Bhattacharya, Rabi

Majumdar, Mukul

English

Research Papers in Economics

ISSN 1936-5098
CAE Working Paper #11-01
Failure of a Credit System: Implications of the
Large Deviation Theory
                                                   
by
Rabi Bhattacharya
and
Mukul Majumdar
February 2011
  Failure of a Credit System: Implications of the
Large Deviation Theory
Rabi Bhattacharyaand Mukul Majumdar†
April 2010
1 Introduction
 Suppose each of NA = N=A customers (n  1) receives a loan of A dollars from
a bank, and the probability is p that the customer will return RA dollars to the
bank at the end of the year (R > 1), and the probability is 1  p that he or she will
default, returning no money to the bank (0 < p < 1). Assume also that apart from
the N dollars the bank lends each year, it has a backup asset of M dollars per year
(M  0). We will say that the bank fails at the end of T years if
RA
T
å
j=1
Sj  NT  MT; (1)
where Sj is the number of customers in the jth period returning RA dollars to
the bank. We assume Sj, 1  j  T, are independent, as are the NA customers
in each period. Then Sj has the binomial distribution Binom(NA;p), 1  j  T,
and å
T
j=1Sj is also Binom(NAT;p). The probability of bank failure at the end of
period T is then
Q(T)  P
 
T
å
j=1
Sj 
(N M)T
RA
!
; (2)
Department of Mathematics, University of Arizona
†Department of Economics, Cornell University
1which may also be represented as
P
 
NAT
å
i=1
Yi 
(N M)T
RA
!
; (3)
whereYi, 1iNAT, are i.i.d., with P(Yi = 1)= p, and P(Yi = 0)=1 p. Note
that EYi = p 8i.
Consider two cases.
Case I: p < 1
R
 
1  M
N

, Case II : p > 1
R
 
1  M
N

.
Case I. In this case, writing d = 1
R
 
1  M
N

  p, d > 0, we may express (3) as
QI(T) = P
 
å
NAT
i=1 Yi
NAT
 p+d
!
(4)
By the theory of large deviations (see, e.g., Bhattacharya and Waymire (2007),
Theorem 4.8, pp. 54-55),
1 QI(T) = e lNAT (1+o(1)) as NAT 
NT
A
! ¥; (5)
where, writing m(h) = EehYi = peh+(1  p), one has
l = c(p+d); c(x) := sup
h2R
fxh lnm(h)g: (6)
Clearly, c(x)  0. Also, for 0 < jxj < 1, xh lnm(h) !  ¥ as h ! ¥. Hence
c(x) may be obtained by solving (for h) the equation
0 =
d
dh
fxh lnm(h)g = x 
peh
m(h)
; or; peh =
1  p
1 x
 (1  p) =
(1  p)x
1 x
: (7)
or h = ln
n
(1 p)x
p(1 x)
o
.
Then
c(x) = xln

(1  p)x
p(1 x)

 ln

1  p
1 x

; (8)
and
l = c(p+d) =
1
R

1 
M
N

ln
(
1  p
p

 
1  M
N

=R
1 
 
1  M
N

=R
)
 ln
(
1  p
1 
 
1  M
N

=R
)
= Bln

1  p
p

B
1 B

 ln

1  p
1 B

;
2B :=

1 
M
N

1
R
: (9)
Thus the probability that the bank does not fail at the end of period T is
e lNT=A(1+o(1)), which goes to zero exponentially fast as NT=A ! ¥. One
may, in this case prove the stronger result that
1 Q
I(T) = fe lNT=Ag(1+o(1)); (10)
Q
I (T) := P(the bank fails at some periodt; 1 t  T): (11)
Remark 1. Note that l (in (9)) does not depend on A. Thus the exponent
lNT=A decreases as A increases, so that e lNT=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 N dollars per year, the same probability 1  p
of default by a customer, and the same expected revenue NRp 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 ef-
fect of higher volatility, since var
 
