Fixed-Rate Mortgages: Calculating Loan Balance and Payments - Prof. Steven Hackman, Assignments of Systems Engineering

Download Fixed-Rate Mortgages: Calculating Loan Balance and Payments - Prof. Steven Hackman and more Assignments Systems Engineering in PDF only on Docsity! Fixed-Rate Mortgages 1 Mortgage Basics We adopt the following continuous-time notation. Let LB(t) denote the outstanding loan bal- ance at time t. Let IP (t) and PP (t) denote the interest and principal payments on the loan made at time t. The annual loan interest rate is i, and the (remaining) life of the loan is T . All of the continuous-time formulas have discrete-time counterparts, which we include as we go. Let c(t) denote the rate of cash flow paid to the mortgage holder (i.e. the bank). With an outstanding loan balance of LB(t) at time t, in the next ∆t units of time the bank increases the loan balance by iLB(t)∆t, the interest it expects the mortgagee to pay, but will subtract the payment c(t)∆t it receives. Therefore, the change in loan balance from t to t + ∆t is LB(t + ∆t)− LB(t) = iLB(t)∆t− c(t)∆t, (1) which, as ∆t → 0, implies that d dt LB(t) = iLB(t)− c(t) (2) The solution to the differential equation (2) may be readily verified as1 LB(t) = eit{LB(0)− ∫ t 0 e−isc(s)ds}. (3) Note that we can “invert” (3) to reveal that LB(0) = e−itLB(t) + ∫ t 0 e−isc(s)ds, (4) which merely states that present value of all cash the bank receives up to time t including repayment of the outstanding loan balance at time t (which pays off the loan) must equal the original loan balance, i.e., the bank is indifferent to these two cash flow streams. A conventional fixed-rate mortgage requires the mortgagee to pay a fixed total payment of M per month for a duration of T years. In continuous-time, c(s) = M and LB(T ) = 0. Substituting these values into (4), LB(0) = ∫ T 0 e−isMds = M i (1− e−iT ) (5) 1Use the product rule of differentiation and the Fundamental Theorem of Calculus. 1 or that M = iLB(0) 1− e−iT = ( iLB(0) 1− (1 + i/12)−12T ) . (6) The value for M is, of course, equal to “Annual Equivalent” of LB(0) over 12T monthly periods with a rate of interest of i/12 per month. Both the continuous-time and discrete-time formulas (the one in parentheses) represent the annual payment; to obtain the monthly payment, simply divide by 12. After replacing c(s) = M in (3) with its solution (6), and performing a little algebra2 LB(t) = LB(0) { 1− e−i(T−t) 1− e−iT } . (7) Observe that the expression in braces equals the proportion LB(t)/LB(0) of the original loan balance remaining, and T − t represents the time (or number of periods) remaining on the loan. This is a very useful fact to remember. Remark 1. Suppose it is time t and we wish to take out a loan in the amount of LB(t) for a period of T − t at the same rate of interest i. What would be the “new” (monthly) mortgage payment Mt→T ? Intuition suggests the process is regenerative, namely, the “process starts over” and this claim is easily verified. From (6) and (7), Mt→T = iLB(t) 1− e−i(T−t) = iLB(0) 1− e−iT = M. (8) We now proceed to calculate the principal and interest payments at time t. The total payment M = IP (t) + PP (t) (9) is the sum of the principal and interest payments, and since IP (t) = iLB(t), PP (t) = M − iLB(t). (10) Since we now know how to compute M and LB(t) we can compute PP (t), too. However, we can do better. Since iLB(t) = IP (t) and c(t) = IP (t)+PP (t), it follows immediately from (2) that d dt LB(t) = −PP (t). (11) Now take the time derivative of both sides of (10) and use the identity (11) to obtain that d dt PP (t) = iPP (t), (12) or that PP (t) = PP (0)eit. (13) 2To obtain the corresponding discrete-time formula simply replace e−ix with (1+ i/12)−12x, exactly as in (6). 2