On Modeling Economic Default Time: A Reduced-Form Model Approach

Abstract

In the aftermath of the global financial crisis, much attention has been paid to investigating the appropriateness of the current practice of default risk modeling in banking, finance and insurance industries. A recent empirical study by Guo et al. (Rev Deriv Res 11(3): 171–204, 2008) shows that the time difference between the economic and recorded default dates has a significant impact on recovery rate estimates. Guo et al. (http://arxiv.org/abs/1012.0843, 2011) develop a theoretical structural firm asset value model for a firm default process that embeds the distinction of these two default times. In this paper, we assume the market participants cannot observe the firm asset value directly and we develop reduced-form models for characterizing the economic and recorded default times. We derive the probability distributions of these two default times. Numerical experiments with empirical data are given to demonstrate the proposed models. Our approach helps researchers to gain a new perspective for economic and recorded defaults and is more feasible in general practice compared with current method. Our results can also contribute to the understanding of the impacts of various parameters on the economic and recorded default times.

Appendix

1.1 Appendix 1: Proof of Proposition 2

We note that Eq. (2) follows from Eq. (1) by using

$$\begin{aligned} P(\tau _r =N_{i+1}\mid \mathcal {G_{\infty }})=P(\tau _e \in (N_i, N_{i+1}] \mid \mathcal {G_{\infty }}). \end{aligned}$$

Eq. (3) follows from Eq. (1) by using

$$\begin{aligned} P(\tau _r-\tau _e > t \mid \mathcal {G_{\infty }})=\sum _{i=0}^{\infty } P(\tau _e \in (N_i, N_{i+1}-t] \mid \mathcal {G_{\infty }}). \end{aligned}$$

And Eq. (1) follows by

$$\begin{aligned} \begin{array}{l@{\quad }l@{\quad }l} &{}&{}P(\tau _e \in (N_i, N_i +t] \mid \mathcal {G_{\infty }})\\ &{}&{}\quad =\displaystyle \sum _{n_i=1}^{K-1}P(S_{N_1} \ne K, \ldots , S_{N_{i-1}} \ne K, S_{N_i} =n_i \mid \mathcal {G_{\infty }})\nonumber \\ &{}&{}\quad \quad \times P(\tau _e \in (N_i, N_i +t] \mid S_{N_i} =n_i, \mathcal {G_{\infty }})\\ &{}&{}\quad =\displaystyle \sum _{n_i=1}^{K-1}\sum _{n_{i-1}=1}^{K-1}\ldots \sum _{n_{1}=1}^{K-1}P(S_{N_1} =n_1, \ldots , S_{N_{i-1}} =n_{i-1}, S_{N_i} =n_i \mid \mathcal {G_{\infty }})\\ &{}&{}\quad \quad \times P_X(N_i, N_i +t)_{n_i, K}\exp \left\{ -\int _{N_i+t}^{N_{i+1}} \lambda _K(X_u) du\right\} \\ &{}&{}\quad =\displaystyle \sum _{n_i\!=\!1}^{K-1}\sum _{n_{i-1}\!=\!1}^{K-1}\ldots \sum _{n_{1}=1}^{K\!-\!1}P_X(N_0, N_1)_{S_0, n_1}\ldots P_X(N_{i-1}, N_i)_{n_{i\!-\!1}, n_i} P_X(N_i, N_i \!+\!t)_{n_i, K}\\ &{}&{}\quad \quad \times \exp \left\{ -\int _{N_i+t}^{N_{i+1}} \lambda _K(X_u) du\right\} \\ &{}&{}\quad =\left( \displaystyle \prod _{j=0}^{i-1} P^{**}_X(N_j, N_{j+1})\cdot P^*_X(N_i, N_i +t)\right) _{S_0, K}\exp \left\{ -\int _{N_i+t}^{N_{i+1}} \lambda _K(X_u) du\right\} . \end{array} \end{aligned}$$

