Dear R helpers
 
I am working on the credit risk default data and am referring to "An 
introduction to Credit Risk Modelling" by Christian Bluhm. The literature 
affirms that 'the default frequencies grow exponentially with decreasing credit 
worthiness'.  

I have a data of rating wise default frequencies. The idea is to fit 
exponentail of the type 
Y = a * exp( b*X ) where Y is (dependent variable) default frequency and X is 
rating class code. So e.g. suppose I have rating classes AAA, AA, A.... and so 
on coded as 1, 2, 3  etc. and suppose the observed default frequencies are  say 
0.00011, 0.0029, 0.0083 respectively, then my values of X and Y are as follows.
 
X                     Y
1                   0.0001
2                   0.0029 
3                   0.0083
..............................
..............................
 
etc and to this data I am fitting Y = a * exp(b*X)
 
My quereis are 
(1) is there any R function which will find out the estimated values of a and b 
instead of taking logaritham and converting both sides into linear equation and 
solve for a and b;
 
(2) How do I find out out Statistically that this is the best fit? e.g. had it 
been linear regression, the R^2 gives me some idea about the fitted linear 
line. In otehr words, is there any way of comparing the observed values of Y 
against the estimated values of Y i.e. values obtained from the equation Y = a 
* exp(b*X). Will the Chi-Square will be a good choice? 
 
I was trying the t-test as given below and am not sure whether I am right to do 
so. 
For each rating class, I have the observed value of Y and estimates value. Say 
for AAA, the observed value is 0.0001 whereas the estimated value is say 
0.0017. I have tested the following hypothesis 
 
Ho : Y(AAA) = 0.0017
H1 : Y(AAA) <not equal> 0.0017
 
So this is a two tailed case and I have applied the 't' test as 
 
tcal (AAA) = [Y(obse) - Y(estim)]/(s / sqrt(n)).
 
Probelm is in some cases, I have observed significant difference ie. the 
t(calculated value) falling in Critical region. Thus, the need to find out 
whetehr my fit is correct or not.
 
Thanking you in advance
 
Regards
 
Julia Cains, Brisbane
 
 
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