Why Bond Prices Fall When Yields Rise

Why the Price-Yield Relationship Matters

A bond’s price is the present value of its future payments. If those payments and their dates are fixed, a higher yield discounts each payment more heavily. Differentiation makes this intuition precise: it gives the direction of the price change and measures the local dollar response to a small yield change.

The result is useful when checking a calculated price, comparing how two bonds respond to a market-yield movement, or estimating a price change without fully repricing the bond. The estimate is local: for a larger yield change, the exact present-value equation should be evaluated again.

Model: Suppose a bond has fixed cash flows $C_k$ at times $t_k>0$, where every $C_k>0$. Let $i>-1$ be the effective annual yield, and suppose each cash flow is discounted consistently using that yield. The price is $$ P(i)=\sum_{k=1}^{n} C_k(1+i)^{-t_k}. $$ For annual cash-flow dates, the $t_k$ are positive integers. The same formula applies to fractional times if the stated yield convention supports fractional-year discounting. If a bond is priced using a periodic yield, use the number of periods and the yield per period consistently instead.

This model treats the cash flows and payment dates as known and unchanged while the yield varies. It therefore describes a fixed-cash-flow bond scenario. It does not by itself account for a change in expected cash flows, a change in the bond’s redemption terms, or an embedded option whose exercise changes when yields move.

Differentiating the Price Equation

For each payment, $C_k(1+i)^{-t_k}$, the cash amount and payment time are fixed. Differentiating with respect to $i$ gives

\frac{d}{di}\left[C_k(1+i)^{-t_k}\right] =-t_kC_k(1+i)^{-t_k-1}.

Adding the derivatives of all the payments gives the price sensitivity:

P'(i)=-\sum_{k=1}^{n}\frac{t_kC_k}{(1+i)^{t_k+1}}.

Every term in the sum after the minus sign is positive: $t_k>0$, $C_k>0$, and $1+i>0$. Therefore $P'(i)<0$. The price is strictly decreasing as the yield increases. This is a statement about the direction of the relationship; the magnitude of $P'(i)$ measures the local dollar price response to a change in yield.

To see why the derivative has this form, focus on one payment. Raising the yield slightly increases its discount denominator. The derivative’s factor $t_k$ also shows why, all else equal, a payment farther in the future can contribute more to the price response. The contributions from the individual payments add because the bond price is the sum of their present values.

Local price-change approximation: If the yield changes from $i$ to $i+\Delta i$, with $\Delta i$ small, then $$ \Delta P \approx P'(i)\Delta i. $$ The yield change is entered as a decimal change. For example, a rise from 4.0% to 4.5% is $\Delta i=0.005$, not $0.5$ or $0.05$. The approximation is not an exact repricing formula; its accuracy generally improves as the yield change becomes smaller.

Worked Applications

In each calculation below, the yield and cash-flow timing use the same annual basis. A derivative is quoted in dollars per unit change in yield: “per unit” means per 1.00 change in the decimal yield, not per one percentage point.

Worked Example: A Single Payment and a Small Yield Increase

A hypothetical zero-coupon bond pays $300 at the end of year 3. Its effective annual yield is 4%. Find its price, its derivative with respect to yield, and the local estimated price change if the yield rises by 0.5 percentage point. Give dollar amounts to the nearest cent.

Method. Treat the payment as one fixed positive cash flow at $t=3$, discount it at the stated annual yield, and differentiate the same present-value expression.

Price. $$ P(0.04)=\frac{300}{(1.04)^3} =\frac{300}{1.124864} =266.6989\ldots $$ The price is $266.70.

Derivative. For one payment at time 3, $P'(i)=-3(300)/(1+i)^4$. Thus $$ P'(0.04)=-\frac{900}{(1.04)^4} =-\frac{900}{1.16985856} =-769.3238\ldots $$ The local slope is −$769.32 per unit yield.

