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[Source:] Generally speaking, for every percentage point that rates rise, a bond’s value will decline by its duration (stated in years). So if rates climb by one percentage point, the value of a bond fund with an average duration of two years will drop by 2 percent.

Why's the above true? If the proof requires more than second-year multivariable calculus and linear algebra, then please omit the proof, and expose and and explain only the intuition.

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It's not true, it's only an approximation. It can be explained as follows.

Let $P(y)$ be the price of the bond as a function of the yield. Then $-P'(y)/P(y)$ is called the "modified duration" of the bond. Even though it is given by a formula in terms of the price, the units it is measured in is years. By properties of the derivative (e.g. the mean value theorem), it can be used to approximate the percentage change in price as the yield changes.

Let us consider the case of a zero coupon bond with face amount $100$ maturing in $t$ years with yield $y$. Then by definition $$P(y)=100/(1+y)^t$$ So $$P'(y)=-100t/(1+y)^{t+1}$$ Dividing by $-P(y)$ gives us that the modified duration is $$t/(1+y)$$ Thus the modified duration, which is approximately equal to the percentage change in price when the yield is changed by $0.01$, is approximately equal to the time to maturity of a zero coupon bond, justifying its name. (If we measured the duration using the "Macaulay duration" instead, it would be exactly $t$.)

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Take a zero coupon bond due in $t$ years, when it will pay $100$. If the present interest rate is $r$ per year, it is now worth $\frac {100}{(1+r)^t}$ If interest rates change to $s$ per year, it will then be worth $\frac {100}{(1+s)^t}$. The new value is $\frac {(1+r)^t}{(1+s)^t}$ of the old one. As long as $rt, st \ll 1$ we can keep only the first term in the binomial and the new value is $\frac {1+rt}{1+st}\approx 1-(s-r)t$ of the old value. This shows that if interest rates rise, the fall in bond value is $t$ times the change in interest rate.

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  • $\begingroup$ Thanks, but would you please explain why: $\dfrac {1+rt}{1+st}\approx 1-(s-r)t$ ? $\endgroup$
    – NNOX Apps
    Jul 9 '15 at 18:30
  • $\begingroup$ Would you please respond in your answer, which is easier to read than comments? $\endgroup$
    – NNOX Apps
    Jul 9 '15 at 18:31
  • $\begingroup$ As long as $x \ll 1, \frac 1{1+x} \approx 1-x$ I just used that with $x=st$ $\endgroup$ Jul 9 '15 at 18:40
  • $\begingroup$ Is your approximation related to this? $\endgroup$
    – BCLC
    Aug 26 '15 at 9:56
  • $\begingroup$ @BCLC: Not exactly. I am expanding each of the numerator and denominator separately, but the result is the same. $\endgroup$ Aug 26 '15 at 13:39

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