J. Stat. Appl. Pro. Lett. 2, No. 2, 97-103 (2015) 97 Journal of Statistics Applications & Probability Letters An International Journal http://dx.doi.org/10.12785/jsapl/020201 A Merton Model of Credit Risk with Jumps Hoang Thi Phuong Thao1,∗ and Vuong Quan Hoang2 1 Hanoi University of Sciences, Vietnam National University, Vietnam 2 Centre Emile Bernheim, Université Libre de Bruxelles, Belgium Received: 13 Oct. 2014, Revised: 25 Nov. 2014, Accepted: 29 Nov. 2014 Published online: 1 May 2015 Abstract: In this note we consider a Merton model for default risk, where the firm's value is driven by a Brownian motion and a compound Poisson process. Keywords: Merton model, default risk, default probability, processes with jumps 1 Introduction Various models of Merton's type for credit risk have been studied so far (refer [1] to [8]). This paper aims to our recent results, where a model driven by a jump process is studied in [9] and another model governed by a jumps-diffusion is investigated in [10]. Suppose that the asset value Vt of a company, under a risk neutral measure, is given by the following differential equation dVt = (r−β λ )Vtdt +σVtdWt +Vt−dQt , (1.1) where Wt is a standard Brownian motion, Q(t) = ∑ N(t) i=1 Yi is a compound Poisson process, N(t) is a Poisson process with intensity λ > 0, Yi's are independent and identically distributed random variables with E(Yi) = β . All of these processes are supposed to be considered under the risk neutral measure. In (1.1), r is the interest rate, σ > 0 is a constant and Nt expresses the number of jumps of Qt while Yi is the i-th jump size of Q(t). The model (1.1) reflects a fact that, the firm's value can change randomly not only in a continuous way but also in a cumulatively discrete fashion. We will study on the probability of default of the company when its value Vt is less than some debts. 2 Case of one debt L A bankruptcy situation will occur at some time t when the company asset value is less than a debt L. And the problem is how to calculate the default probability P(Vt < L). It is known that the solution of (1.1) is given by (see [7]) Vt =V0 exp[σWt +(r−β λ − σ2 2 )t] Nt ∏ i=1 (Yi + 1). (2.1) We see that lnVt = lnV0 +σWt +(r−β λ − σ2 2 )t + Nt ∑ i=1 ln(Yi + 1). And the event {Vt < L} or {lnVt < lnL} means that σWt +Zt < xt , (2.2) ∗ Corresponding author e-mail: phuongthao09@mail.com c© 2015 NSP Natural Sciences Publishing Cor. 98 H. T. P. Thao, V. Q. Hoang: A Merton Model of Credit Risk with Jumps where xt = lnL− (r−β λ − σ2 2 )t − lnV0, (2.3) Zt = Nt ∑ i=1 Ui, with Ui = ln(1+Yi). (2.4) Zt is also a compound Poisson process where Ui's are i.i.d. random variables. We calculate first the characteristic function ΨZt (s) of Zt . Zt is also a compound Poisson process where Ui are i.i.d. random variables. We recall first the characteristic function ΨZt (s) of Zt : ΨZt (s) = E(e isZt ) = ∞ ∑ j=0 E ( eisZt |Nt = j ) P(Nt = j) = ∞ ∑ j=0 E ( eis(U1+...