A 0.5-percentage-point increase is $\Delta i=0.005$. The local estimate is $$ \Delta P\approx(-769.3238\ldots)(0.005) =-3.8466\ldots $$ So the estimated price change is −$3.85, and the estimated new price is $266.6989\ldots-3.8466\ldots=262.8523\ldots$, or $262.85.

Check and mistake avoided. The negative sign agrees with the inverse price-yield relationship. The yield change was entered as 0.005, not 0.5. The result is a local estimate; it is not the exact price at 4.5%.

Worked Example: Adding the Sensitivity Contributions

A hypothetical bond pays $104 at the end of year 1 and $216.32 at the end of year 2. Its effective annual yield is 4%. Find its price and derivative, then estimate the price change for a 0.5-percentage-point yield increase. Give dollar amounts to the nearest cent.

Method. Calculate each payment’s present value to obtain the price, then differentiate each term and add the two derivative contributions.

Price. The year-1 payment has present value $104/1.04=100$. Since $1.04^2=1.0816$, the year-2 payment has present value $216.32/1.0816=200$. Therefore, $$ P(0.04)=100+200=300.00. $$

Derivative. The year-1 contribution is $$ -\frac{1(104)}{(1.04)^2} =-\frac{104}{1.0816} =-96.1538\ldots. $$ The year-2 contribution is $$ -\frac{2(216.32)}{(1.04)^3} =-\frac{432.64}{1.124864} =-384.6154\ldots. $$ Adding them gives $P'(0.04)=-480.7692\ldots$, or −$480.77 per unit yield.

A 0.5-percentage-point rise means $\Delta i=0.005$, so $$ \Delta P\approx(-480.7692\ldots)(0.005) =-2.4038\ldots. $$ The estimated price decrease is $2.40, giving an estimated new price of $297.60.

Check and mistake avoided. The two derivative terms are both negative, so adding them must give a negative slope. Do not use only one payment’s derivative or omit the time factor $t_k$; the year-2 contribution includes a factor of 2.

Worked Example: Comparing the Dollar Response of Two Bonds

At an effective annual yield of 5%, compare two hypothetical bonds: Bond A pays $525 in one year, and Bond B pays $578.8125 in two years. Find each price and derivative. Which has the larger local dollar price decrease for a 1-percentage-point increase in yield? Give prices and estimated changes to the nearest cent, and derivatives to four decimal places.

Method. Price each fixed payment at the same yield, differentiate with respect to that yield, and compare the resulting dollar slopes.

For Bond A, $$ P_A(0.05)=\frac{525}{1.05}=500.00, \qquad P_A'(0.05)=-\frac{525}{(1.05)^2} =-\frac{525}{1.1025} =-476.1905\ldots. $$ Thus Bond A’s derivative is −$476.1905 per unit yield.

For Bond B, $(1.05)^2=1.1025$, so $$ P_B(0.05)=\frac{578.8125}{1.1025}=525.00. $$ Its derivative is $$ P_B'(0.05)=-\frac{2(578.8125)}{(1.05)^3} =-\frac{1157.625}{1.157625} =-1000.0000. $$ Thus Bond B’s derivative is −$1,000.0000 per unit yield.

A 1-percentage-point rise is $\Delta i=0.01$. The estimated changes are $$ \Delta P_A\approx(-476.1905\ldots)(0.01)=-4.7619\ldots, \qquad \Delta P_B\approx(-1000)(0.01)=-10.00. $$ Bond A’s estimated decrease is $4.76; Bond B’s is $10.00. Bond B therefore has the larger local dollar decrease, even though its price at the stated yield is also higher.

Check and mistake avoided. Rechecking Bond B’s price is important: its present value is $525, not $500. The two bonds have different prices and different payment dates, so the comparison is specifically about the dollar decrease, not a claim that their percentage price responses are the same.

What Determines the Size of the Response?

