Penalized Independent Factor

Abstract

We propose a penalized independent factor (PIF) method to extract independent factors via a sparse estimation. Compared to the conventional independent component analysis, each PIF only depends on a subset of the measured variables and is assumed to follow a realistic distribution. Our main theoretical result claims that the sparse loading matrix is consistent. We detail the algorithm of PIF, investigate its finite sample performance and illustrate its possible application in risk management. We implement the PIF to the daily probability of default data from 1999 to 2013. The proposed method provides good interpretation of the dynamic structure of 14 economies’ global default probability from pre-Dot Com bubble to post-Sub Prime crisis.

Appendix

1.1 Proof of Theorem 1

Proof

The explicit form of the density function $g_j$ is not required, as long as the two conditions are fulfilled. Under condition C1 and C2, Equation $\Vert \hat{B}-B\Vert =O_P(n^{-1/2}+a_n)$ is equivalent to proof that for any given $\epsilon >0$, there exist a large C s.t.

$$\begin{aligned} P\{\sup _{\Vert u\Vert =C}Q(B+\alpha _nu)<Q(B)\}\ge 1-\epsilon \end{aligned}$$

(10.11)

where Q(B) is the penalized likelihood and u is a p-by-p matrix.

Let $D_n(u)=Q(B+\alpha _nu)-Q(B)$

$I_u(B)=-E(tr(\nabla _Btr(\nabla \frac{1}{n}l(B)^{\top }u)^{\top }u))=-E(tr(\nabla _Bd_u\frac{1}{n}l(B)^{\top }u))>0$ for any $y\in R^{p*p}$ based on condition (B)

If $D_n(u)<0$ by choosing a sufficiently large C, then the proof is done.

$$\begin{aligned} D(u)&=l(B+\alpha _nu)-l(B)-n\sum \{\rho _{\lambda _n}(|B_{jk}+\alpha _nu_{jk}|)-\rho _{\lambda _n}(|B_{jk}|)\}\nonumber \\&\le l(B+\alpha _nu)-l(B)-n\sum _{B_{jk}\ne 0}\{\rho _{\lambda _n}(|B_{jk}+\alpha _nu_{jk}|)-\rho _{\lambda _n}(|B_{jk}|)\} \nonumber \\&\le \alpha _ntr(\nabla l(B)^{\top }u)+\frac{1}{2}\alpha _n^2tr(\nabla _Bd_ul(B)^{\top }u)\{1+o_P(1)\} \nonumber \\&-\sum _{B_{jk}\ne 0}[n\alpha _n\rho _{\lambda _n}^{'}(|B_{jk}|)sgn(B_{jk})u_{jk}+n\alpha _n^2\rho _{\lambda _n}^{''}(|B_{jk}|)u_{jk}^2\{1+o(1)\} \nonumber \\&\le \alpha _ntr(\nabla l(B)^{\top }u)-\frac{1}{2}n\alpha _n^2I_u(B)\{1+o_P(1)\} \nonumber \\&-\sum _{B_{jk}\ne 0}[n\alpha _n\rho _{\lambda _n}^{'}(|B_{jk}|)sgn(B_{jk})u_{jk}+n\alpha _n^2\rho _{\lambda _n}^{''}(|B_{jk}|)u_{jk}^2\{1+o(1)\} \end{aligned}$$

(10.12)

The first inequality is because $\rho _{\lambda _n}(0)=0$ and $\rho _{\lambda _n}(\beta )\ge 0$. The next inequality is Taylor expansion. Then substitute $I_u(B)$ into the equation.

Base on condition (A), $n^{-1/2}tr(\nabla l(B)^{\top }u)=O_P(1)$, thus the first term of (8) is of order $O_P(n^{1/2}\alpha _n)=O_P(n\alpha _n^2)$. By choosing a sufficiently large C, the second term dominates the first term in $\Vert u\Vert =C$.

The last term in (8) is bounded by

$$\begin{aligned} \sqrt{s}n\alpha _na_n\Vert u\Vert +n\alpha _n^2max\{\rho _{\lambda _n}^{''}(|B_{jk}|):B_{jk}\ne 0\}\Vert u\Vert ^2 \end{aligned}$$

(10.13)

The first part of (9) is dominated by the second term in (8) when choosing a sufficiently large C. The second term in (9) is also dominated by the second term in (8) as $\max \{\rho _{\lambda _n}^{''}(|B_{jk}|):B_{jk}\ne 0\}\rightarrow 0$

Proof is completed.