1.2 Appendix 2.1: Proof of Proposition 3

Proof

Eq.s (12) and (13) are obvious and it suffices to show Eq. (11). Now we have

$$\begin{aligned} \begin{array}{l@{\quad }l@{\quad }l} &{}&{}P(\tau _e \in (N_i, N_i+t]\mid \mathcal {G_{\infty }})\\ &{}&{}\quad =\displaystyle \prod _{j=0}^{i-1}\left( m_1\exp \left[ \int _{N_j}^{N_{j+1}} \mu _1(X_u) du\right] +m_2 \exp \left[ \int _{N_j}^{N_{j+1}} \mu _2(X_u) du\right] \right) \\ &{}&{}\quad \quad \displaystyle \times \left( n_1\exp \left[ \int _{N_{i}}^{N_i+t} \mu _1(X_u) du\right] +n_2 \exp \left[ \int _{N_{i}}^{N_i+t} \mu _2(X_u) du\right] \right) \\ &{}&{}\quad \quad \displaystyle \times \exp \left[ \int _{N_{i}+t}^{N_{i+1}} p_1\mu _1(X_u)+p_2\mu _2(X_u)du\right] \\ &{}&{}\quad = \displaystyle \sum _{\mathbf{e} \in \hat{E}_i} \hat{m}(\mathbf{e})\exp \left[ \int _{N_0}^{N_{i+1}} \hat{\mu }(\mathbf{e},u)du\right] . \end{array} \end{aligned}$$

Hence

$$\begin{aligned} P(\tau _e \in (N_i, N_i+t]) =\sum _{\mathbf{e} \in \hat{E}_i} \hat{m}(\mathbf{e}) E\left( \exp \left[ \int _{N_0}^{N_{i+1}} \hat{\mu }(\mathbf{e},u)du\right] \right) . \end{aligned}$$

(15)

For a fixed $\mathbf{e} \in \hat{E}_i$, let

$$\begin{aligned} \begin{array}{l@{\quad }l@{\quad }l} R_{i+1}(\mathbf{e})&{}=&{}p_1 \mu _1+p_2\mu _2\\ R_j(\mathbf{e})&{}=&{}\mu _{e_j}, j=0,1,\ldots , i\\ w_{i+1}(\mathbf{e})&{}=&{}0\\ w_i(\mathbf{e})&{}=&{}\beta (N-t; R_{i+1}(\mathbf{e}), w_{i+1}(\mathbf{e}))\\ w_{i-1}(\mathbf{e})&{}=&{}\beta (t; R_{i}(\mathbf{e}), w_{i}(\mathbf{e}))\\ w_j(\mathbf{e})&{}=&{}\beta (N; R_{j+1}(\mathbf{e}), w_{j+1}(\mathbf{e})), j=0, 1, \ldots , i-2\\ v_{i+1}(\mathbf{e})&{}=&{}\exp [\alpha (N-t; R_{i+1}(\mathbf{e}), w_{i+1}(\mathbf{e}))]\\ v_{i}(\mathbf{e})&{}=&{}\exp [\alpha (t; R_{i}(\mathbf{e}), w_{i}(\mathbf{e}))]\\ v_{j}(\mathbf{e})&{}=&{}\exp [\alpha (N; R_{j}(\mathbf{e}), w_{j}(\mathbf{e}))], j=0, 1, \ldots , i-1.\\ \end{array} \end{aligned}$$

Then we can rewrite $\hat{\mu }(\mathbf{e}, s)$ as

$$\begin{aligned} \hat{\mu }(\mathbf{e}, s)&= 1_{\{s \in [N_i+t, N_{i+1})\}}(R_{i+1}(\mathbf{e}) X_s)+1_{\{s \in [N_i, N_i+t)\}}(R_{i}(\mathbf{e})X_s) \\&+ \sum _{j=0}^{i-1} 1_{\{s \in [N_j, N_{j+1})\}}(R_j(\mathbf{e}) X_s). \end{aligned}$$

