How to solve simultaneous equations?

Hello, I would like to solve simultaneously the following system of 2 equations for S and T: (fyi, these equations are an application of the Merton model for credit risk)

The first equation is:

P = N(d1) - (1-P)*exp(r*T)*N(d2)

where N(d1) and N(d2) are the cumulative normal function (cdfnorm) as in Black & Scholes with

d1 = (-ln(1-P)-(r-S^2/2)*T)/(S*sqrt(T))

d2 = d1 - S*sqrt(T)

The second equation is:

E = S*N(d1)/P

Is there a way in matlab to find values for S and T (volatility and maturity) that are consistent with the parameter P, E, and r (respectively, the leverage, the volatility of equity and the interest rate) that are known parameters (for instance, P=0.7, E=0.3, and r=0.03)?

Thank you very much for your help!

[bym]

do you have the symbolic toolbox and/or statistic toolbox

Answer 1

[bym]

I tried

P = .7; E = .3; r = .03;

syms S T real

d1 = (-1*log(1-P)-(r-S^2/2)*T)/(S*sqrt(T));
d2 = d1 - S*sqrt(T);

N1 = 1/2*(1+erf(d1/sqrt(2))); %your N(d1)
N2 = 1/2*(1+erf(d2/sqrt(2))); %your N(d2)

sol = solve(S*N1/(N1-(1-P)*exp(r*T)*N2)-E);

with the following result

Warning: Explicit solution could not be found.

Looks like it needs to be solved numerically w/ fzero. Don't have time tonight, maybe tomorrow

[andreas patsikos]

i have the same problem!!!i want EDF???

DO u know the matlab code for EDF?

Answer 2

[Walter Roberson]

Using Andrew's code as a base, I found an expression for S in terms of T, but solving for T in terms of S was not feasible in the time I spent.

I am currently running a plot over T from 0 to 1000, and it is taking a fair bit of time. The expression for S pretty much has to be solved numerically. Ummm, long enough that I quit out of the plot and now am seeing if I can find a more tractable expression for S.

[Walter Roberson]

I worked with the symbolic solution, but the amount of time it takes to come up with any given answer doesn't make it feasible.

You are better off taking S*N1/(N1-(1-P)*exp(r*T)*N2)-E and substituting your T and asking for a numeric solution with a guess of S=0.2 . For sufficiently large T, S rises approximately linearly.

Question: does a negative S make sense? For a range of maturities (up to roughly 10 with the given parameters) there are two real solutions, one negative and one positive.

[Walter Roberson]

Follow up: with the given parameters, it appears for any given negative S below a certain value, there is a corresponding T such that the pair form a solution. The T is close to being linear in log(S) for sufficiently large (magnitude) of S. With the given parameters, the boundary limit for existence is somewhere between -1/100 (exists) and -1/1000000 (does not)

Answer 3

[Fred]

Looks more complicated than I thought. To answer your question, a negative S would not make sense as it represents a volatility. A typical value would be between 0.1 to 0.3 (but it will always be lower than E which is 0.3 in our example).

In order to assure a starting point for which the numerical method will quickly yield a solution, would it be possible to solve the equations separately for S for a few fixed values of T? Then we can use the intersection of the solution curves S(T)as a starting point for the system of equations.

[Walter Roberson]

It turns out that as the time gets longer, more significant digits are required in order to get an answer that is not numeric nonsense. In some cases I needed more than 50 digits. (I might have found a transformation that reduces that requirement.)

What time scales are being considered?

For the particular example parameters given earlier, here are the S values corresponding to T=1:60 :

[.2072590928183311, .2044393299409702, .2016129607336800, .1989365188764057, .1965116411993847, .1943628414410152, .1924755755786901, .1908221440403525, .1893731280492593, .1881015672429913, .1869840994556714, .1860009411324904, .1851354816011146, .1843737955734661, .1837041854608264, .1831167873386968, .1826032442154253, .1821564396382651, .1817702819421725, .1814395297312950, .1811596504570243, .1809267054062888, .1807372557388538, .1805882853256427, .1804771370389273, .1804014598566101, .1803591647003469, .1803483873624612, .1803674572161160, .1804148706684868, .1804892685245436, .1805894165925309, .1807141889913439, .1808625537224179, .1810335601503601, .1812263281018650, .1814400383449575, .1816739242529594, .1819272644918681, .1821993765976739, .1824896113327941, .1827973477292435, .1831219887411948, .1834629574418072, .1838196937091338, .1841916513539489, .1845782956487980, .1849791012227362, .1853935502903118, .1858211311865575, .1862613371822442, .1867136655555605, .1871776168978373, .1876526946320406, .1881384047236041, .1886342555638410, .1891397580067379, .1896544255404400, .1901777745752472, .1907093248304763]

This is almost quadratic looking.

