Normalizing the Direction Vector in Stratified Gaussian Sampling
Summary
The document concerns a covariance matrix used in a Gaussian construction for stratified Monte Carlo sampling, motivated by simulating a barrier option under Black–Scholes dynamics. The answer identifies a normalization error: the stated result assumes the direction vector has unit Euclidean norm. A vector of all ones does not meet that condition when it has more than one component, so the proposed matrix formed by subtracting its outer product from the identity is not positive semidefinite as required for a covariance matrix.
For a two-dimensional example, the response derives eigenvalues showing that the matrix has a negative eigenvalue when the vector is unnormalized. It then uses the unit-norm condition to establish the covariance factorization needed by the result. The key lesson is to normalize the stratification direction before applying the construction. The answer does not cover the full daily barrier simulation, variance reduction performance, or implementation details, and its displayed notation has some transpose inconsistencies that warrant checking against the original reference.
Key ideas
- The Gaussian stratification result assumes a direction vector with unit Euclidean norm.
- An unnormalized vector of ones can make the proposed covariance matrix non-positive-semidefinite.
- The eigenvalues in the two-dimensional case expose the failure of the unnormalized construction.
- Normalize the direction vector before applying the covariance construction.
- The response addresses the matrix condition, not the complete barrier-option simulation procedure.
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# Implementation of Stratified Sampling in Monte Carlo
# Implementation of Stratified Sampling in Monte Carlo
Background
I am trying to implement Monte Carlo Simulation with Stratified Sampling for barrier option under Black Scholes Model. I understand there is an analytic formula for this instrument and we can directly simulate the integration from time 0 to maturity because we have the distribution of stock price under this model. However, I would like to simulate it with daily step, i.e looping $S_{t_i} = S_{t_{i-1}}e^{(r-\frac{1}{2}\sigma^2)(t_i - t_{i-1})+\sigma\sqrt{(t_i - t_{i-1})}X}, X\sim N(0,1)$
Lecture notes found on google
I am trying to implement Martin Haugh's guideline. When applying the "Result 2" on page 52, we have
$\vec{a} = (1,1,...,1)^T$(column vector), Then we have $\vec{V} = w\vec{a} + MVN(\vec{0},I_m - \vec{a}\vec{a}^T)$
Question
- $I_m - \vec{a}\vec{a}^T$ is not symmetric positive semi-definite.
- Why do we have $\Sigma = I_m - \vec{a}\vec{a}^T$?
Thanks!
## Answer by ir7 (score 2, accepted)
https://quant.stackexchange.com/a/57051
Your vector $a=(1,\ldots,1)^T$ does not satisfy
$$\| a \|^2=a_1^2 + \ldots + a_m^2=1, $$
as assumed by authors in Result 2.
(Q1) For $m=2$, we see it is needed when computing the eigenvalues of $I_2 - aa^T$, that is the roots $\lambda$ of equation
$$ 0=\det \begin{pmatrix} 1-\lambda- a_1^2 & -a_1 a_2 \\ -a_1 a_2 & 1-\lambda- a_2^2 \end{pmatrix} = (1-\lambda)(1-\lambda - a_1^2 -a_2^2).$$
We get $\lambda_1 = 1$ and $\lambda_2 = 1- a_1^2 -a_2^2$, which must be non-negative (as $I_2 - aa^T$ is positive semi-definite). With your vector you would get $\lambda_2 = -1$.
(Q2) Same property, $a_1^2 +a_2^2=1$, allows for:
$$ (a^Ta)^2 = \begin{pmatrix} a_1^4+a_1^2 a_2^2 & -a_1^3 a_2 -a_1 a_2^3 \\ -a_1^3 a_2 -a_1 a_2^3 & a_1^4+a_1^2 a_2^2 \end{pmatrix}$$ $$=(a_1^2 +a_2^2)\begin{pmatrix} a_1^2 & -a_1 a_2 \\ -a_1 a_2 & a_2^2 \end{pmatrix} =a^Ta$$
This in turn gives:
$$ (I_2-a^Ta)(I_2-a^Ta)^T = (I_2-a^Ta)(I_2-aa^T) $$
$$= I_2 - a^Ta -a^Ta + (a^Ta)^2 = I_2 - a^Ta$$
So, for $\Sigma:= I_2 - a^Ta$ we have:
$$ \Sigma = \Sigma \Sigma^T$$
which provides one of the matrices respecting equality:
$$ \Sigma = CC^T.$$ That is $\Sigma$ itself: $$C= \Sigma.$$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.