The derivative shows that sensitivity depends on both the size and timing of the cash flows. A larger payment has a larger contribution to the slope, all else equal. A later payment also has a larger time multiplier $t_k$, although discounting reduces its present-value contribution. Consequently, the full comparison must use the derivative terms rather than timing alone.

The derivative is a local rate of change. If $P'(i)=-500$ at a particular yield, then a small increase of 0.01 in yield gives an estimated change of $-500(0.01)=-$5$. It does not mean that a one-percentage-point increase always changes the price by exactly $5. The derivative itself varies with the yield, so for a substantial movement, calculate the price at the new yield from the original present-value equation.

For positive fixed cash flows, the price is also convex as a function of yield. Differentiating again gives

P''(i)=\sum_{k=1}^{n} \frac{t_k(t_k+1)C_k}{(1+i)^{t_k+2}}>0.

Every term is positive under the stated assumptions. Thus the slope becomes less negative as the yield rises, and a straight-line estimate based on the initial slope need not equal the exact change over a finite interval. This second derivative is a mathematical property of the fixed-positive-cash-flow model; it does not remove the need to reprice when the yield movement is not small.

Common Exam Traps

Using the wrong sign. The derivative has a minus sign. A positive fixed payment loses present value when yield rises, so a positive price response is inconsistent with the model.
Entering percentage points as whole numbers. A move from 5% to 6% has $\Delta i=0.01$. Multiplying the derivative by 1 instead would overstate the local change by a factor of 100.
Dropping the time multiplier. Differentiating $(1+i)^{-t}$ produces the factor $-t$. It is not enough to attach a negative sign to the original present value.
Confusing slope with a price change. $P'(i)$ is measured per unit yield. To estimate a price change, multiply it by the decimal yield change.
Calling a tangent estimate exact. $P'(i)\Delta i$ is a local approximation. An exact price change is $P(i+\Delta i)-P(i)$, evaluated from the price equation at both yields.
Ignoring the model’s assumptions. The proof uses fixed positive cash flows and unchanged dates. If cash flows or redemption timing change with yield, this derivative does not describe the whole pricing response.

Key takeaway: For a bond with fixed positive cash flows, differentiating present value gives a strictly negative price-yield derivative. Its value measures the local dollar sensitivity; multiply by a yield change expressed as a decimal to estimate the corresponding local price change.

Check Your Understanding

Use the fixed-cash-flow present-value model. Treat yield changes as decimal changes, and round dollar answers to the nearest cent.

A hypothetical bond pays $500 at the end of year 2 and has an effective annual yield of 4%. Find its price, derivative, and local estimated price change for a 0.25-percentage-point yield increase. (Answer: Price $462.28; derivative −$889.00 per unit yield; estimated change −$2.22.)
A bond pays $104 at the end of year 1 and $216.32 at the end of year 2. At a 4% effective annual yield, find its price and derivative, then estimate the price change for a 0.5-percentage-point yield increase. (Answer: Price $300.00; derivative −$480.77 per unit yield; estimated change −$2.40.)
At a 10% effective annual yield, Bond A pays $110 in one year and Bond B pays $146.41 in four years. Both have the same price at that yield. Find their derivatives and identify which has the larger local dollar decrease for a 2-percentage-point yield increase. (Answer: Both prices are $100.00; derivatives are −$90.91 and −$363.64 per unit yield, respectively; Bond B has the larger estimated decrease, $7.27 versus $1.82.)
A hypothetical bond pays $100 at the end of year 2. Calculate its exact price change when the effective annual yield rises from 5% to 6%, and compare it with the local estimate based on the derivative at 5%. (Answer: Exact change approximately −$1.70; local estimate approximately −$1.73.)
For one fixed payment of $100 at the end of year 2, find the first and second derivatives of price with respect to yield at $i=0$. What do their signs imply locally? (Answer: $P'(0)=-$200$ per unit yield and $P''(0)=$600$ per squared unit yield; the price is decreasing and convex at $i=0$.)

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