Asset Allocation and Portfolio Management HW2

• Problem 1
  • Brief Proof of Block Matrix Inversion Formula
• Problem 2
• Reference

Problem 1

From

$\begin{pmatrix} P & A^T \\ A & 0 \\ \end{pmatrix} \begin{pmatrix} x^* \\ v^* \\ \end{pmatrix} = \begin{pmatrix} -q \\ b \\ \end{pmatrix} \tag{1}$

According to block matrix inversion formula, if we want to inverse $\begin{pmatrix} P & A^T \\ A & 0 \\ \end{pmatrix}$, $P$ and $0$ should be invertible which is false. So we rewrite (1) as

$\begin{pmatrix} A & 0 \\ P & A^T \\ \end{pmatrix} \begin{pmatrix} x^* \\ v^* \\ \end{pmatrix} = \begin{pmatrix} b \\ -q \\ \end{pmatrix} \tag{2}$

Then we need to inverse $\begin{pmatrix} A & 0 \\ P & A^T \\ \end{pmatrix}$, the requirement is changed to $A$ and $A^T$ should invertible which we can assume to be true. So we apply block matrix inversion formula to $\begin{pmatrix} A & 0 \\ P & A^T \\ \end{pmatrix}$ and get

$\begin{aligned} \begin{pmatrix} A & 0 \\ P & A^T \\ \end{pmatrix}^{-1} &= \begin{pmatrix} (A - 0(A^T)^{-1}P)^{-1} & - A^{-1}_{11}0(-PA^{-1}_{11}0 + (A^T))^{-1} \\ - (A^T)^{-1}P(A - 0(A^T)^{-1}P)^{-1} & (-PA^{-1}_{11}0 + (A^T))^{-1} \end{pmatrix} \\ &= \begin{pmatrix} A^{-1} & 0 \\ - (A^T)^{-1}P(A)^{-1} & (A^T)^{-1} \end{pmatrix} \\ \end{aligned}$

$\begin{aligned} (2) \Rightarrow \begin{pmatrix} x^* \\ v^* \\ \end{pmatrix} &= \begin{pmatrix} A & 0 \\ P & A^T \\ \end{pmatrix}^{-1} \begin{pmatrix} b \\ -q \\ \end{pmatrix} \\ &= \begin{pmatrix} A^{-1} & 0 \\ - (A^T)^{-1}P(A)^{-1} & (A^T)^{-1} \end{pmatrix} \begin{pmatrix} b \\ -q \\ \end{pmatrix} \\ &= \begin{pmatrix} A^{-1}b \\ - (A^T)^{-1}P(A)^{-1} b - (A^T)^{-1}q \\ \end{pmatrix} \\ \end{aligned}$

So

$x^* = A^{-1}b$

And we must assume $A$ and $A^T$ are invertible. What’s more, we know that $A$ is invertible if and only if $A^T$ is invertible. So actually we only must assume $A$ is invertible.

And if $P$ is positive semi-definite but has a non-trival null space, then since $x^*$ is only depends on $A$ and $b$, so the solutions to the original equality-constrained optimization problem are still existing and unique.

Brief Proof of Block Matrix Inversion Formula

Please see the complete proof here. [1]

$A = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \\ \end{pmatrix}, \quad B = \begin{pmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \\ \end{pmatrix}$

Let $B = A^{-1}$, then

$AB = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \\ \end{pmatrix} \begin{pmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \\ \end{pmatrix} = \begin{pmatrix} I_k & O_{k,n-k} \\ O_{n-k,k} & I_{n-k} \end{pmatrix}$

$\Rightarrow \begin{cases} A_{11}B_{11} + A_{12}B_{21} = I_{k} \\ A_{11}B_{12} + A_{12}B_{22} = O_{k,n-k}\\ A_{21}B_{11} + A_{22}B_{21} = O_{n-k,k}\\ A_{21}B_{12} + A_{22}B_{22} = I_{n-k}\\ \end{cases} \tag{1}$

$(1)\Rightarrow \begin{cases} B_{12} = - A^{-1}_{11}A_{12}B_{22} \\ B_{21} = - A_{22}^{-1}A_{21}B_{11} \\ \end{cases} \tag{2}$

$(1)\Rightarrow \begin{cases} A_{11}B_{11} - A_{12}A_{22}^{-1}A_{21}B_{11} = I_{k} \\ (A_{11} - A_{12}A_{22}^{-1}A_{21})B_{11} = I_{k} \\ -A_{21}A^{-1}_{11}A_{12}B_{22} + A_{22}B_{22} = I_{n-k}\\ (-A_{21}A^{-1}_{11}A_{12} + A_{22})B_{22} = I_{n-k}\\ \end{cases} \tag{3}$

