# Is Duration really the slope of the Price-Yield curve?

When looking at the Price-vs-Yield graph for a fixed rate instrument, we are often told that the duration is the slope of that curve. But is that really right?

Duration is (change in price) divided by (price times change in yield). That's hardly the slope of the curve which would be (change in price) divided by (change in yield). Yield is expressed in percentage terms which makes it look relative, but going from 1% to 2% is a relative increase of 100%, because it's a 1% increase only in absolute terms.

That added factor of price is not constant and so the slope and duration differ by different ratios for different prices!?

-

The Macaulay duration is a measure of how sensitive a bond's price is to changes in interest rates. Duration is related to, but differs from, the slope of the plot of bond price against yield-to-maturity. The slope of the price-yield curve is $-\frac{D}{1+r}P,$ where $D$ is Macaulay duration, $P$ is bond price, and $r$ is yield.
Here's how the definition of duration arises. Let's expand the price of a bond, $P$, in terms of the yield-to-maturity, $r$, using Taylor's theorem: $$\Delta P=P(r+\Delta r)-P(r)\approx\frac{\partial P(r)}{\partial r}\Delta r+\frac{1}{2}\frac{\partial^2 P(r)}{\partial r^2}(\Delta r)^2.$$ Since $$P(r)=\sum_{t=1}^{T}\frac{C_t}{(1+r)^t},$$ where $C_t$ are the cash flows, we have that $$\Delta P\approx -\frac{\Delta r}{1+r}\sum_{t=1}^{T}\frac{t\ C_t}{(1+r)^t}+\frac{(\Delta r)^2}{2(1+r)^2}\sum_{t=1}^{T}\frac{t(t+1)C_t}{(1+r)^t},$$ and dividing both sides by $P$, we arrive at the expression $$\frac{\Delta P}{P}\approx -\frac{D}{1+r}\Delta r+\frac{\mathcal C}{2}(\Delta r)^2.$$ Here $$D=\frac{1}{P} \sum_{t=1}^{T}\frac{t\ C_t}{(1+r)^t}$$ is the Macaulay duration, and $$\mathcal C= \frac{1}{P(1+r)^2}\sum_{t=1}^{T}\frac{t(t+1)C_t}{(1+r)^t}$$ is a measure of curvature, or convexity, in the plot of bond price against yield-to-maturity.