# Can the duration of a floating rate bond with yield spread be negative?

Summary

In my calculations below I find that the effective duration(not spread duration, but interest duration) of a floating rate bond with yield spread can become negative. Do you see if they are correct? Does it happen in the real world that the effective interest duration of a floating rate bond can become negative?

Calculations

I assume that we have a bond paying out at $$K$$ periods during the year, it has face value $$N$$ it has price $$P$$, we also assume that the current forward rates are all the same not denoted $$r$$. The first payment has been set to $$c$$, and it is $$\Delta t_1$$ time left ot the first payment, the are currently $$V+1$$ payments dates left.

We find the yield to maturity y, by solving the equation:

$$P = \frac{c}{(1+\frac{y}{K})^{\Delta t_1 K}}+\sum\limits_{i=1}^{V-1}\frac{\frac{Nr}{K}}{(1+\frac{y}{K})^{\Delta t_1 K+i}}+\frac{N(1+\frac{r}{K})}{(1+\frac{y}{K})^{\Delta t_1 K+V}}.$$

We can now calculate the yield-spread by $$s=y-r$$.

We now treat $$r$$ as a variable, but keep the yield-spread $$s$$ fixed. We denoted $$G(r)$$ as the price of the bond right after the first reset date. We then get

$$G(r)=\sum\limits_{i=1}^{V-1}\frac{\frac{Nr}{K}}{(1+\frac{r+s}{K})^{i}}+\frac{N(1+\frac{r}{K})}{(1+\frac{r+s}{K})^{V}}.$$

Our aim is to find a better expression for $$G(r)$$, by adding and subtracting $$Ns/K/(1+(r+s)/K)^V$$ we get

$$G(r)=\sum\limits_{i=1}^{V-1}\frac{\frac{Nr}{K}}{(1+\frac{r+s}{K})^{i}}+\frac{N(1+\frac{r+s}{K})}{(1+\frac{r+s}{K})^{V}}-\frac{\frac{Ns}{K}}{(1+\frac{r+s}{K})^V}\\=\sum\limits_{i=1}^{V-1}\frac{\frac{Nr}{K}}{(1+\frac{r+s}{K})^{i}}+\frac{N}{(1+\frac{r+s}{K})^{V-1}}-\frac{\frac{Ns}{K}}{(1+\frac{r+s}{K})^V}.$$

By doing then adding and subtracting $$Ns/K/(1+(r+s)/K)^{V-1}$$ we simplify it even further. By doing this trick again and again we end up with

$$G(r)=N-\sum\limits_{i=1}^V\frac{\frac{Ns}{K}}{(1+\frac{r+s}{K})^i}.$$

We write $$G(r)=N-F(r)$$, where

$$F(r)=\sum\limits_{i=1}^V\frac{\frac{Ns}{K}}{(1+\frac{r+s}{K})^i}$$.

Note that $$F'(r)<0$$.

Also note that the present value of the price is

$$\frac{(c+G(r))}{(1+\frac{r+s}{K})^{\Delta t_1 K}}.$$

Now we can calulate the duration

$$-\frac{1}{P}\frac{dP}{dr}=-\frac{\frac{d}{dr}\left(\frac{(c+G(r))}{(1+\frac{r+s}{K})^{\Delta t_1 K}}\right)}{\frac{(c+G(r))}{(1+\frac{r+s}{K})^{\Delta t_1 K}}}\\=-\frac{\frac{-\Delta t_1(c+G(r))}{(1+\frac{r+s}{K})^{\Delta t_1 K+1}}+\frac{G'(r)}{(1+\frac{r+s}{K})^{\Delta t_1 K}}}{\frac{(c+G(r))}{(1+\frac{r+s}{K})^{\Delta t_1 K}}}\\=\frac{\Delta t_1}{(1+\frac{r+s}{K})}-\frac{-F'(r)}{c+G(r)}\\=\frac{\Delta t_1}{(1+\frac{r+s}{K})}+\frac{F'(r)}{c+G(r)}.$$

Now since $$F'(r)$$ is negative, we see that when $$\Delta t_1$$ is small, this entire expression can be negative.

Do you see anything wrong with my argument, or is it correct?