New Video: Rebalancing a Bond Portfolio for a New Macaulay Duration

Rebalancing a portfolio is an important kind of financial planning action. It can be done to increase your rate of return or to change the susceptibility of the value of your investments to changes in yield rates.

For a coupon bond, the “duration” or “length” of the bond is a way to measure its susceptibility to changes in yield rates. Bonds with high duration are more susceptible than bonds with low duration. The two main measures of duration that are used are Macaulay Duration and Modified Duration.

The Macaulay Duration of a coupon bond is the quantity that is most clearly a measure of the “average length” of time of payments to you (the coupons and the redemption value) from the bond. For an arbitrary periodic yield rate $j,$ and times of payment at $t=1,2,3,\ldots,n,$ it can be represented as a weighted average of the times of payment:

$D_{\mbox{mac}}(j)=w_{1}\cdot 1+w_{2}\cdot 2+\cdots+w_{n}\cdot n=\displaystyle\sum_{t=1}^{n}w_{t}\cdot t.$

In this sum, the weights are $w_{t}=w_{t}(j)=\frac{K_{t}(1+j)^{-t}}{\sum_{m=1}^{n}K_{m}(1+j)^{-m}},$ where the payments (the coupons and redemption value) are $K_{1},K_{2},\ldots,K_{n}.$

For example, if a 2 year bond with face amount $100 has a 2% semiannual coupon rate (so that the coupons are $2) and an effective semiannual yield rate $j=3\%=0.03,$ then the price (present value) of the bond is

$\displaystyle\sum_{m=1}^{4}K_{m}(1+j)^{-m}=2\cdot 1.03^{-1}+2\cdot 1.03^{-2}+2\cdot 1.03^{-3}+102\cdot 1.03^{-4}\approx 1.942+1.885+1.830+90.626=\$96.28.$

This gives weights equal to $w_{1}\approx \frac{1.94}{96.28}\approx 0.0201,$ $w_{2}\approx \frac{1.89}{96.28}\approx 0.0196,$ $w_{3}\approx \frac{1.83}{96.28}\approx 0.0190,$ and $w_{4}\approx \frac{90.63}{96.28}\approx 0.9413$ . The Macaulay duration is therefore approximately

$0.0201\cdot 1+0.0196\cdot 2+0.0190\cdot 3+0.9413\cdot 4\approx 3.88$ years.

We can algebraically rearrange the formula for the Macaulay duration (and change an index of summation from $m$ to $t$ in the bottom of the fraction) to write

$D_{\mbox{mac}}(j)=\frac{\sum_{t=1}^{n}tK_{t}(1+j)^{-t}}{\sum_{t=1}^{n}K_{t}(1+j)^{-t}}=\frac{\sum_{t=1}^{n}tK_{t}(1+j)^{-t}}{P(j)},$

where $P(j)=\displaystyle\sum_{t=1}^{n}K_{t}(1+j)^{-t}$ is the price (present value) of the bond at yield rate $j.$

On the other hand, the Modified duration of the bond is defined to be its relative rate of decay: $D_{\mbox{mod}}(j)=-\frac{P'(j)}{P(j)}.$ Note that, since bond prices decrease as yield rates increase, $P'(j)< 0$ so that $D_{\mbox{mod}}(j)>0.$

It is not completely obvious at first, but these two measures are related. This can be confirmed by differentiating $P(j)=\displaystyle\sum_{t=1}^{n}K_{t}(1+j)^{-t}$ to get, by the Power Rule, $P'(j)=\displaystyle\sum_{t=1}^{n}-tK_{t}(1+j)^{-t-1}=-(1+j)^{-1}\displaystyle\sum_{t=1}^{n}tK_{t}(1+j)^{-t}.$

This implies that $D_{\mbox{mod}}(j)=-\frac{P'(j)}{P(j)}=\frac{(1+j)^{-1}\sum_{t=1}^{n}tK_{t}(1+j)^{-t}}{P(j)}=\frac{D_{mac}(j)}{1+j}.$ Of course, we could then also write $(1+j)D_{\mbox{mod}}(j)=D_{\mbox{mac}}(j)$ and $-(1+j)P'(j)=P(j)D_{\mbox{mac}}(j).$

Relationship to Portfolio Rebalancing

In the video embedded below, an investor has a portfolio of two bonds of different present values (prices) and different Macaulay durations. The investor wishes to rebalance the portfolio to obtain a new portfolio with the same total present value but a smaller duration, making it less prone to duration risk. This reduction of the duration would be a good thing if yield rates go up in the near future, because increasing yields lead to decreasing prices, which is bad for the investor if they need to resell their bonds. By reducing the duration of the portfolio, the decline in value of the portfolio will be less extreme as yield rates increase.

