CAE Working Paper # Failure of a Credit System: Implications of the Large Deviation Theory. Rabi Bhattacharya and Mukul Majumdar.

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1 ISS CAE Working Paper #11-01 Failure of a Credit System: Implications of the Large Deviation Theory by Rabi Bhattacharya and ukul ajumdar February 011

2 Failure of a Credit System: Implications of the Large Deviation Theory Rabi Bhattacharya and ukul ajumdar April Introduction Suppose each of A = /A customers n 1 receives a loan of A dollars from a bank, and the probability is p that the customer will return dollars to the bank at the end of the year R > 1, and the probability is p that he or she will default, returning no money to the bank 0 < p < 1. Assume also that apart from the dollars the bank lends each year, it has a backup asset of dollars per year 0. We will say that the bank fails at the end of T years if T S j T T, 1 j=1 where S j is the number of customers in the jth period returning dollars to the bank. We assume S j, 1 j T, are independent, as are the A customers in each period. Then S j has the binomial distribution Binom A, p, 1 j T, and T j=1 S j is also Binom A T, p. The probability of bank failure at the end of period T is then T T QT P S j, j=1 Department of athematics, University of Arizona Department of Economics, Cornell University 1

3 which may also be represented as A T T P Y i, 3 where Y i, 1 i A T, are i.i.d., with PY i = 1 = p, and PY i = 0 = 1 p. ote that EY i = p i. Consider two cases. Case I: p < R 1, Case II : p > 1 R. Case I. In this case, writing δ = R 1 p, δ > 0, we may express 3 as AT Q I T = P Y i A T p + δ 4 By the theory of large deviations see, e.g., Bhattacharya and Waymire 007, Theorem 4.8, pp , Q I T = e λ AT 1 + o1 as A T T A where, writing mh = Ee hy i = pe h + p, one has, 5 λ = c p + δ, c x := sup{xh lnmh}. 6 h R Clearly, c x 0. Also, for 0 < x < 1, xh lnmh as h ±. Hence c x may be obtained by solving for h the equation 0 = d peh {xh lnmh} = x dh mh, or, peh = p x { } or h = ln 1 px p1 x. Then and λ = c p + δ = 1 R p = Bln p c x = xln { } px ln p x { p ln p B ln B p B, p = px x. 7 p, 8 x } { } /R p ln /R /R

4 B := 1 R. 9 Thus the probability that the bank does not fail at the end of period T is e λt /A 1 + o1, which goes to zero exponentially fast as T /A. One may, in this case prove the stronger result that Q I T = {e λ T /A}1 + o1, 10 Q I T := Pthe bank fails at some period t, 1 t T. 11 Remark 1. ote that λ in 9 does not depend on A. Thus the exponent λt /A decreases as A increases, so that e λt /A increases as A increases. This shows that, in Case I, the probability of bank failure increases as A increases. That is, with the same capital outlay of dollars per year, the same probability p of default by a customer, and the same expected revenue Rp per year i.e., the same interest rate R 1 charged to a customer, the probability of bank failure rises as the amount of loan per customer rises. One may think of this as the effect of higher volatility, since var S j = R A A p p = R p pa although ES j = R p is not affected by A. Remark. ote that d dx c x = ln { } px, 0 < x < 1 1 p x Hence d/dxc x > 0 if px > p x, and d/dxc x < 0 if px < p x. Since pp + δ > p p δ, 1 λ = c p + δ c decreases as R increases. 13 R In other words, the chance of bank failure decreases as R or the interest rate increases - a rather obvious conclusion, but with a precise calculation of the rates. Case II. Assume now that p > R 1. Then one may rewrite 3 as A T T Q II T P Z i, 14 where Z i = Y i, 1 i A T, are i.i.d., PZ i = 1 = p, PZ i = 0 = p, EZ i = p, mh = Ee hz i = pe h + p. One now has 1 = p δ,δ := p R R 3