+U j) ) P(Nt = j) = ∞ ∑ j=0 ( EeisU1 ...EeisU j ) P(Nt = j) = ∞ ∑ j=0 (ψU(s)) j (λ t) j j! e−λ t = exp[λ t(ψU(s)− 1)] (2.5) where ψU (s) is the common characteristic function of Ui's. It is known also that, for a compound Poisson process as Zt we have μ(t) = EZt = λ tE(Ui) = λ tE ln(1+Yi) = λ tm; σ2(t) =VarZt = λ tE(U2i ) = λ tE[ln(1+Yi)]2 = λ tγ2, where E ln(1+Yi) = m and E[(ln(1+Yi))2] = γ2. Denote by Zt the normalization of Zt Zt = Zt − μ(t) σ(t) . And we will show that Zt has an approximately normal distribution. Indeed, according to the Taylor expansion for characteristic function ψU(s) = ∞ ∑ k=0 (is)k k! E|U |k, we can write ψU(s) = 1+ ism− γ2 2 s2 + o(s2). (2.6) Now we compute the characteristic function of Zt = 1σ(t)Zt − μ(t) σ(t) , ΨZt (s) = e −is μ(t)σ(t)ΨZt (s/σ(t)). Taking account of (2.5) and (2.6) we have ΨZt (s) = e −is μ(t)σ(t) exp[λ t(ψU(s/σ(t))− 1)] = e −is μ(t)σ(t) exp[iλ tm s σ(t) − λ tγ 2 σ(t)2 s2 2 + o( s2 t )] = e −is μ(t)σ(t) exp[isμ(t)/σ(t)− σ 2(t) σ2(t) s2 2 + o( s2 t )] = exp( −s2 2 + o( s2 t )), as t → ∞. c© 2015 NSP Natural Sciences Publishing Cor. J. Stat. Appl. Pro. Lett. 2, No. 2, 97-103 (2015) / www.naturalspublishing.com/Journals.asp 99 Then Zt ≃ N (0,1) or Zt ≃ N (μ(t),σ(t)2), where μ(t) = λ tE ln(1+Yi), σ(t) = √ λ tE[ln(1+Yi)]2 = √ λ tγ. Now we can consider σWt +Zt as a sum of two independent normal random variables for each t large enough, so it has also a normal distribution with mean μ∗(t) = μ(t) = λ tE ln(1+Yi) and variance σ∗(t) = σ2 +σ2(t) = σ2 +λ tE[ln(1+Yi)]2, where σ > 0 is a known constant as in (1.1). And P(σWt +Zt < xt)≈ Φ( xt − μ∗(t) σ∗(t) ), (2.7) where Φ(x) is the standard normal distribution function. We are now in the position to state the follow theorem. Theorem 2.1 The default probability can be approximated by Pde f ault ≈ 1 σ∗(t) √ 2π ∫ xt −∞ e−(u−μ ∗(t))2/2σ∗2(t)du, (2.8) where xt = lnL− (r−β λ −σ2/2)t − lnV0 μ∗(t) = λ tE ln(1+Yi), σ∗(t) = λ tE[ln(1+Yi)2]. 3 Case of many liabilities L1,L2, ...,Lm Now we consider the case where the company faces up numerous debts L1,L2, ...,Lm that should be paid at times t1, t2,...,tm respectively, with t1 < t2 < ... < tm = T . The company will jump into default position before the time T if and only if at one of time ti (i = 1,2, ...,m), it happens that Vti < Li. So the probability of default before T is Pde f ault(0,T ) = 1−P(Vti > Li,∀ti). Denote L = max{L1, ...,Lm} It is easy to see that for all ti(i = 1, ...,m) we have (Vti > Li)⊃ (Vti > L). Then Pde f ault(0,T )≤ 1−P(Vti > L,∀ti). (3.1) Put Xt = σWt +Zt , where, as before Zt = ∑Nti=1 Ui, Ui = ln(1+Yi). The inequality Vti > L is equivalent to Xti = σWti +Zti > lnL− lnV0 − (r−β λ − σ2 2 )ti := xti . Consider the event A = {Vti > L,∀ti}= m ⋂ i=1 {Xti > xti}. (3.2) Then Pde f ault(0,T )≤ 1−P(A). It is known that a compound Poisson process is a process of independent increments. The processes (Wt) and (Zt) are independent and both are of independent increments, so is the process Xt = σWt +Zt . Denoting by Ai the event {Xti > xi}, i = 1,2, ...,m we can see that A1 = {Xt1 > xt1}= {Xt1 −X0 > xt1}, c© 2015 NSP Natural Sciences Publishing Cor. 100 H. T. P. Thao, V. Q. Hoang: A Merton Model of Credit Risk with Jumps A2 = {Xt2 > xt2}= {Xt2 −Xt1 > xt2 −Xt1} ⊃ {Xt2 −Xt1 > xt2 − xt1}, if A1 occurs. . . . Am = {Xtm > xtm}= {Xtm −Xtm−1 > xtm −Xtm−1} ⊃ {Xtm −Xtm−1 > xtm − xm−1}, if A1, ...Am−1 occur. Put Bi = {Xti −Xti−1 > xti − xti−1} for i = 1,2, ...,m and x0 = 0 by convention. It follows that m ⋂ i=1 Bi ⊂ m ⋂ i=1 Ai = A. Because of the independence of increments we have P(A)≥ P( m ⋂ i=1 Bi) = m ∏ i=1 P(Bi), (3.3) And by definition of Bi, P(Bi) = P(Xti −Xti−1 > xti − xti−1) = P(σ(Wti −Wti−1)+ (Zti −Zti−1)> xti − xti−1). (3.4) Put X i = Xti −Xti−1 , W i = σ(Wti −Wti−1) and Zi = Zti −Zti−1 , where Zt is defined as in (2.4). The random variable W i has normal distribution N (0,σ2(ti − ti−1)). The random variable Zi = ∑ Nti k=Nti−1+1 Uk has the same distribution with that of ∑ Nti−ti−1 k=1 Uk since Ui's are i.i.d and Nt is a process of stationary and independent increments. We can see that the distribution of Zi is given by FZi(z) = P(Zi ≤ z) = ∞ ∑ n=0 P(Nti−ti−1 = n)P(Zi ≤ z/Nti−ti−1 = n) = ∞ ∑ n=0 λ n(ti − ti−1)n n! e−λ (ti−ti−1)P(Zi ≤ z/Nti−ti−1 = n) = ∞ ∑ n=1 λ n(ti − ti−1)n n! e−λ (ti−ti−1)P( n ∑ k=1 Ui ≤ z) = ∞ ∑ n=1 λ n(ti − ti−1)n n! e−λ (ti−ti−1)F∗nU (z), (3.5) where F∗nU is the n fold convolution of common distribution of U ′ ks. Suppose now that Ui's are continuous random variables, so are Zi's and Zi's. Then the density function of X i =W i +Zi is fX i(x) = fW i ⋆ fZi(x) = ∫ ∞ −∞ fW i(x− z) fZi(z)dz = 1 σ √ 2π(ti − ti−1) ∫ ∞ −∞ exp [ − (x− z) 2 2σ2(ti − ti−1) ] fZi(z)dz, (3.6) where fZi(z) = d dz FZi(z) is the density function of Zi. Now we have P(Bi) = 1− ∫ xti−xti−1 −∞ fX i(x)dx, where fX i(x) is defined by (3.6). And so, the following assertion is ready to be stated: c© 2015 NSP Natural Sciences Publishing Cor. J. Stat. Appl. Pro. Lett. 2, No. 2, 97-103 (2015) / www.naturalspublishing.com/Journals.asp 101 Theorem 3.1 If Ui's are continuous random variables then the probability of default before T is estimated by Pde f ault(0,T )≤ 1− m ∏ i=1 ( 1− ∫ xti−xti−1 −∞ [ 1 σ √ 2π(ti− ti−1) × × ∫ ∞ −∞ exp [ − (x− z) 2 2σ2(ti − ti−1) ] fZi(z)dz ] dx ) , (3.8) where xti = lnL− lnV0 − (r−β λ − σ2 