Stock Market Information and the Merton Structural Model Assignment PLEASE follow all instructions and requirements in the word docdata needed is uploaded

Stock Market Information and the Merton Structural Model Assignment PLEASE follow all instructions and requirements in the word docdata needed is uploaded for (Mathematica)Sample as well uploaded for guidance Computational Finance Using Excel and Mathematica
© Problem Set 4: Merton Model of Asset Value and Its Volatility Using Equity Market Valuation
Merton Model of Asset, Debt and Equity Valuation
In this problem, worth 20 points, the Merton model of Asset, Debt and Equity Valuation will be
employed to estimate a firm’s:
1. market asset values (as opposed to book asset values),
2. the market implied volatility of asset returns,
3. risk premium on its debt,
4. solvency ratio,
5. distance to default (as defined by KMV), and
6. probability of insolvency computed assuming V is a lognormal variable.
Data that is needed will come from the firm’s balance sheet and includes:
1.
2.
3.
4.
5.
The book value of assets as of the last annual report to start the estimation,
The sum of debt of the company separated into secured debt and subordinated debt,
The risk free interest rates for 1, 2 and 3 years’
The market capitalization of the firm near the lasts annual report, and
An estimate of the volatility over 1 year of the firm’s stock returns (implied volatility can be
used for a 1 year call option).
Choose two firms: one that seems to be near bankruptcy and another that is not.
Computational Requirements
Use Mathematica to perform your calculations. Two approaches will be used:
1. The Mathematica function of FindRoot or Nsolve using the two equations in the Merton Model
as shown below, and
2. Newtons’s iterative method as shown in the Matlab and Mathematica coding added to
BlackBoard.
Equations
The underlying assets are assumed to be stochastic and generated by an Ito process in continuous
time. Consistent with this model, equity value of a firm can be considered as a call option on its assets
with a strike price being its total promised debt, B, is:
E = VN (d1 ) ? (B ) exp (? R f T )N (d 2 )
(1)
where
d1
?
ln (V ) + (R
B
=
f
)
+ 0.5? V2 T
?
?V T
and d 2 = d1 ? ? V T
and
E=
the market value of equity (stock price times number of shares outstanding,
V=
the market value of assets,
B=
the promised value of bank liabilities discounted at the risk-free rate to time T,
Rf =
the risk-free rate with a maturity consistent with the time to asset valuation (bank examination),
?=
the time to expiration of the option and time to maturity of the debt,
?V =
the standard deviation (volatility) of the rate of return on assets,
ln(x) = the natural logarithm of x,
exp(x) = the value e raised to the power of x, and
N(x) = the cumulative standard normal distribution.
Our objective is to estimate two parameters of the contingent claims model of pricing: the market
value of assets, V, and the volatility of asset returns, ?V. To solve for two variables a second equation is
necessary. Ronn and Verma (1986) and Hull (2000, p.630-631) show that by applying Ito’s lemma to
the generating process for the value of assets, the following relationship with observable market value of
equity and its volatility can be used as the second equation in our system:
?EE = N(d1)?VV, and by rearranging, ? V = E? E
VN (d1 )
.
(2)
where ?E is the volatility of the return on equity as computed from the market value of equity and all
other variables are defined as above. Equation (2) shows that asset volatility is derived from leverage
weighted observable volatility of equity. However, it needs to be emphasized that d1 has the asset value
and asset volatility as arguments (see equation (1)).
Nonlinear equations (1) and (2) are used to solve for the implied values of V and ?V via an
iterative process such as that of Newton.
Also,
Write 3-4 pages analysis/summary based on the results and findings
Using Stock Market Information and the Merton Structural Model to
Estimate the Probability of Insolvency for Banking Companies
Background
The objective of this study is to estimate two separate parameters that will allow different, but
related, estimates of the likelihood of banking company insolvency and failure. The method used is the
structural model of Merton (1974) and Black and Cox (1973) in a number of formulations that allows
estimation of intermediate parameters that can then be used to estimate the probability of insolvency.
