Brownian Bridge™

1. The Cholesky factorisation $A = LL^\top$ exists if and only if $A$ satisfies which condition?

A is symmetric and positive definiteA is square and invertibleA is symmetric and has positive traceA has positive diagonal entries

2. Given covariance $\Sigma = \begin{pmatrix} \sigma^2 & \rho\sigma^2 \\ \rho\sigma^2 & \sigma^2 \end{pmatrix}$ with $\sigma = 0.20$ and $\rho = 0.60$ , the Cholesky factor $L$ satisfies $L_{22} =$

0.160.120.200.08

3. To generate correlated samples $X \sim \mathcal{N}(0, \Sigma)$ using Cholesky $\Sigma = LL^\top$ , the correct procedure is:

Draw ε ~ N(0, I), then set X = LεDraw ε ~ N(0, I), then set X = L⁻¹εDraw ε ~ N(0, Σ) directly using rejection samplingDraw ε ~ N(0, I), then set X = LᵀLε

4. Compared to LU factorisation, Cholesky factorisation for an $n \times n$ SPD matrix requires approximately how many floating-point operations?

Half as many — n³/3 vs 2n³/3The same — both require n³/3Twice as many — due to the square root stepsOne third as many — n³/9

5. You call `numpy.linalg.cholesky(Sigma)` and receive `LinAlgError: Matrix is not positive definite`. The most informative diagnostic step is:

Compute the eigenvalues of Σ and identify which are ≤ 0Check if the diagonal entries of Σ are all positiveVerify that Σ is squareRerun with float32 instead of float64 for higher precision

6. A yield curve bootstrapping problem has been factorised as $PA = LU$ . A trader requests 200 different rate scenarios, each requiring a new right-hand side $b_1, \ldots, b_{200}$ . What is the correct computational strategy?

Factorise PA = LU once, then solve each scenario with two triangular sweeps costing O(n²) eachFactorise PA = LU for each scenario independently, as the matrix A changes per scenarioCompute A⁻¹ once and multiply each bᵢ by A⁻¹ to get xᵢUse only backward substitution since forward substitution is not needed after LU

7. A covariance matrix estimated from $T = 30$ daily returns on $n = 100$ assets is passed to Cholesky. What should you expect?

Cholesky fails — the matrix has rank ≤ 30, so at least 70 eigenvalues are zeroCholesky succeeds — sample covariance matrices are always positive definiteCholesky fails — sample covariance matrices are never symmetricCholesky succeeds if all 100 assets have non-zero variance

8. On a rates desk, a quantitative developer proposes solving the curve bootstrapping system with `x = inv(A) @ b`. A senior quant objects. The strongest objection is:

Computing inv(A) accumulates more rounding error and costs O(n³) — the same as factorisation — so nothing is gained and stability is lostinv(A) does not exist for non-square matricesThe result will have the wrong units if A is not normalisedMatrix inversion is undefined when A has complex eigenvalues

Quiz: LU and Cholesky Factorisation

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