2 )ti (3.9) and fZi(z) = ∞ ∑ n=0 d dz λ n(ti − ti−1)n n! e−λ (ti−ti−1)P(Zi ≤ z/Nti−ti−1 = n). (3.10) 4 Particular cases of Theorem 3.1 We consider some particular cases for distribution of Uk's. 4.1. Case of normal random variables Suppose that U =Uk ∼ N (0,1) then we have ∑nk=1 Uk ∼ N (0,n) with density function 1√2πn e −z2/2n and the density of Zi is fZi(z) = 1√ 2πn ∞ ∑ n=1 λ n(ti − ti−1)n n! e−λ (ti−ti−1)e−z 2/2n (4.1) From (3.8) and (4.1) we have Pde f ault(0,T )≤ 1− m ∏ i=1 ( 1− ∞ ∑ n=1 λ n n! (ti − ti−1)ne−λ (ti−ti−1) 1 2πσ √ n(ti − ti−1) × × ∫ xti−xti−1 −∞ ∫ ∞ −∞ exp [ − (x− z) 2 2σ2(ti − ti−1) − z 2 2n ] dzdx ) . (4.2) 4.2. Case of exponential random variable Uk with parameter ν > 0 We know that if Uk ∼ exp(ν) then ∑nk=1 Uk ∼ Gamma(n,ν) with the density function zn−1e−z/ν νnΓ (n) , where Γ is Gamma function. Then fZi(z) = ∞ ∑ n=1 λ n(ti − ti−1)n n! e−λ (ti−ti−1) zn−1e−z/ν νnΓ (n) . We can see the estimation in (3.8): Pde f ault(0,T )≤ 1− m ∏ i=1 ( 1− ∫ xti−xti−1 −∞ 1 σ √ 2π(ti − ti−1) ∫ ∞ 0 exp [ − (x− z) 2 2σ2(ti − ti−1) ] × × ∞ ∑ n=1 λ n(ti − ti−1)n n! e−λ (ti−ti−1) zn−1e−z/ν νnΓ (n) ]dzdx ) = 1− m ∏ i=1 ( 1− ∞ ∑ n=1 λ n n! (ti − ti−1)ne−λ (ti−ti−1) 1 σ √ 2π(ti − ti−1) × × ∫ xti−xti−1 −∞ ∫ ∞ 0 exp [ − (x− z) 2 2σ2(ti − ti−1) − z ν ] zn−1 νnΓ (n) dzdx ) . (4.3) c© 2015 NSP Natural Sciences Publishing Cor. 102 H. T. P. Thao, V. Q. Hoang: A Merton Model of Credit Risk with Jumps 5 When U =Uk's are general discrete random variables In this case we have P(Zi = z) = P( Nti−ti−1 ∑ k=1 Uk = z) = ∞ ∑ n=1 P(Nti−ti−1 = n)P( Nti−ti−1 ∑ k=1 Uk = z/Nti−ti−1 = n) = ∞ ∑ n=1 P(Nti−ti−1 = n)P( n ∑ k=1 Uk = z) = ∞ ∑ n=1 λ n(ti − ti−1)n n! e−λ (ti−ti−1)P( n ∑ k=1 Uk = z). (5.1) Denote by L the set of all possible values of Zi ≡d ∑ Nti−ti−1 k=1 Uk. So that P(X i < x) = P(σW i +Zi < x) = ∑ z∈L P(σW i < x− z)P(Zi = z) = ∑ z∈L ∞ ∑ n=1 ∫ x−z −∞ 1 σ √ 2π(ti − ti−1) exp[− u 2 2σ2(ti − ti−1 )]× × λ n(ti − ti−1)n n! e−λ (ti−ti−1)P( n ∑ k=1 Uk = z)du. (5.2) The default probability in this case is estimated by Pde f ault(0,T )≤ 1− m ∏ i=1 ( 1− ∑ z∈L ∞ ∑ n=1 1 σ √ 2π(ti − ti−1) ∫ xti−xti−1 −∞ exp[− (x− z) 2 2σ2(ti − ti−1) ]dx× × λ n(ti − ti−1)n n! e−λ (ti−ti−1)P( n ∑ k=1 Uk = z) ) . (5.3) 6 U is Poisson random variable with parameter β > 0 If U =Uk ∼ Poisson(β ) then n ∑ k=1 Uk ∼ Poisson(nβ ) with mass probability pz = P( n ∑ k=1 Uk = z) = e −nβ (nβ )z z! , z = 0,1,2, ... Then Pde f ault(0,T )≤ 1− m ∏ i=1 ( 1− ∞ ∑ z=0 ∞ ∑ n=1 1 σ √ 2π(ti − ti−1) × × ∫ xti−xti−1 −∞ exp[− x 2 2σ2(ti − ti−1) − nβ ]dxλ n(ti − ti−1)n n! e−λ (ti−ti−1) (nβ )z z! ) = 1− m ∏ i=1 ( 1− ∞ ∑ z=0 ∞ ∑ n=1 λ n(ti − ti−1)n n! (nβ )z z! 1 σ √ 2π(ti − ti−1) × × ∫ xti−xti−1 −∞ exp[− x 2 2σ2(ti − ti−1) − nβ ]dx ) . 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