The Merton model shows that the value of a firm’s equity can be expressed in terms of an European
call option on the assets of the firm owned by the shareholders with a strike value of the promised
value of the liabilities of the firm. In the Merton model of the valuation of the equity and debt of the
firm, the firm is assumed to have liabilities that are zero coupon debt with a single maturity. The
underlying assets are assumed to be stochastic and generated by an Ito process in continuous time.
Consistent with this model, equity value of a firm can be considered as a call option on its assets with a
strike price being its total promised debt, B, is:
E = VN (d1 ) ? (B ) exp (? R f T )N (d 2 )
(1)
where
d1
?ln (V B)+ (R
=
f
)
+ 0.5? V2 T
?
?V T
and d 2 = d1 ? ? V T
and
E=
the market value of equity (stock price times number of shares outstanding,
V=
the market value of assets,
B=
the promised value of bank liabilities discounted at the risk-free rate to time T,
Rf =
the risk-free rate with a maturity consistent with the time to asset valuation (bank examination),
?=
the time to expiration of the option and time to maturity of the debt,
?V =
the standard deviation (volatility) of the rate of return on assets,
ln(x) = the natural logarithm of x,
exp(x) = the value e raised to the power of x, and
N(x) = the cumulative standard normal distribution.
Our objective is to estimate two parameters of the contingent claims model of pricing: the
market value of assets, V, and the volatility of asset returns, ?V. To solve for two variables a second
equation is necessary. Ronn and Verma (1986) and Hull (2000, p.630-631) show that by applying Ito’s
lemma to the generating process for the value of assets, the following relationship with observable
market value of equity and its volatility can be used as the second equation in our system:
E? E
?EE = N(d1)?VV, and by rearranging, ? V =
VN (d1 )
.
(2)
where ?E is the volatility of the return on equity as computed from the market value of equity and all
other variables are defined as above. Equation (2) shows that asset volatility is derived from leverage
weighted observable volatility of equity. However, it needs to be emphasized that d1 has the asset value
and asset volatility as arguments (see equation (1)).
Nonlinear equations (1) and (2) are used to solve for the implied values of V and ?V via an
iterative process such as that of Newton. An algorithm to solve nonlinear equations based on Newton’s
method with numeric derivatives is presented in the Appendix. This algorithm is programmed in
Matlab and Visual Basic for Applications (VBA) in Excel. Included are sample results to benchmark
programming the process in SAS.
2
Using the Bank Holding Company and Bank Database to Estimate the Parameters
Monthly Balance Sheet Data from Quarterly Balance Sheet Data
The data for the BHC database and the bank database has previously computed variables that
can be used to compute these estimates. However, to increase the usefulness of these estimates, they
need to be computed on a monthly basis and not a quarterly basis. Since the BHC and bank Y-9C and
Call Reports on quarterly, monthly balance sheet data will need to be computed using a linear
interpolation of successive end of quarter values. An example is given below where September and
December data are known and values for October and November are to be computed.
Month
September (Vt)
October (Vt+1)
November (Vt+2)
December (Vt+3)
Known
Values
100
?
?
130
Interpolated
Difference
((Vt+3 – Vt )/3)
10
10
Interpolated
Values
100
110
120
130
The formula for the interpolation is: Vt+1 = Vt + ((Vt+3 – Vt )/3) and Vt+2 = Vt+1 + ((Vt+3 – Vt )/3).
The balance sheet variables that will be used in the models that will need interpolation are:
Total Liabilities,
Total Assets,
Total Deposits,
Subordinated Debt,
Total Deposits + Fed Funds Purchased + All Other Debt Securities of the BHC or Bank
Stock Return Volatility (?E)
This variable may need to be taken from the weekly file for each company. In this case, the
observation for the last week in the month should be used. This variable should not be the result of
interpolation of the standard deviations of returns on the quarterly files.
3
Risk Free Rate Rf
This variable is from the Federal Reserve Board H.15 releases for the last week in each month.
The data are for the 1-year constant maturity Treasury bill yield to maturity. The data are the same for
each banking company and a file will be available on the shared drive in the Hanweck folder and with
a file name of T-bill1-yr monthly.xls.
