Who is Dongxu?

I am living in New York now and I watch a lot movies and I used to run a weekly movie night.

I have a more detailed résumé if you’re into that sort of thing.

I am into mathematics, coding and finance.

• I have some collections of pet projects:
  • Endogenous stock market participation and wealth accumulation: a life-cycle model perspective. Paper Link
  • Portfolio simulation project based on the data from yahoo finance. Still under developement
  • Reduce dimensionality of US Treasury dataset. link
  • Two approaches to price binary options. link
  • Thoughts on stock market. link
  • Risk Monitor Program Structure for Trading Strategy. Still under improvememt
  • Review of two C++ projects on image editing and service simulation.
    • image editing
    • service simulation
  • A dive into the banks balance sheet.
    • Asset Part
    • Liability Part

Solved a mathematical problem I found interesting:

• Qustion:
  • \(0 < t_1 < t_2 < t_3 < ... < t_n\), show that n by n matrix \(\Sigma\) with components \(\Sigma_{i,j} = min(t_i,t_j)\) is a positive definite matrix.

  • Proof: The definition of positive definite: for any vector \(x \in R^n\) we have \(x^T \Sigma x > 0\). To begin with, define the brownian motion \(B_{t}\), then \(B_{t}\) follows a normal distibution with mean 0 and variance of t.

    From the memoryless property of brownian motion, we have \(Cov(B_{t_{i}},B_{t_{j}})=min(t_{i},t_{j})\) , then we cound write the \(\Sigma\) as \(Cov(B,B)\), where B is a n by 1 vector:

    \[\begin{array}{cccccccc} B & = & \left[\begin{array}{c} B_{t_{1}}\\ B_{t_{2}}\\ \vdots\\ B_{t_{n}} \end{array}\right] & = & \left[\begin{array}{ccccc} 1\\ 1 & 1\\ 1 & 1 & 1\\ \vdots & \vdots & \vdots & \ddots\\ 1 & 1 & 1 & \cdots & 1 \end{array}\right] & \left[\begin{array}{c} B_{t_{1}}\\ B_{t_{2}}-B_{t_{1}}\\ B_{t_{3}}-B_{t_{2}}\\ \vdots\\ B_{t_{n}}-B_{t_{n-1}} \end{array}\right] & = & A\tilde{B}\end{array}\]

    Where:

    \[\begin{array}{c} A\end{array}=\left[\begin{array}{ccccc} 1\\ 1 & 1\\ 1 & 1 & 1\\ \vdots & \vdots & \vdots & \ddots\\ 1 & 1 & 1 & \cdots & 1 \end{array}\right],\tilde{B}=\left[\begin{array}{c} B_{t_{1}}\\ B_{t_{2}}-B_{t_{1}}\\ B_{t_{3}}-B_{t_{2}}\\ \vdots\\ B_{t_{n}}-B_{t_{n-1}} \end{array}\right]\]

    So we could rewrite \(Cov(B,B)\) as:

    \[\Sigma=Cov(B,B)=Cov(A\tilde{B},A\tilde{B})=ACov(\tilde{B},\tilde{B})A^{T}\]

    From the independent property of Brownian motion, \(B_{t_{1}}, B_{t_{2}}-B_{t_{1}},B_{t_{3}}-B_{t_{2}}, ... , B_{t_{n}}-B_{t_{n-1}}\) are multually independent, then:

    \[\Sigma=ACov(\tilde{B},\tilde{B})A^{T}=A\cdot diag\{t_{1},t_{2}-t_{1},t_{3}-t_{2},...,t_{n}-t_{n-1}\}\cdot A^{T}\]

    Finally for any vector \(x\in R^{n}\), we have:

    \[x^{T}\Sigma x=x^{T}(A\cdot diag\{t_{1},t_{2}-t_{1},t_{3}-t_{2},...,t_{n}-t_{n-1}\}\cdot A^{T})x\]

    Which is:

    \[x^{T}\Sigma x=(A^{T}x)^{T}\cdot diag\{t_{1},t_{2}-t_{1},t_{3}-t_{2},...,t_{n}-t_{n-1}\}\cdot(A^{T}x)\]

    We could calculate:

    \[A^{T}x=\left[\begin{array}{c} x_{1}+x_{2}+...+x_{n}\\ x_{2}+...+x_{n}\\ \vdots\\ x_{n} \end{array}\right]\]

    To further simplify:

    \[x^{T}\Sigma x=\sum_{k=1}^{n}(t_{k+1}-t_{k})(\sum_{j=k}^{n}x_{k})^{2}\]

    Obviously for any vector \(x\in R^{n}\), we have: \(x^{T}\Sigma x>0\). Hence the matrix \(\Sigma\) is positive definite.