Duration and Modified Duration

Apparently people have been wondering about this for a long time.

“Hey!! Let’s just say a person’s required return is basically the bank’s interest rate, OK?

I mean, if the bank’s rate goes up, your required return goes up. Goes down, it goes down. They move together, right?????”

“Yeah yeah. So what I’m curious about is — how does the bond price react when the interest rate moves??!!”

That, apparently, was the question.

Alright, let’s go.

First, a quick reminder about what calculus actually means.

Our man Leibniz cooked up calculus, and the whole vibe is this:

When $x$ changes by thiiiis much, how much does $y$ change??!?!! →

$$\frac{dy}{dx}$$

OK. Now apply that to our situation.

When $r$ changes by thiiiis much, how much does Price change?!?!!~ →

$$\frac{dP}{dr}$$

Yep. Differentiate the price equation with respect to $r$ — and we should be able to answer the question.

Before we do that, let me clean up the price equation a tiny bit. Just to make it easier to push around.

$$P = \frac{c_1}{(1+r)^1} + \frac{c_2}{(1+r)^2} + \cdots + \frac{c_t}{(1+r)^t}$$

I’ll call the numerators a sequence: $c_1, c_2, \ldots, c_t$.

$$P = \sum_{t} \frac{c_t}{(1+r)^t}$$

Boom. Sigma. Everything packed up to the last term in one go.

$$P = \sum_{t} c_t (1+r)^{-t}$$

OK, with that set up — one step toward the actual question.

Goal: “$dP/dr$.”

$$\frac{dP}{dr} = \sum_{t} c_t \cdot (-t)(1+r)^{-t-1}$$

Super basic differentiation, done!!!

So we wanted to know the tiny change in bond price for a tiny change in $r$ — the sensitivity. We differentiated $P$, got $dP/dr$ — and now I’ll do one more little massage.

$$\frac{dP}{dr} = \frac{-1}{(1+r)} \sum_{t} \frac{t \cdot c_t}{(1+r)^t}$$

And one more idea I want to slip in here: “how much did it change relative to PV~~~~?”

$$\frac{1}{P} \cdot \frac{dP}{dr} = \frac{-1}{(1+r)} \cdot \frac{1}{P} \sum_{t} \frac{t \cdot c_t}{(1+r)^t}$$

Writing it like this — we’ve kind of set a yardstick, right?!?!

But here’s the thing.

The definition of duration we want is

$$D = \frac{1}{P} \sum_{t} \frac{t \cdot c_t}{(1+r)^t}$$

If you take that and multiply by $(-1)$, then multiply by $(1+r)$ — that is the definition of duration.

heh heh heh…

$$D = \frac{1}{P} \sum_{t} \frac{t \cdot c_t}{(1+r)^t}$$

hahahahahaha.

OK — I think I just spoke in a way that might have made you slightly mistake duration for sensitivity. I phrased it a little loosely on purpose, just to keep the original goal in front of us.

From here on out, we’re gonna look at why this is the definition of duration. And to explain that, I’m gonna detour for a second and then come back.

Imagine a class with students of all sorts of ages.

20% are 20 years old 20% are 21 years old 20% are 22 years old 20% are 23 years old 20% are 24 years old

What’s the average age in this class?!?

$$\text{Average age} = 0.2 \times 20 + 0.2 \times 21 + 0.2 \times 22 + 0.2 \times 23 + 0.2 \times 24$$

Get a feel for that. Now back to duration.

$$D = \frac{1}{P} \sum_{t} \frac{t \cdot c_t}{(1+r)^t}$$

I’ll just shuffle it a bit so it’s easier on the eyes.

$$D = \sum_{t} t \cdot \frac{c_t / (1+r)^t}{P}$$

Hmm, still doesn’t quite click. Let me unpack the sigma — write out every term. And while I’m at it, let me unpack PV too.

$$D = 1 \cdot \frac{c_1/(1+r)^1}{PV} + 2 \cdot \frac{c_2/(1+r)^2}{PV} + \cdots + t \cdot \frac{c_t/(1+r)^t}{PV}$$

Still not clicking, right?

I once tried to draw out what PV is made of.

PV (the bar)

is all those little discounted cash flows, each one shrunk down by its own discount factor, then stacked up.