Example Graphical Results For the Bank of New England Before Failure
The monthly value of probability of insolvency is shown in Figure 1. These results demonstrate
that after December 1987, the Bank of New England had substantially increased its likelihood of
insolvency.
Figure 1
Probability
1
Bank of New England
Probability of Insolvency
(monthly)
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Aug-90
Jul-90
Jun-90
May-90
Apr-90
Mar-90
Feb-90
Jan-90
Dec-89
Nov-89
Oct-89
Sep-89
Aug-89
Jul-89
Jun-89
May-89
Apr-89
Mar-89
Feb-89
Jan-89
Dec-88
Nov-88
Oct-88
Sep-88
Aug-88
Jul-88
Jun-88
May-88
Apr-88
Mar-88
Feb-88
Jan-88
Dec-87
Nov-87
Oct-87
Sep-87
Aug-87
Jul-87
Jun-87
May-87
Apr-87
Mar-87
Feb-87
Jan-87
Dec-86
Nov-86
Oct-86
Sep-86
Aug-86
Jul-86
Jun-86
May-86
Apr-86
Mar-86
*Bank of New England failed as of January 30, 1991. Its probability of insolvency one year before was between 0.96 and 0.99.
Date
4
The distance to default for the Bank of New England is shown in Figure 2. These data confirm
that, after December 1987, the Bank of New England experienced a significant deterioration in its
distance to default and, by implication, its probability of insolvency. A negative value shows severe
deteerioration.
Figure 2
Bank of New England
Distance to Default
Distance to Default
(monthly)
2.5
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-2.5
-3
-3.5
-4
-4.5
-5
Aug-90
Jul-90
Jun-90
May-90
Apr-90
Mar-90
Feb-90
Jan-90
Dec-89
Nov-89
Oct-89
Sep-89
Aug-89
Jul-89
Jun-89
May-89
Apr-89
Mar-89
Feb-89
Jan-89
Dec-88
Nov-88
Oct-88
Sep-88
Aug-88
Jul-88
Jun-88
May-88
Apr-88
Mar-88
Feb-88
Jan-88
Dec-87
Nov-87
Oct-87
Sep-87
Aug-87
Jul-87
Jun-87
May-87
Apr-87
Mar-87
Feb-87
Jan-87
Dec-86
Nov-86
Oct-86
Sep-86
Aug-86
Jul-86
Jun-86
May-86
Apr-86
Mar-86
*Bank of New England failed as of January 30, 1991. Its distance to default one year before was between -1.7 and -2.5.
Date
5
Appendix
Algorithm to solve the two nonlinear equations in Matlab coding.
% Newton Iteration 2D
% clear, clg, fprintf(‘n’)
% Bank of America 9/30/98
% v = 552500;
% v = 242954;
% v = 594673;
% sigma = .05;
% prem = 0.012061;
% dep = 345756;
% ltdebt = 180345;
% rf = 0.0463;
% xt = 1.0;
% stkval = 53963.01;
% stkvol = 0.25;
% Bank of New England 12/29/1989 (1 year before failure). Values in $millions
v = 24147.6601; %Initial guess for V from balance sheet
sigma = .025; %Initial guess for sigmaV from nothing
prem = 0.047885;
dep = 24487.13;
ltdebt = 253.1;
rf = 0.0825;
xt = 1.0;
stkval = 650.288;
% stkvol = 1.1362;
stkvol = .531332;
% nrows=size(data,1);
% ncols=size(data,2);
% for n2=228:nrows
% v=data(n2,4)/1000;
% sigma=0.05;
% prem=data(n2,13)/100;
% dep=(data(n2,5)-data(n2,6))/1000;
% ltdebt=data(n2,6)/1000;
% rf=data(n2,14)/100;
xt=0.5;
% stkval=data(n2,11);
% stkvol=data(n2,12);
n1=0;
% for xt = .25:.25:2.25
dxv = 1000;
dysigma = 0.001;
dxt=.01;
n1=n1+1;
% xv = input(‘Initial guess for v? ‘);
% ysigma = input(‘Initial Guess for sigma? ‘);
xv = v;
ysigma = sigma;
zxt = xt;
6
%BEGIN ITERATIVE SOLUTION
for n=1:200
s = [xv,ysigma];
xp = xv+dxv;
yp = ysigma+dysigma;
% xtp = zxt+dxt;
% if((xtp < 0) | (xtp > 2))
%
zxt = xt+dxt;
%
xtp=zxt;
% end;
% j() is a matrix of first derivatives, numerically evaluated.
