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Interest Rate Conventions
The notion of compounding frequency
Take three fixed income investments that all promise interest payments of 12% per year:
- Investment (1) makes one interest payment of 12% of the principal after one year
- Investment (2) pays 6% of the amount invested every 6 months
- Investment (3) makes an interest payment of 1% every month
Obviously, all of these indeed pay, in total, 12% interest per year. Does that mean that they are equivalent?
The answer is no. Assume that you invest $ 100, and that your investment horizon is 1 year, so you reinvest all interest payments until that horizon. Let’s consider each of the three investments above. How much money do you end up with after 1 year?
- Investment (1): easy... $
- Investment (2): you receive $6 in interest after 6 months. You reinvest everything (principal and the $6 interest, $106 in total) for another 6 months, and end up with: 106(1 + 6%) = 106(1.06) = 100(1.06)^2 = $112.36 after 1 year.
- Investment (3): you receive $1 in interest after 1 month, so at this point you have $101. You reinvest all of this over the next month, and end up with 101(1 + 1%) = 101(1.01) = 100(1.01)^2 = $102.01 after 2 months... And so on... you have $100(1.01)n^ after n months, and 100(1.01)^12 = $112.68 after 12 months.
So, even though all of these investments claim to pay a total interest of 12% per year, they are not equivalent. Does that mean that anybody lied to you?
No. The difference is due to something that is known as the compounding frequency. The compounding frequency, in a fixed income investment, refers to how often the interest is computed and paid (or added to the principal and reinvested). In the example above, the three investments are not equivalent because they have different compounding frequencies: annual with Investment (1), semi-annual with Investment (2) and monthly in the case of Investment (3).
This example shows that, when an interest rate is specified, this does not mean much unless the compounding frequency is specified as well. The interest rate really is the same (12%) with the three investments above, but clearly Investment (3) is more attractive. When you compare the yields on different investments, you need to make sure that they have the same compounding frequency, otherwise the yields are not comparable.
This is important in practice because, in the real world, different fixed income securities have different compounding frequencies: for example, for corporate bonds and US Treasury Bonds semi-annual compounding is normally used (since these bonds make two coupon payments per year); for Eurobonds, annual compounding is typically used; but a home mortgage may have monthly or bi-monthly payments, and then the compounding frequency will be monthly or bi- monthly. You have to be detail-oriented if you want to deal with bonds!
Compounding m times a year
The formal definition is as follows: receiving an annual rate equal to r, compounded m times a year, means receiving an interest equal to r/m (times the amount invested) m times a year.
For example, with all three investments in the above example, r = 12% = 0.12, but m equals 1 with Investment (1), 2 with Investment (2) and 12 with Investment (3).
By generalizing the computations we did above, it is easy to see that your payoff per $ invested (given that the interest is reinvested) is equal to: ( 1 + r m
)m after 1 year ( 1 + r m
)mt after t year(s)
In the last formula, t can be less than 1. For example, t = 0.5 means 6 months.
To reduce the likelihood of confusion, we will usually denote an interest rate with compound- ing m times a year by r(m).
Converting rates with different compounding frequencies
With compounding once a year, your payoff is (1 + r) after 1 year, and (1 + r)t^ after t year(s). So you see that if r and r(m) are related by the following formula: ( 1 + r( mm)
)m = 1 + r,
which is equivalent to r(m) = m
[ (1 + r)^1 /m^ − 1
] ,
time period t is expressed in year(s) (e.g., t = 0.5 means 6 months, t = 1.5 means 1 year and 6 months, etc.)
The effect of changing the compounding frequency
Let us fix the interest rate (to, say, 12%) and study the effect of changing the compounding frequency. The table below reports, for different compounding frequencies, the annually com- pounded equivalent rate. In other words, we vary m, but r(m) remains equal to 12%, and we look at the equivalent annually compounded rate r =
( 1 + r( mm)
)m − 1.
m Compounding periodicity Annually compounded equivalent (r) 1 annual 12% 2 semi-annual 12.36% 4 quarterly 12.55% 12 monthly 12.68% 52 weekly 12.73% 8760 hourly 12.7496% 525,600 every minute 12.7497%
What we see is that the higher the compounding frequency, the better for the investor, because he ends up with more money after 1 year (per $1 invested, $1.12 with annual compound- ing, $1.1236 with semi-annual compounding, and so on). This is easy to understand: a higher compounding frequency means that interest is paid sooner. So it can be reinvested sooner and the investor ends up receiving more “interest on interest,” or compound interest with a higher compounding frequency.