Financial Math for Actuarial Exam 2 (FM), Video #168. Exercise *7.1.12 (modified) from “The Mathematics of Investment and Credit”, 7th Edition, by Samuel A. Broverman.

Let $D_{\mbox{mac},1}(j)$ and $D_{\mbox{mac},2}(j)$ be the Macaulay durations of bonds 1 and 2, respectively; let $P_{1}(j)$ and $P_{2}(j)$ be the present values (prices) of bonds 1 and 2, respectively; and let $P(j)=P_{1}(j)+P_{2}(j)$ be the total present value of the portfolio.

If $D_{\mbox{mod,tot}}(j)$ represents the duration of the entire portfolio, the key fact needed in the video to solve the problem is that $D_{\mbox{mod,tot}}(j)=\frac{P_{1}(j)}{P(j)}D_{\mbox{mac},1}(j)+\frac{P_{2}(j)}{P(j)}D_{\mbox{mac},2}(j).$

Notice that this is a weighted average of $D_{\mbox{mac},1}(j)$ and $D_{\mbox{mac},2}(j)$ , where the weights are the relative sizes of the present values of each bond with respect to the overall present value of the portfolio.

In the notation used in Chapter 7 of the 7th Edition of “Mathematics of Investment and Credit”, by Samuel A. Broverman, this can be generalized to a portfolio of $m$ bonds, with present values $X_{k}$ and total present value $X$ to say that $D_{\mbox{mod,tot}}(j)=\displaystyle\sum_{k=1}^{m}u_{k}\cdot D_{\mbox{mac},k}(j),$ where $u_{k}=\frac{X_{k}}{X}.$

Verification of the Key Fact

Let us now mathematically verify the key fact from above (in the situation with two bonds). This verification is also presented in the second part of the video above. To complete the verification, we need the facts from further above relating Macaulay and Modified durations, as well as the linearity of the derivative operator.

By the facts further above,

$D_{\mbox{mac,tot}}(j)=(1+j)D_{\mbox{mod,tot}}(j)=\frac{-(1+j)P'(j)}{P(j)}=\frac{-(1+j)\frac{d}{dj}\left(P_{1}(j)+P_{2}(j)\right)}{P(j)}.$

By linearity, and the facts that $-(1+j)P_{1}'(j)=P_{1}(j)D_{\mbox{mac},1}(j)$ and $-(1+j)P_{2}'(j)=P_{2}(j)D_{\mbox{mac},2}(j)$ , this becomes

$D_{\mbox{mac,tot}}(j)=\frac{-(1+j)P_{1}'(j)-(1+j)P_{2}'(j)}{P(j)}=\frac{P_{1}(j)}{P(j)}D_{\mbox{mac,1}}(j)+\frac{P_{2}(j)}{P(j)}D_{\mbox{mac,2}}(j).$

But this is what we wanted to show.

Generalization and Functional Thinking

To become highly skilled at mathematics, it is important to be able to generalize. This usually involves working with symbolic quantities — sometimes thinking of them as constants, and sometimes thinking of them as variables. Even when we think of these quantities as “constants”, we might want to consider what happens as these constants change. In such a situation, they are often called parameters instead. In all this, the main thing to realize is that no matter what you name these symbols, the key is how you think about them.

This point of view has many benefits, among them: 1) it improves symbolic manipulation skills, 2) it allows you to solve many (even infinitely many) problems at once, 3) it can help you see connections between subjects, which improves both understanding and problem-solving skills, and 4) it often allows you to bring graphs into play, which can help you make both qualitative and quantitative predictions.

In the present situation, besides allowing for more bonds in our portfolio as mentioned above, we can generalize by allowing for arbitrary symbols to represent various values. Instead of solving the specific problem in the video, let us solve a more general problem (still with two bonds as in the video).

To simplify the notation, let us also say the yield rate $j$ is fixed and not refer to it in our notation. Suppose we have a portfolio of two bonds with present values $P_{1}$ and $P_{2},$ and Macaulay durations $D_{1}$ and $D_{2}.$ Suppose further that $D_{1}<D_{2}$ and that $c$ is a number with $D_{1}<c<D_{2}.$

Let $x$ represent the monetary amount to sell of bond 2 (with the higher duration) and to buy of bond 1 (with the lower duration) so that the Macaulay duration of the new portfolio equals $c.$

Note that we are actually allowing $x$ to be negative here. In the case where $x$ is negative, we are really selling some of bond 1 to buy some of bond 2. To be more precise, $x$ will be positive when the value of $c$ is less than the “current” value of the Macaulay duration and $x$ will be negative when the value of $c$ is greater than the “current” Macaulay duration.

