}I\), which gives \(\fbox{\(\Sigma = \tfrac{1}{18}I\)}\).

Without rigorously defining the uniform distribution over the Sierpinski triangle, we can at least recognize that there are certain properties it should have. The first relevant property is that it's invariant under rotation by 120°; it follows that \(\Sigma = kI\) for some \(k\). (Consider: The ellipse \(\mathbf{x}^*\Sigma^{-1} \mathbf{x} = 1\) will be invariant under this rotation, and the only ellipses invariant under this rotation are circles.) We have \(k = \Sigma_{11} = \V(X_1)\), so we have reduced the problem to finding \(\V(X_1)\).

The next relevant property of the distribution is that it's equal to a mixture of 3 equally weighted copies of itself, with each copy scaled by 1/2 and translated by \((1/4,\text{something}),(-1/4,\text{something})\), or \((0,\text{something})\). (The somethings are respectively \(-\sqrt{3}/12,-\sqrt{3}/12,\sqrt{3}/6\), but we don't need them.) In particular, if \(T\) is uniform over \(\{-1/4,0,1/4\}\) (and independent of \(X_1\)), then \(X_1\) has the same distribution as \(\tfrac{1}{2}X_1 + T\). We can directly compute \(\V(T) = 1/24\), and we then have \[ \begin{align} k = \V(X_1) &= \V(\tfrac{1}{2}X_1 + T)\\ &= \tfrac{1}{4}\V(X_1) + \V(T)\\ &= \tfrac{1}{4}k + 1/24, \end{align} \] and solving for \(k\) now yields \(k = 1/18\), so \(\Sigma = \tfrac{1}{18}I\).

A similar approach is to write \(X_1 = \sum_{n=0}^\infty \frac{1}{2^n}T_n\) where the \(T_n\) are independent copies of \(T\). Then \(\V(X_1) = \sum_{n=0}^\infty \V(\frac{1}{2^n}T_n) = \frac{1}{24} \sum_{n=0}^\infty \frac{1}{4^n} = 1/18\).

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