While the dollar payoff increases with the compounding frequency, it does not seem to “ex- plode” (i.e., become infinite when the compounding frequency is very large). This observation is confirmed by calculus, which tells us that the limit of the formula for the payoff after 1 year, per $1 invested,
( 1 + r( mm)
)m , as m becomes very large, is equal to exp(r(m)) = er(m). You may know that this is known as the exponential of r(m), and is equal to e to the power r(m), where e = 2.71828 (approximately).
In other words, this means that if we could compound “continuously” (i.e., compute the inter- est and add it to the principal every “instant”), $1 invested at 10% would yield exp(0.10) = e^0.^10 = $1.1052 after 1 year. Of course, this is an abstraction; it is not actually possible to compound continuously (and there are no bonds that actually pay interest continuously!) Nonetheless, the user-friendly properties of the exponential make continuous compounding useful.
Continuous compounding
Our formal definition is as follows: $1 invested at the continuously compounded rate r(∞)
yields $ exp(r(∞)) = $er(∞)^ after 1 year, and $ exp(r(∞)t) = $er(∞)t^ after t year(s) (where t can be less than 1).
Given a certain continuously compounded rate r(∞), we can compute an equivalent annually compounded rate r. What we have to do is pick the two rates so that the payoffs after 1 year (per $1 invested) are the same, i.e.:
exp(r(∞)) = (1 + r)
So the equivalent annually compounded rate is equal to exp(r(∞))−1. To go in the other direction and find the continuously compounded rate that is equivalent to the annually compounded rate r, we use algebra, which tells us that the above equation is equivalent to: r(∞) = ln(1 + r), where ln denotes the natural logarithm function. The natural logarithm is the inverse of the exponential (i.e., for any number x, ln(exp(x)) = x and exp(ln(x)) = x).
Why use continuous compounding? It turns out that the exponential function has some nice properties that make continuous compounding particularly convenient. For any two numbers x and y: e−x^ = 1/ex, ex+y^ = exey^ , ex−y^ = ex/ey^ and (ex)y^ = exy.
How are these properties useful to us finance people? Let’s take an example. Think of an investment that lasts two years, yields 10% in year 1 and 12% in year 2. What is your total rate of return after 2 years? What is your average rate of return per year? These computations are much easier to do if we’re dealing with continuously compounded rates.
First, assume that these are annually compounded rates. For $1 invested, you receive (1.10)(1.12) = $1.232 after 2 years, so your total return over the investment period is 23.2% and your average annual rate of return (the IRR – internal rate of return – of your investment) is y such that (1 + y)^2 = (1.10)(1.12), which implies: y = [(1.10)(1.12)]^1 /^2 − 1 = 10.9955%.
Now, assume that we’re dealing with continuously compounded rates. For $1 invested, you receive (e^0.^10 )e^0.^12 = e^0 .10+0.^22 = e^0.^22 = $1.246 so your continuously compounded total return over two years is simply 10% + 12% = 22%, and your average continuously compounded annual rate of return is y such that e^2 y^ = e^0.^22 , implying y = 0.11 = 11%. We see that the average annual rate of return over the two years is just a simple average of the annual rates of return during year 1 and year 2 (10% and 12%) – a much easier calculation than with annually compounded rates!
These results are true in general: with continuously compounded interest rates, your total rate of return is the sum of the rates of return over the subperiods, and you average rate of return over the period is the simple (arithmetic) average of the rates of return over the subperiods.
We will see later in the course that there are other cases in which continuously compounded rates prove much more convenient to use (for computing forward rates, in risk management). One could say that continuous compounding is hard to understand but easy to use, while annual compounding is easy to understand but hard to use.
We can use continuously compounded rates to compute present values. The present value of
∑T t=
(1 + y)t^
1 + y
1 − (^) (1+^1 y)T 1 − (^) 1+^1 y
1 − (^) (1+^1 y)T y
Price and yields quotes: the case of US Treasury bonds
Fixed income markets participants usually quote yields or rates, rather than prices. They use different conventions for different types of securities. We could probably spend the whole class on these issues, but let’s look at only one example, that of US Treasury bonds.