$(3)\Rightarrow \begin{cases} B_{11} = (A_{11} - A_{12}A_{22}^{-1}A_{21})^{-1} \\ B_{22} = (-A_{21}A^{-1}_{11}A_{12} + A_{22})^{-1}\\ \end{cases} \tag{4}$

$\text{Substituting (4) in (2)} \Rightarrow \begin{cases} B_{12} = - A^{-1}_{11}A_{12}(-A_{21}A^{-1}_{11}A_{12} + A_{22})^{-1} \\ B_{21} = - A_{22}^{-1}A_{21}(A_{11} - A_{12}A_{22}^{-1}A_{21})^{-1} \\ \end{cases} \tag{5}$

$(4) \,\&\, (5) \Rightarrow A^{-1} = \begin{pmatrix} (A_{11} - A_{12}A_{22}^{-1}A_{21})^{-1} & - A^{-1}_{11}A_{12}(-A_{21}A^{-1}_{11}A_{12} + A_{22})^{-1} \\ - A_{22}^{-1}A_{21}(A_{11} - A_{12}A_{22}^{-1}A_{21})^{-1} & (-A_{21}A^{-1}_{11}A_{12} + A_{22})^{-1} \end{pmatrix}$

This formula requires $A_{11}$ and $A_{22}$ to be invertible as $A_{11}^{-1}$ and $A_{22}^{-1}$ are used in the formula.

Problem 2

The Lagrangian for the perturbed problem is

$L(x,\lambda, \nu, u, v) = f_0(x) + \sum_{i=1}^{m}\lambda_i[f_i(x) - u_i] + \sum_{j=1}^{p}\nu_j[h_j(x) - v_j]$

The Lagrange dual function is

$g(\lambda, \nu, u, v) = \inf_{x\in \mathcal{D}} L(x,\lambda, \nu, u, v)$

$\text{strong duality holds}\Rightarrow d^* = p^*$

$p^*(u,v) = \inf\{f_0(x) \mid f_i(x) - u_i \leq 0, \, h_j(x) - v_i= 0 \, (\forall i,j)\}$

$\lambda^*_i = -\frac{\partial p^*}{\partial u_i}(0,0),\quad \nu^*_j=-\frac{\partial p^*}{\partial v_i}(0,0)$

Reference

[1] http://www.math.chalmers.se/~rootzen/highdimensional/blockmatrixinverse.pdf

• The utility of a market buy order is ($\beta_2\,V - A$).

• The utility of a limit buy order is $(A - \beta_2\,V) \,{\mathbb{P} }^B_3$.

Thus, submit a MS only if $B - \beta_2\,V \geq (A - \beta_2\,V) \,{\mathbb{P} }^B_3$ or equivalently

$\beta_2 \leq \frac{B-A\,{\mathbb{P} }^B_3}{V\,(1-{\mathbb{P} }^B_3)} = \frac{B-A\,{\mathbb{P} }^B_3}{V\,(1-{\mathbb{P} }^B_3)} =\frac{160}{187}.$

Continuing in this way, we compute that a seller will choose as follows:

Action	Condition	Numerically
Marketsell(MS)	$B-\beta_2\,V \geq (A-\beta_2\,V)^+\,{\mathbb{P} }^B_{3}$	$$0 \leq \beta_2<\frac{160}{187}$$
Limitsell(LS)	$$(A-\beta_2,V),{\mathbb{P} }B_{3}>(B-\beta_2,V)+$$	$$\frac{160}{187}\leq \beta_2<\frac{12}{11}$$
Do nothing	otherwise	$$\frac{12}{11}\leq \beta_2 \leq 2$$

Thus, conditional on an empty book and the trader being a seller,

\begin{eqnarray*} {\mathbb{P} }_2^{MS}&=&{\mathbb{P} }\left(\beta_2<\frac{160}{187}\right)=\frac{80}{187}\ {\mathbb{P} }2^{0}&=&{\mathbb{P} }\left(\beta_2>\frac{12}{11}\right)=\frac{5}{11}\ {\mathbb{P} }2^{LS}&=&1-{\mathbb{P} }{MS}-{\mathbb{P} }{0}=1-\frac{80}{187}-\frac{5}{11}=\frac{2}{17}. \end{eqnarray*}