j(1,1) = (f_stkx2d(xp,ysigma,prem,dep,ltdebt,rf,xt,stkval)- f_stkx2d(xv,ysigma,prem,dep,ltdebt,rf,xt,stkval))/dxv;
j(1,2) = (f_stkx2d(xv,yp,prem,dep,ltdebt,rf,xt,stkval) – f_stkx2d(xv,ysigma,prem,dep,ltdebt,rf,xt,stkval))/dysigma;
% Using the interest rate spread or premium as the second equation.
% j(2,1) = (f_prem2d(xp,ysigma,prem,dep,ltdebt,rf,xt,stkval) – f_prem2d(xv,ysigma,prem,dep,ltdebt,rf,xt,stkval))/dxv;
% j(2,2) = (f_prem2d(xv,yp,prem,dep,ltdebt,rf,xt,stkval) – f_prem2d(xv,ysigma,prem,dep,ltdebt,rf,xt,stkval))/dysigma;
% Using volatility of stock returns as the second equation.
j(2,1) = (f_stkv2D(xp,ysigma,dep,ltdebt,rf,xt,stkval,stkvol) – f_stkv2D(xv,ysigma,dep,ltdebt,rf,xt,stkval,stkvol))/dxv;
j(2,2) = (f_stkv2D(xv,yp,dep,ltdebt,rf,xt,stkval,stkvol) – f_stkv2D(xv,ysigma,dep,ltdebt,rf,xt,stkval,stkvol))/dysigma;
% j(1,3) = (f_stkx2d(xv,ysigma,prem,dep,ltdebt,rf,xtp,stkval)- f_stkx2d(xv,ysigma,prem,dep,ltdebt,rf,zxt,stkval))/dxt;
% j(2,3) = (f_prem2d(xv,ysigma,prem,dep,ltdebt,rf,xtp,stkval) – f_prem2d(xv,ysigma,prem,dep,ltdebt,rf,zxt,stkval))/dxt;
% j(3,3) = (f_stkv2d(xv,ysigma,dep,ltdebt,rf,xtp,stkval,stkvol)- f_stkv2d(xv,ysigma,dep,ltdebt,rf,zxt,stkval,stkvol))/dxt;
% j(3,1) = (f_stkv2d(xp,ysigma,dep,ltdebt,rf,zxt,stkval,stkvol)- f_stkv2d(xv,ysigma,dep,ltdebt,rf,zxt,stkval,stkvol))/dxv;
% j(3,2) = (f_stkv2d(xv,yp,dep,ltdebt,rf,zxt,stkval,stkvol) – f_stkv2d(xv,ysigma,dep,ltdebt,rf,zxt,stkval,stkvol))/dysigma;
f(1) = f_stkx2d(xv,ysigma,prem,dep,ltdebt,rf,zxt,stkval);
% f(2) = f_prem2d(xv,ysigma,prem,dep,ltdebt,rf,zxt,stkval);
f(2) = f_stkv2d(xv,ysigma,dep,ltdebt,rf,zxt,stkval,stkvol);
ds = -jf’;
xv = xv + ds(1);
ysigma = ysigma + ds(2);
% zxt = zxt + ds(3);
% fprintf(‘ n=%2.0f, xv=%12.5f, ysigma=%12.5f, xt=%12.5f’, n,xv,ysigma,xt)
% if((zxt < 0) | (zxt > 2))
% zxt=xt;
% end;
% if (~isreal(xv) | ~isreal(xp) | ~isreal(ysigma) | ~isreal(yp)), break;
% end
if (abs(f(1)) < 1.0e-9 & abs(f(2)) Purchase answer to see full attachment