I decided to have a look to see how sensitive the solution is to the other parameters. The first one I looked at was P, but for whatever reason, the symbolic package I was using was unable to isolate S for a symbolic P.

The second one I tried was "r" -- and it turns out the solution is very sensitive to r.

Below, I have used NaN to indicate there is no (real-valued) solution.

Over r=0:0.0025:0.1, with T=25, S is

[.2367869077192663, .2334774001956742, .2299868419834384, .2263010809075183, .2224039543449438, .2182768445411721, .2138980991245670, .2092422631355814, .2042790415947689, .1989718669475625, .1932758698934157, .1871349179656345, .1804771370389273, .1732078401278894, .1651977495076567, .1562619958566648, .1461191355663390, .1343004223448158, .1199064674372540, .1006878125253988, 0.3419789628568295e-1, NaN, 0.8831092249680447e-3, 0.5260340496896748e-3, 0.1079583955841704e-2, 0.1788956732959545e-2, 0.2287174855937050e-2, 0.4388587665304212e-2, 0.1379940252933243e-2, 0.6399063398417710e-3, 0.2419694982600019e-2, 0.1031180824683200e-2, 0.1583211803026529e-2, 0.4178098067335098e-2, 0.2577486078312967e-2, 0.1556143097211814e-2, 0.5175808895800029e-2, 0.4113080725843936e-2, 0.6998707813988716e-2, 0.5442013674273168e-2, 0.4215142482463095e-2]

Zeroing in, over r=0.0475:.0001:0.0550, S is

[.1006878125253988, 0.9974259341536431e-1, 0.9877724005446151e-1, 0.9779052860862733e-1, 0.9678110540419658e-1, 0.9574746677162544e-1, 0.9468793467747240e-1, 0.9360062701619107e-1, 0.9248342105753623e-1, 0.9133390801740799e-1, 0.9014933596584605e-1, 0.8892653718916388e-1, 0.8766183449599609e-1, 0.8635091848826506e-1, 0.8498868397759661e-1, 0.8356900758188733e-1, 0.8208443837911595e-1, 0.8052575607154917e-1, 0.7888131990197861e-1, 0.7713607271755982e-1, 0.7526994658544382e-1, 0.7325516120267416e-1, 0.7105129925951827e-1, 0.6859540577945885e-1, 0.6577910566449783e-1, 0.3419789628568295e-1, 0.3893049485913096e-1, NaN, NaN, NaN, NaN, NaN, 0.1843937661456162e-3, NaN, 0.1275394361371588e-3, NaN, NaN, NaN, 0.2239007614006089e-3, NaN, NaN, NaN, NaN, 0.3999523364767706e-3, NaN, NaN, NaN, 0.7500544155135456e-3, NaN, 0.8716666575457631e-3, NaN, 0.7348839946741266e-3, NaN, NaN, NaN, NaN, NaN, 0.2960633693250945e-3, NaN, NaN, 0.1944568319446463e-3, NaN, NaN, 0.8734373143513995e-3, 0.4175360507625134e-3, 0.5940891524851323e-3, 0.1103532033792065e-2, 0.7104593149570572e-4, 0.2998481217286259e-3, 0.3305483919817678e-3, 0.1136433557052380e-2, 0.6861591898190689e-3, 0.1023329090028364e-2, 0.1434638051142099e-2, 0.1347607839143373e-2, 0.8831092249680447e-3]

Yes, there is a big drop from 0.0499 to 0.0500, and Yes, 0.0501 is slightly higher -- but then there is a gap of values and then a mix of values that exist and that do not.

As I was using a symbolic math package, the exact values at which problems occurred likely depended upon exactly what the decimal r value got reduced to as a fraction.

Answer 4

[J.]

Send me an e-mail and I will send you an m-file for it. [e-mail].