Assuming that I have means of determining and calculating the following metrics:

  1. Risk (i.e. probability*) of a default to a particular borrower as P
  2. Profit margin of X%

The profit margin is taken to mean that "irrespective of defaults that might occur, in the long run, I expect to make X% for lending to this particular (class of) borrower.

Thinking it through (from first principles):

Expectation[given loans to borrower with P% of default at a rate of R%] = X%

For the sake of simplicity, lets assume that a default implies the entire lent out capital is lost, so then:

( (100 - P)/100 ) * (1+R) = X

We then trivially, solve for R.

Somehow, I think I've missed something. Can anyone shed some light on if this is a good (correct?) way to solve for R the interest rate to charge the borrower.

Note: I am aware that I'm using a slightly different definition of risk premium from that used in textbooks.

I'm using the frequentist interpretation of probability, where P denotes the number of occurrences (defaults) in a sequence of 100 "runs".

  • $\begingroup$ I'd write either $(1-P)(1+R)=(1+X)$, either $\frac{100-P}{100}\frac{100+R}{100}=\frac{100+X}{100}$. Everything as perunages or everything as percentages. $\endgroup$ Jan 25, 2018 at 13:50
  • $\begingroup$ Why do you feel that you've missed anything? $\endgroup$
    – Lipton
    Jan 25, 2018 at 21:54
  • $\begingroup$ @Lipton I haven't done this sort of thing for a while - (I work in a different field now), so I wanted to double check with industry practitioners that my thinking through it was sound. $\endgroup$ Jan 26, 2018 at 10:49
  • $\begingroup$ @Raskolnikov do you want to submit that as an answer? $\endgroup$ Jan 26, 2018 at 10:50

1 Answer 1


I think your idea is right. But I would just stay consistent with the representation of the numbers and not mix percentages and perunages. Either write


in which case the numbers are perunages. Either you write


and all the numbers are interpreted as percentages. It's really not a fundamental distinction and of course solving using one of the equations, you can always rapidly get the values of the other by multiplying/dividing by 100.


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