Question: Testing the Distribution Let X be a random variate with density f and let a1 < a2 < ... < al define a partition of
Testing the Distribution Let X be a random variate with density f and let a1 < a2 < ... < al define a partition of the support of f into subintervals, including the unbounded intervals xal. Recall from (B1.1), (B1.2) that the probability of a realization of X falling into ak ≤ x
b) Let τj be the exponentially distributed jump instances of a Poisson experiment, see Section 1.9 and Property 1.20e. How should the jump intensity λ be chosen such that the expectation of the Δτ is Δt? Implement and test the algorithm, and visualize the results. Experiment with several values of the jump intensity λ. Exercise 2.18 Spectral Decomposition of a Covariance Matrix For symmetric positive definite n × n matrices Σ there exists a set of orthonormal eigenvectors v(1),...,v(n) and eigenvalues λ1 ≥ ... ≥ λn > 0 such that Σv(j) = λjv(j) , j = 1, . . . , n. Arrange the n eigenvector columns into the n×n matrix B := (v(1),...,v(n)), and the eigenvalues into the diagonal matrices Λ := diag(λ1,...,λn) and Λ1 2 := diag(√λ1,..., √λn).
a) Show ΣB = BΛ.
b) Show that A := BΛ1 factorizes Σ in the sense Σ = AAtr .
c) Show AZ = n j=1 λj Zj v(j)
d) And the reversal of Section 2.3.3 holds: For a random vector X ∼ N (0, Σ) the transformed random vector A−1X has uncorrelated components: Show Cov(A−1X) = I and Cov(B−1X) = Λ.
e) For the 2 × 2 matrix Σ = 5 1 1 10 calculate the Cholesky decomposition and BΛ1 2 . Hint: The above is the essence of the principal component analysis. Here Σ represents a covariance matrix or a correlation matrix. (For an example see Figure 2.10.) The matrix B and the eigenvalues in Λ reveal the structure of the data. B defines a linear transformation of the data to a rectangular coordinate system, and the eigenvalues λj measure the corresponding variances. In case λk+1 λk for some index k, the sum in
c) can be truncated after the kth term in order to reduce the dimension. The computation of B and Λ (and hence A) is costly, but a dominating λ1 allows for a simple approximation of v(1) by the power method.
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