We denote the coupon rate by c, the maturity date by M , the par value by F. US Treasury bonds make two coupon payments a year, so the bond under consideration makes a payment of $cF/2 twice a year (at dates that are exactly 6 months apart), plus $F at maturity (reimburse- ment of the face value). Prices are quoted per $100 of par, so they’re really expressed in per- centages of the face value. To express the quantity of bonds under consideration, traders will use the total par value (e.g., when a trader says he’s buying $20,000,000 of a certain bond, it doesn’t mean he’s paying $20,000,000, it means that the total par value of the bonds is $20,000,000).
US Treasury bonds usually settle “next day,” i.e., one business day after the trade date – here, the trade date is the date at which the terms of the transaction are fixed, and the settlement date is the date at which the bonds and the money are exchanged. When doing the computations, you should assume you’re at the settlement date.
The bond’s quoted yield (y) is its semi-annually compounded internal rate of return. In the case when the settlement date coincides with a coupon date (in which case the seller receives the coupon), the formula that relates y and the bond’s price per $100 par, P , is as follows:
P = 100
[( (^) ∑T
t=
c/ 2 (1 + y/2)t
)
] ,
where T is the number of 6-month between settlement and the maturity date M.
In the case of a settlement date between two coupon dates, the formula is as follows:
P =
(1 + y/2)nSN^ /nLN
T∑NM
t=
c/ 2 (1 + y/2)t
(^) + 1 (1 + y/2)TNM
(^) ,
where TN M denotes the number of 6-month periods between the next coupon and maturity, nSN is the actual number of days between the settlement date and the next coupon, nLN is the num- ber of days between the last and the next coupon. Basically, the square bracket is the present value of the bond’s cash flows at the next coupon date; this present value is then discounted back to the settlement date based on the actual number of days remaining until the next coupon.
US Treasury bonds: an example
Let’s take an example: let’s price the “8 5/8’s of 8/15/03 at 3.21% for settlement 9/8/02,” which means that the coupon rate c = (8 + 5/8)/100 = 8.625% = 0.08625, the maturity date M is August 15, 2003, the yield y = 3.21% and the settlement date is September 8, 2002. Since the bonds matures on August 15,2003, and US Treasury bonds make two coupon payments per year, that occur exactly 6 months apart, this means that the bond makes a coupon payment twice a year, on February 15 on August 15. At the settlement date, the bond has two coupon payments left, that will occur on February 15, 2003 and August 15, 2003, and TN M = 1. Settlement is not on a coupon date so we need to use the second, more complicated formula.
How many days are there between settlement (September 8) and the next coupon (February 15)? We need to take into account the actual number of days in each month, and find that nSN = 22 + 31 + 30 + 31 + 31 + 15 = 160. Similarly, nLN , the number of days between the last coupon (August 15) and the next coupon, equals 16 + 30 + 31 + 30 + 31 + 31 + 15 = 184. The formula then yields the bond price:
P =
( 1 + 0.^03212
) (^160184)
[ (^) ∑ 1
t=
(1 + 0. 0321 /2)t^ +^
(1 + 0. 0321 /2)^1
]
( 1 + 0.^03212
) (^160184)
[ 0. 08625
] = 105. 5062627.
Recall that this price is expressed in percentage of the par value. For example, if the size of the trade is $20,000,000 (expressed in par value), the price paid will be 20, 000 , 000(105. 5062627 /100) = $21, 101 , 252 .54.
The price we just computed, P , is known as the full price of the bond. The price that is quoted in markets is the flat price, p, which is equal to the full price minus the accrued interest (p = P − a). The accrued interest, a, is given by the following formula: a = 100(c/2) (nLS /nLN ), where nLS is the number of days between the last coupon and settlement. The accrued interest is an approximation of the interest corresponding to the period between the last coupon and settlement.
In our example, nLS , the number of days between the last coupon (August 15) and settlement (September 8), equals 16 + 8 = 24, and a = 100(0. 08625 /2)(24/184) = 0.5625. The flat price is p = P − a = 105. 5062627 − 0 .5625 = 104.9438.
A last complication is that prices are not quoted using “regular” decimal numbers, but in 32nds. For example, the price quote for the bond in our example would be 104:30, meaning 104 + 30/32, which is approximately equal to the bond’s flat price.
Finally, note that these formulas are only used for bonds that have at least two coupon payments left. For bonds near the end of their life whose maturity is less than one coupon period, the yield is computed using the simple interest technique (meaning that, in the computation, the yield is multiplied by the remaining maturity). In that case, the formula relating the full price