Using the iterated expectation and Eq. (10) we obtain

$$\begin{aligned} \begin{array}{lll} &{} &{}E\left( \exp [\int _{N_0}^{N_{i+1}} \hat{\mu }(\mathbf{e},u)du]\right) \\ &{}&{}\quad =E\left( \exp [\int _{N_0}^{N_{i}+t} \hat{\mu }(\mathbf{e},u)du]E(\exp [\int _{N_{i}+t}^{N_{i+1}} R_{i+1}(\mathbf{e})X_u du] \mid \mathcal {G}_{N_i+t})\right) \\ &{}&{}\quad =v_{i+1}(\mathbf{e})E\left( \exp [\int _{N_0}^{N_{i}+t} \hat{\mu }(\mathbf{e},u)du] \exp [w_i(\mathbf{e}) X_{N_i+t}]\right) \\ &{}&{}\quad =v_{i+1}(\mathbf{e}) E\left( \exp [\int _{N_0}^{N_{i}} \hat{\mu }(\mathbf{e},u)du]E(\exp [\int _{N_{i}}^{N_{i}+t} R_{i}(\mathbf{e})X_u du+w_i(\mathbf{e}) X_{N_i+t}] \mid \mathcal {G}_{N_i})\right) \\ &{}&{}\quad =v_{i+1}(\mathbf{e})v_{i}(e) E\left( \exp [\int _{N_0}^{N_{i}} \hat{\mu }(\mathbf{e},u)du] \exp [w_{i-1}(\mathbf{e})X_{N_i}]\right) \\ &{}&{}\quad \!\!=v_{i\!\!+\!\!1}(\mathbf{e}) v_{i}(\mathbf{e}) E\left( \exp [\int _{N_0}^{N_{i\!\!-\!\!1}} \hat{\mu }(\mathbf{e},u)du]E(\exp [\int _{N_{i-1}}^{N_{i}} R_{i-1}(\mathbf{e})X_u du\!\!+\!\!w_{i-1}(\mathbf{e})\right. \\ &{}&{}\quad \left. X_{N_i}] \mid \mathcal {G}_{N_{i-1}})\right) \\ &{}&{}\quad =v_{i+1}(\mathbf{e}) v_{i}(\mathbf{e}) v_{i-1}(\mathbf{e}) E\left( \exp [\int _{N_0}^{N_{i-1}} \hat{\mu }(\mathbf{e},u)du] \exp [w_{i-2}(\mathbf{e})X_{N_{i-1}}]\right) \\ &{}&{}\quad =\left( \prod _{j=0}^{i+1} v_j(\mathbf{e})\right) \exp [\beta (N; R_0(\mathbf{e}), w_0(\mathbf{e})) X_0] \ \mathrm{(by \ iteration)} \end{array} \end{aligned}$$

Hence Eq. (11) follows. $\square$

1.3 Appendix 2.2: Proof of Proposition 4

Proof

We let

$$\begin{aligned} H_i(X_0, t):=P(\tau _e \in (N_i, N_i+t]) \end{aligned}$$

then by the proof of Proposition 3, for $i\ge 1$,

$$\begin{aligned} P(\tau _e \in (N_i,N_i+t]\mid \mathcal {F}_{N_1})&= \left( m_1\exp \left[ \int _{N_0}^{N_1}\mu _1(X_u) du\right] \right. \nonumber \\&\left. +\,m_2\exp \left[ \int _{N_0}^{N_1}\mu _2(X_u) du\right] \right) H_{i-1}(X_{N_1}, t)\nonumber \\ H_i(X_0, t)&= E\left[ \left( m_1\exp \left[ \int _{N_0}^{N_1}\mu _1(X_u) du\right] \right. \right. \nonumber \\&\left. \left. +\,m_2\exp \left[ \int _{N_0}^{N_1}\mu _2(X_u) du\right] \right) H_{i-1}(X_{N_1}, t)\right] \end{aligned}$$

(16)

By Proposition 3, we obtain that

$$\begin{aligned} H_0(x, t)=a_{0,1}\exp (b_{0,1}x)+a_{0,2}\exp (b_{0,2}x). \end{aligned}$$

Combining Eqs. (16) and (10), Proposition 4 follows. $\square$

1.4 Appendix 3

Let $\delta =18$ days, $t_i=\delta i, i=0, 1, \ldots , 10$. Let $N_i$ denote the number of firms whose time difference of economic and recorded default date is inside the interval $(t_{i-1}, t_i]$. Then the log-likelihood function is given by

$$\begin{aligned} \begin{array}{l@{\quad }l@{\quad }l} \displaystyle \mathcal {L}(\lambda _1,\lambda _2) &{}=&{} \displaystyle \sum _{i=1}^{10} N_i \left( \ln \left[ (e^{-\lambda _2 t_{i-1}}-e^{-\lambda _2 t_{i}})-e^{-(\lambda _1+\lambda _2)N}(e^{\lambda _1 t_{i-1}}-e^{\lambda _1 t_{i}})\right] \right. \\ &{}&{} \displaystyle \left. -\ln \left[ 1-e^{-(\lambda _1+\lambda _2)N} \right] \right) \end{array} \end{aligned}$$

By setting

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l@{\quad }l} \displaystyle \frac{\partial \mathcal {L}(\lambda _1,\lambda _2)}{\partial \lambda _1}=0\\ \displaystyle \frac{\partial \mathcal {L}(\lambda _1,\lambda _2)}{\partial \lambda _2}=0, \end{array} \right. \end{aligned}$$

we have two nonlinear equations for $\lambda _1$ and $\lambda _2$. Solving these equations numerically yields $\lambda _1=0.3631$ and $\lambda _2=0.0238$.