RASj

= R2A2NAp(1  p)=
 
R2Np(1  p)A

(although ERASj = RNp is not affected by A).
Remark 2. Note that
d
dx
c(x) = ln

(1  p)x
p(1 x)

; (0 < x < 1) (12)
Hence(d=dx)c(x)>0if(1  p)x> p(1 x), and(d=dx)c(x)<0if(1  p)x<
p(1 x). Since (1  p)(p+d) > p(1  p d),
l = c(p+d)  c

1
R

1 
M
N

decreases as R increases: (13)
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 > 1
R
 
1  M
N

. Then one may rewrite (3) as
QII(T)  P
 
NAT
å
i=1
Zi   
(N M)T
RA
!
; (14)
where Zi =  Yi, 1  i  NAT, are i.i.d., P(Zi =  1) = p, P(Zi = 0) = 1  p,
EZi =  p, m(h) = EehZi = pe h+1  p. One now has
1
R

1 
M
N

= p d;d := p 
1
R

1 
M
N

: (15)
3Then, by the large deviation principle,
QII(T)  P
 
å
NAT
i=1 Zi
NAT
  p+d
!
= e lNAT (1+o(1)) (16)
= e lNT=A(1+o(1)), as NAT ! ¥,
where
l = c( p+d), c(x) := sup
h
h
xh ln

pe h+1  p
i
. (17)
By symmetry, or by direct calculation as in (7), (8), one may show that, in this
case,
l  c( p+d) = (p d)ln

1  p
p

p d
1  p+d

 ln
1  p
1  p+d
= Bln

1  p
p

B
1 B

 ln

1  p
1 B

= Bln

B
p

 (1 B)ln

1  p
1 B

; (18)
B : =
1
R

1 
M
N

.
Remark 3. Since l in (18) does not involve A, it follow that the (exponen-
tially small) probability of bank failure, as given by (16), increases as A increases
(showing the effect of volatility). Also, as in Remark 2, if R increases then the
probability of bank failure decreases, since the revenue grows (given that p re-
mains the same). The relation (18), however, reﬁnes this obvious fact.
A numerical illustration.
Case II. N = 1000, T = 5, p = 0:9, R = 1:2
(a) A = 10 [NA = 100]. Then
l =
1
1:2
ln

1
(1:2)(0:9)

 
2
1:2
ln

(0:1)(1:2)
0:2

= 0:02101
QII(T)  e (0:02101)500 = e 10:505 = 0:00027
(b) A = 100 [NA = 10]. Then
QII(T)  e 1:0505 = 0:35
.
4The 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
Normal approximation to the probability is
P(Z > 1:58) = 0:057;
where Z is a standard Normal random variable.
In Case II, the probability of bank failure before or in period T is (for M = 0)
Q
I(T) = P(Bank failure occurs at the end of period 1)
+P(First failure occurs at the end of period 2)
++P(First failure occurs at the end of period T)
6 e lNA(1+o(1))+e 2lNA(1+o(1))++e TlNA(1+o(1))
= e lNA +e 2lNA ++e TlNA +o(e lNA)
 e lNA
(19)
On the other hand, obviously,
Q
I(T)  P(Bank fails at the end of period 1) = e lNA(1+o(1)): (20)
It follows from (19) and (20) that
Q
I(T) = e lNA(1+o(1)) (21)
We consider next the more realistic model in which the probability p depends on
the state q of nature. Given the state q that obtains, the customers behave indepen-
dentlywithregardtoloanrepayment, withacommonprobability pq ofrepayment.
The distribution of customers is thus exchangeable. It is also assumed that the se-
quence qn : n > 1 of values of q. For simplicity, let q have two possible values
q = a1 (e.g., ’normal rainfall’) and q = a2 (’drought’). Let p(ai) = Prob(q = ai),
i = 1;2. Assume pa1 > 1
R
 