5 Then, by the large deviation principle, AT Q II T P Z i A T p + δ = e λat 1 + o1 16 where = e λt /A 1 + o1, as A T, [ ] λ = c p + δ, c x := sup h xh ln pe h + p. 17 By symmetry, or by direct calculation as in 7, 8, one may show that, in this case, { } p λ c p δ p + δ = p δln ln p p p + δ p + δ { } p B p B p = Bln ln = Bln Bln 18, p B B p B B : = 1. R Remark 3. Since λ in 18 does not involve A, it follow that the exponentially small probability of bank failure, as given by 16, increases as A increases showing the effect of volatility. Also, as in Remark, if R increases then the probability of bank failure decreases, since the revenue grows given that p remains the same. The relation 18, however, refines this obvious fact. A numerical illustration. Case II. = 1000, T = 5, p = 0.9, R = 1. a A = 10 [ A = 100]. Then λ = 1 1. ln ln 0. = Q II T e = e = b A = 100 [ A = 10]. Then Q II T e =

6 The calculation in b for the approximate probability of ruin is better done using the central limit theorem, rather than large deviations. For in this case the ormal approximation to the probability is PZ > 1.58 = 0.057, where Z is a standard ormal random variable. In Case II, the probability of bank failure before or in period T is for = 0 Q I T = PBank failure occurs at the end of period 1 + PFirst failure occurs at the end of period + + PFirst failure occurs at the end of period T e λ A 1 + o1 + e λ A 1 + o1 + + e T λ A 1 + o1 = e λ A + e λ A + + e T λ A + oe λ A e λ A 19 On the other hand, obviously, Q I T PBank fails at the end of period 1 = e λ A 1 + o1. 0 It follows from 19 and 0 that Q I T = e λ A 1 + o1 1 We consider next the more realistic model in which the probability p depends on the state θ of nature. Given the state θ that obtains, the customers behave independently with regard to loan repayment, with a common probability p θ of repayment. The distribution of customers is thus exchangeable. It is also assumed that the sequence θ n : n 1 of values of θ. For simplicity, let θ have two possible values θ = a 1 e.g., normal rainfall and θ = a drought. Let πa i = Probθ = a i, i = 1,. Assume p a1 > 1 R, pa < 1 R. In one period i.e., T = 1, the probability of bank failure is Q1 = πa 1 P By the preceding see 6,10, and 16, 18, S 1 θ = a i. Q1 = πa 1 e λ a 1 /A 1 + o1 + πa e λ a T /A 1 + o1, 3 5

7 where, with B = 1 R as in 10, one has see 10 and 18. B pa1 λ a1 = Bln Bln p a1 B pa λ θ = Bln pa p a B ln B B For the case T =, the corresponding failure probability is Q = πa i πa j P S 1 + S i, j=1 = πa 1 e λ a 1 A + πa 1 πa P,. 4 θ 1 = a i,θ = a j 1 + o1 + πa [ e λ a A S 1 + S 1 + o1] θ 1 = a i,θ = a j. The last summand on the right side in 5 may be expressed as πa 1 πa A y=0 A y p y a 1 p a1 A y P S 5 m θ = a. 6 For asymptotics, one may use a number of approximations to 6 or the last term in 5. pa1 +p Consider two cases. First, suppose a > R 1. Then, by Bernstein s inequality, we can show that PS 1 + S θ 1 = a 1,θ = a is exponentially small, namely, Oexp c A for some positive constant c. In this case, Q = πa + 0e c A 7 for some constant c > 0 Thus Q is essentially πa. Secondly, suppose < R 1. Then, again by Bernstein s inequality, one can show that, pa1 +p a P S 1 + S θ 1 = a 1,θ = a = δ A, where δ A 0 exponentially fast with A. In this case, for same positive constant c. Q = πa + πa 1 πa + oe c A 8 6

8 In the general case of T periods T > 1, one may express the failure probability as T QT = πa T + π T 1 a πa 1 1 T m P S S T θ 1 = a 1,θ i = a for i T T + π T a π a 1 T m P S S T θ 1 = a 1,θ = a 1,θ i = a for 3 i T T r + + P S S T π T r a π r a 1 T m θ i = a 1, for 1 i r, θ i = a for r + 1 i T + + π T a 1 T m P S S T θ i = a 1, for 1 i T. 9 Assume, for simplicity, that rp a1 + T rp a does not equal T for any r. Again we consider several cases. Suppose r, 0 r T 1, is the largest integer such that, Then case r : rp a1 + T rp a < QT = r j=0 T r = 0,1,,T T π j a 1 π T j a + o1 for r = 0,1,,T j The error o1 is of the order exp c r A, where c r > 0 can be estimated using Bernstein s inequality. ote c r is increasing in r. 7