Calculating spot rates from forward rates

Question

I am working on a problem where I am trying to calculate the forward rates from two different spot rates. I have the following:

1 Year Spot Rate = 1%

2 Year Spot Rate = 2%

Specifically, I would like to find the forward rate between the first and second year. (Using semiannual compounding).

My thoughts are to use the following:

Forward Rate = $(1 + r_a)^{ta} \over (1 + r_b)^{tb}$ - 1

Which in my case would look like:

Forward Rate = $(1 + .01)^{1} \over (1 + .02)^{2}$ - 1

Is this the correct approach? It seems like using this method might not account for semiannual compounding., Any thoughts or advice would be greatly appreciated.

Answer 1

Forward Rate = $\frac {(1+(0.5) 2\%)^{2 * 2}} {(1+(0.5) 1\%)^{2 *1}} -1$

The above works fine when the day count convention is 30/360.

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General formula -

$F(t,t+1,t+2)= \frac {P(t,t+1) - P(t,t+2)} {\tau P(t,t+2)}$

where $F(t,t+1,t+2)$ is the forward rate between $t+1$ and $t+2$, as seen at $t$

$P(t,t+1)$ is the price of zero-coupon bond with maturity $t+1$, as seen at $t$

$\tau$ is the accrual fraction between $t+1$ and $t+2$

The formula is easy to understand by reading it as -

$F(t,t+1,t+2){\tau P(t,t+2)} = {P(t,t+1) - P(t,t+2)} $

so that

on LHS: $F(t,t+1,t+2)\tau$ is the interest rate earned by \$1 between $t+1$ and $t+2$, and this interest discounted to $t$ by multiplying by $P(t,t+2)$

on RHS: the difference between the two zero-coupon bonds, which return \$1 at $t+1$ and $t+2$ respectively is the discounted value of interest earned by re-investing maturity proceeds of "$t+1$"-maturity bond (i.e. \$1) between $t+1$ and $t+2$ at the forward rate.

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With numbers in the question above:

(assuming 30/360 day count convention)

$P(t,t+1) = (1+\frac {1\%} {2})^{-2}$

$P(t,t+2) = (1+\frac {2\%} {2})^{-4}$

$\tau = yearfrac(t+1,t+2) = 1$

Answer 2

If you want to calculate the forward rate given semi-annual compounding then the answer should be:

\begin{equation} F(0,t_a,t_b)=\Bigg(\sqrt[2*(t_b-t_a)]{\frac{(1 + \frac{r_b}{2})^{2*t_b}}{(1 + \frac{r_a}{2})^{2*t_a}}}-1\Bigg)*2 \end{equation}

This is derived by the fact that : \begin{equation} \Bigg(1+\frac{r_b}{2}\Bigg)^{2*t_b} = \Bigg(1+\frac{r_a}{2}\Bigg)^{2*t_a}*\Bigg(1+\frac{F(0,t_a,t_b)}{2}\Bigg)^{2*(t_b-t_a)} \end{equation}

If you rearrange the terms in the last formula then you get the first equation. Thank you for the question, I hope this helps.