Hussman Strategic Advisors has a bunch of pre-interview questions for Financial Analyst / Financial Engineer position. I will try my best to answer some of those questions. I will tackle the 1st question in this post.
What is the payment on a 12‐year, $400,000 mortgage that compounds on the payment date every 73 days at 6%? (show your formula and calculations, and ignore leap year)
I do not really understand the question so here is my best guess. I am assuming I do not have to worry about individual days in a month. Also, I am assuming 6% is the annual rate.
Step 1: Calculate relationship between annual rate, 73 days rate and monthly rate.
365 Days = 12 months => 73 Days = 12*(73/365) months = 2.4 months
Let R be the annual rate in decimal (0.06 in this case); r be the monthly rate in decimal and R* is the rate every 73 days (or every 2.4 months)
To convert rates lets start with something familiar.
If a loan compounds every 6 months (half yearly) then 6 months rate is R/2 i.e. R/(12/6)
If a loan compounds every 3 months (quarterly) the 3 months rate is R/4 i.e. R/(12/3)
If a loan compounds every month then monthly rate is R / 12 i.e. R/(12/1)
So, if a loan compounds ever 2.4 months (73 days) the 73 day rate is R/(12/2.4) = R/5 = R*
Now, monthly rate r compounded for 2.4 months = R* i.e. (1+r)2.4 = 1 + R*
r = (1+R*)(1/2.4) – 1 = (1+(R/5))(1/2.4) – 1 = (1+(0.06/5))(1/2.4) – 1 = 0.004983 = 0.4983%
If the loan had compounded every month instead of 73 days then r = R/12 = 0.5%
Step 2: Derive Payment Formula
Let L be the loan amount ($400,000) and P be the payment every month
At the end of the 1st month:
Outstanding Loan Amount, L1 = L*(1+r) – P
Let (1+r) = X
L1 = L*X – P
At the end of 2nd month:
Outstanding Loan Amount, L2 = L1*X – P = L*X2 – P*X – P
At the end of 3rd month:
Outstanding Loan Amount, L3 = L2*X – P = L*X3 – P*(X2 – X – 1)
At the end of nth month:
Outstanding Loan Amount, Ln = Ln-1*X – P = L*Xn – P*(Xn-1 – Xn-2 – … 1)
Sum of Geometric Progression (GP) 1+x+x2+x3+…xn-1 = (xn – 1)/(x – 1)
Using the GP formula and replacing X by (1+r)
Ln = L*(1+r)n – P*(((1+r)n – 1) / r)
At the end of loan term (n = 144 months in this case) outstanding amount = 0
When Ln = 0 then P = L*r*(1+r)n / ((1+r)n – 1) = L*r / (1 – 1/(1+r)n)
Therefore, P = 400,000*0.004983 / (1-1/(1+0.004983)144) = 3899.39