The following problem links some of the probability and statistics we\'ve covere
ID: 3365708 • Letter: T
Question
The following problem links some of the probability and statistics we've covered with interest rates 1. You are to offer a loan of $100 to potential borrowers who will repay the loan in 1 year from now. If a borrower has a probability p of repaying the loan and q = 1-p of not repaying the loan, then what annually compounded interest rate must you charge to make your expected repayment equal to the original $100 that was borrowed from you. (Assume that the risk-free rate is zero.) a. b. Assuming you charge at a rate r and the borrower has the same profile as in part a), so that you are payed $100(1 + r) at time t = 1 year. What is the variance of your repayment? (Again, assume the risk-free rate is zero.) Now assume that you lend to N borrowers who each has the same rate of default as in part a). At what rate should you lend so that your expected repayment is equal to $100N (the total dollar amount your loaned)? C. What is the percentage variance of the repayment in part c) as a function of the number of borrowers, N? d. Assume N = 1000 (which is large enough so that you can model the distribution of number of defaults as a normal distribution.) Assume that p 0.90. At what rate should you lend so that 95% of the time you recover 100% of the total amount of dollars loaned? e.Explanation / Answer
(a) Let P be the principal amount and r be the interest rate to be charged. So the borrower repays an amount P x (1+r/100) with probability p and repays 0 with probability q.
Expected repayment value = p x P x (1+r/100) + q x 0
This should be equal to the principal amount as required.
Equating, P = p x P x (1+r/100) (A)
Simplifying, 1+r/100 = 1/p
Further simifying, r = 100 x (1-p) / p
(b) This is a binomial distribution with mean 100p. So the variance is 100npq.
In this case, n = 1, so the variance of repayment = 100pq
(c) For the case of N borrowers, n = N. So the equation (A) mentioned in part (a) holds with P replaced by NP
Since N cancels out from both sides, we have the same result for r = 100 x (1-p) / p.
(d) Variance = 100Npq, Mean = 100Np
‰ Variance = q%
(e) Being a normal distribution, the revised equation as mentioned in (A) is as follows :
0.95xNxP = pxNxPx(1+r/100)
0.95 = p x (1+r/100)
1+ r/100 = 0.95/p
r = 100 x (0.95/p-1)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.