By the derivations from earlier, the “current” Macaulay duration is $d=\frac{P_{1}}{P_{1}+P_{2}}D_{1}+\frac{P_{2}}{P_{1}+P_{2}}D_{2}=\frac{P_{1}D_{1}+P_{2}D_{2}}{P_{1}+P_{2}}.$

The goal is to solve the equation $c=\frac{(P_{1}+x)D_{1}+(P_{2}-x)D_{2}}{P_{1}+P_{2}}$ for $x$ . Multiplication of both sides by $P_{1}+P_{2}$ and expansion of the numerator on the right results in $c(P_{1}+P_{2})=P_{1}D_{1}+P_{2}D_{2}+(D_{1}-D_{2})x.$ Using subtraction, division, and a multiplication by -1 on the top and bottom then give the final answer:

$x=\frac{P_{1}D_{1}+P_{2}D_{2}-c(P_{1}+P_{2})}{D_{2}-D_{1}}.$

We have therefore just solved infinitely many problems at one fell-swoop! If you are able to reproduce this result, you have strengthened your symbolic manipulation skills.

Are there any other benefits? Yes, we can think about the last equation as defining the answer, $x$ , as a function of the five quantities (“variables”) $P_{1},P_{2},D_{1},D_{2},$ and $c$ .

Notationally, we could write $x=f(P_{1},P_{2},D_{1},D_{2},c),$ where $f$ is the “name” of this function. We could then, for example, fix (consider to be constant) any four of these variables and graph the result as a function of the remaining (true) variable. It is possible this could provide insights into the nature of the solution to the original problem. It can also lead us to ask new questions and attempt to answer these new questions.

For example, as in the video (with the subscripts swapped), we could take $P_{1}=30000, P_{2}=50000, D_{1}=6,$ and $D_{2}=8$ and consider the function $x=f(30000,50000,6,8,c).$ You should check that its formula simplifies to $x=290000-40000c.$ You should also check that, when the desired duration of the portfolio is $c=7$ , this gives the final answer in the video: to sell amount $x=10000.$

Since the function $x=f(30000,50000,6,8,c)=290000-40000c$ is linear in $c,$ its graph is a straight line.

The graph of $x=f(30000,50000,6,8,c).$ The red dot has coordinates $(c,x)=(7,10000),$ which represents the solution to the problem in the video.

The simplified formula and the graph allows us to see that, for example, as $c$ approaches a duration of 6, the value of $x$ approaches 50000, which happens to be the total amount of bond 2 that is owned; so this is telling us to sell it all when we desire the new duration to be 6!

Being linear, the graph also has a constant slope. The slope in this case is $-40000$ . This is the rate of change of the amount of bond 2 to sell $x$ with respect to the change in desired duration $c$ . It says that, for every small change in $c$ , the value of $x$ changes 40000 times as fast in the opposite direction. For example, if we increase the desired duration by $\Delta c=0.01,$ then the amount of bond 2 to sell changes by $\Delta x=-400$ monetary units (so it goes down by 400 monetary units).

As another example of something we could do, we could let $P_{1}=30000, P_{2}=50000, D_{1}=6,$ and $c=7,$ as in the video, and consider the function $x=f(30000,50000,6, D_{2},7)=\frac{50000D_{2}-380000}{D_{2}-6}.$ The graph of this is shown below.

*The graph of $x=f(30000,50000,6,D_{2},7).$ The red dot has coordinates $(D_{2},x)=(8,10000),$ which represents the solution to the problem in the video.*

We see from this graph that the amount of bond 2 to sell $x$ increases as the duration $D_{2}$ of bond 2 increases, since the graph is rises as we move to the right. Moreover, this graph is concave down, showing that the rate (slope) at which $x$ increases is itself decreasing as $D_{2}$ increases (the slope gets smaller and smaller). This is something you probably would not be able to guess just from the problem statement. New insights can be obtained! Derivatives from calculus could be used to find the precise rates of change and interpret them. I will leave that for you to explore.

Perhaps of most interest to me personally is to make animations showing graphs of one variable change as another quantity changes. Below is an animation of the graph of $x=f(30000,50000,6,D_{2},c)$ as a function of $D_{2},$ as $c$ varies from 6 to 8. The red dot indicates the solution $x$ of the generalized problem in the case where $D_{2}=8.$

Graph of $x=f(30000,50000,6,D_{2},c)$ as a function of $D_{2}$ ,
as $c$ varies from 6 to 8.

Graph of $x=f(30000,50000,6,D_{2},c)$ as a function of $D_{2}$ ,
as $c$ varies from 6 to 8.

Note that the graph shifts downward and becomes more “extreme” in both slope and concavity as $c$ increases from 6 to 8. When $c$ is close to 8, it also gets to the point where all the $x$ values are negative, no matter the value of $D_{2}$ between 7 and 9. It would be a good exercise for you to try thinking through all the financial implications of these facts.

For those who have an interest in the Wolfram Mathematica code used to generate the animation above, it is shown in the figure below.