Then for the question “on average, how many years out???~~~~”

— this might do it:

$$D = 1 \cdot w_1 + 2 \cdot w_2 + \cdots + t \cdot w_t$$

where

$$w_t = \frac{c_t/(1+r)^t}{PV}$$

If you yank out a single term and look at it,

$$w_t = \frac{c_t/(1+r)^t}{PV}$$

— going back to that bar picture — this thing right here is basically a probability weight. Each chunk’s slice of PV.

Which means!!!!!!!! Duration is

$$D = \sum_{t} t \cdot w_t \quad \text{(weighted average maturity)}$$

THAT’S what it is!!!!!!!!!!!!!!!!!!

See it now?

And see why I bailed mid-derivation to talk about the average age of a class.

But — at the start we were clearly chasing sensitivity. We differentiated $P$ with respect to $r$, divided by PV, multiplied by some weird stuff… and then I rolled in going “this is the definition of Duration~.”

OK OK OK OK OK OK — back to the original mission.

How do we actually get the sensitivity?!?!!!!

One more step. Enter Modified Duration.

$$D^* = \frac{D}{1+r}$$$$\frac{1}{P} \cdot \frac{dP}{dr} = -D^*$$

Wait, hold on… why is it

not $dP/dr = $ modified duration

but modified duration $= -(1/P)(dP/dr)$

$$D^* = -\frac{1}{P} \cdot \frac{dP}{dr}$$

like that………. haa, seriously, lol

Because there’s a reason.

$$\frac{dP}{dr} = -D^* \cdot P$$

Let me write it that way.

When you write $dP$, you say “dee-pee,” and it means an infinitesimally small change in $P$. When you write $\Delta P$, you say “delta-pee,” and it just means a change in $P$ — not necessarily tiny.

So I’m going to swap the $d$’s for $\Delta$’s and treat it as an approximation.

$$\Delta P \approx -D^* \cdot P \cdot \Delta r$$

And there was a reason for all this, right?

Here’s the reason. Watch:

$$\frac{\Delta P}{P} \approx -D^* \cdot \Delta r$$

The left side reads as: “how much did $P$ change, relative to the original $P$~~~”

In other words —

$$\frac{\Delta P}{P}$$

— the % change.

And that, apparently, is exactly what modified duration was built for.

How to use $D^*$:

Take a bond with face value $1,000, a 9% coupon rate, a 10% yield to maturity, and 2 years left to maturity. Dur = 1.92 years. Crunch the numbers and the modified duration comes out to 1.75 years.

So:

$$\frac{\Delta P}{P} \approx -D^* \cdot \Delta r$$

If the market interest rate $r$ moves by “1 percentage point (pt),”

$$\frac{\Delta P}{P} \approx -1.75 \times 0.01 = -1.75\%$$

the bond price moves by 1.75%. The other way: if the market rate drops by 1 pt, the bond price climbs by 1.75%~~~

That’s how you read it.

Different bonds have different sensitivities to a move in market yield — depends on maturity, YTM, coupon rate. So if you compute $D^*$ for each bond at each point in time, you can see at a glance how much the price will move when yields move.

(Apparently on the screens of the people who actually do this for a living, modified duration is just there — auto-calculated, sitting in the bottom-left or top-right corner of whatever app, all the time!!!! They say it’s a must-have number!!!!)

To recap:

The thing built around the time-average idea — that’s Duration. The thing built around the sensitivity idea — that’s Modified Duration.

And finally, elasticity.

We did “price elasticity” back in microeconomics. Given a % change in price $P$, what’s the % change in quantity $Q$~~~~ — that’s what price elasticity captured!!

In formula form:

$$\epsilon \;(\text{elasticity}) \;=\; \frac{\frac{dQ}{Q}}{\frac{dP}{P}}$$

Same deal here. Define epsilon as the % change in bond price $P$ per % change in interest rate $r$:

$$\epsilon = \frac{\frac{dP}{P}}{\frac{dr}{r}}$$

Now — this expression. We can rewrite it using the duration / modified duration we just nailed down.

$$D = \frac{1}{P} \sum_{t} \frac{t \cdot c_t}{(1+r)^t}$$$$D^* = \frac{D}{1+r}$$

These were our definitions, right? Solve for $dP/dr$ from there, plug into the elasticity formula above, and:

$$\epsilon = -D^* \cdot r$$

Done.

It’s literally just shuffling an identity around, so… I’m going to bed. heh heh

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Originally written in Korean on my Naver blog (2016-05). Translated to English for gdpark.blog.