1  M
N

, pa2 < 1
R
 
1  M
N

. In one period (i.e., T = 1),
the probability of bank failure is
Q(1) =
2
å
i=1
p(a1)P

S1 6
N M
RA
jq = ai

: (22)
By the preceding (see (6),(10), and (16), (18)),
Q(1) = p(a1)e la1N=A(1+o(1))+p(a2)(1 e la2NT=A(1+o(1)); (23)
5where, with B = 1
R
 
1  M
N

as in (10), one has (see (10) and (18)).
la1 = Bln

B
pa1

 (1 B)ln

1  pa1
1 B

;
lq2 = Bln

1  pa2
pa2

B
1 B

 ln

1  pa2
1 B

: (24)
For the case T = 2, the corresponding failure probability is
Q(2) =
2
å
i;j=1
p(ai)p(aj)P

S1+S2 6
2(N M)
RA
jq1 = ai;q2 = aj

= p(a1)2e
 2la1N
A (1+o(1))+p(a2)2[1 e 
2la2N
A (1+o(1))]
+2p(a1)p(a2)P

S1+S2 6
2(N M)
RA
jq1 = ai;q2 = aj

:
(25)
The last summand on the right side in (25) may be expressed as
2p(a1)p(a2)
NA
å
y=0

NA
y

py
a1 (1  pa1)NA yP

S2 6
2(N m)
RA
jq2 = a2

: (26)
For asymptotics, one may use a number of approximations to (26) (or the last
term in (25)).
Consider two cases. First, suppose

pa1+pa2
2

> 1
R
 
1  M
N

. Then, by Bern-
stein’s inequality( ), we can show that P(S1 +S2 
2(N M)
RA jq1 = a1;q2 = a2) is
exponentially small, namely, O(exp cNA) for some positive constant c. In this
case,
Q(2) = p(a2)2+0(e c0NA) (27)
for some constant c0 > 0 Thus Q(2) is essentially p(a2)2. Secondly, suppose 
pa1+pa2
2

< 1
R
 
1  M
N

. Then, again by Bernstein’s inequality, one can show that,
P

S1+S2 
2(N M)
RA jq1 = a1;q2 = a2

= 1 dNA, where dNA ! 0 exponentially
fast with NA. In this case,
Q(2) = p(a2)2+2p(a1)p(a2)+o(e c00NA) (28)
for same positive constant c00.
6In the general case of T periods (T > 1), one may express the failure proba-
bility as
Q(T) = p(a2)T +

T
1

pT 1(a2)p(a1)
P

S1++ST 6
T(N m)
RA
jq1 = a1;qi = a2 for 2 6 i 6 T

+

T
2

pT 2(a2)p2(a1)
P

S1++ST 6
T(N m)
RA
jq1 = a1;q2 = a1;qi = a2 for 3 6 i 6 T

++

T
r

pT r(a2)pr(a1)
P

S1++ST 6
T(N m)
RA
jqi = a1; for 1 6 i 6 r; qi = a2 for r+1 6 i 6 T

++pT(a1)
P

S1++ST 6
T(N m)
RA
jqi = a1; for 1 6 i 6 T

:
(29)
Assume, for simplicity, that rpa1 +(T  r)pa2 does not equal
T(N M)
RA for any
r. Again we consider several cases. Suppose r, 0r T  1, is the largest integer
such that,
case r : rpa1 +(T  r)pa2 <
T(N M)
RA
(r = 0;1; ;T  1): (30)
Then
Q(T) =
r
å
j=0

T
j

pj(a1)pT j(a2)+o(1) for r = 0;1; ;T  1: (31)
The error o(1) is of the order exp crNA, where cr > 0 can be estimated using
Bernstein’s inequality. Note cr is increasing in r.
7

Failure of a Credit System: Implications of the Large Deviation Theory

http://www.arts.cornell.edu/econ/CAE/11.01.pdf

Failure of a Credit System: Implications of the Large Deviation Theory

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