We need to state whether the given statement is true or false : $$ v(t_2) = v(t_1)v(t_2 - t_1)$$ where $v(t)$ is the discount factor.
I found it ti be true as $v(t) = (1 - d)^{t}$ where $d$ is the effective discount rate , so , $v(t_1)v(t_2 - t_1) = (1 - d)^{t_1}(1 - d)^{t_2 - t_1}= (1 - d)^{t_2} = v(t_2)$ ,
But the solution says the statement is false and the correct representation is : $$ v(t_2) = v(t_1)\dfrac{1}{A(t_1,t_2)}$$ But isn't that the same thing ?
As $A(t_1 , t_2) = (1 + i)^{t_2 - t_1}$ and $ (1+i) = \dfrac{1}{(1-d)}$ , ( $i$ is the effective rate of interest) , which still justifies the statement. Can anyone tell what am I doing wrong ?
Answer:
Let $i_2$ is the interest rate for $t_2$. and let $i_1$ be the interest rate for time $t_1$. Let the forward rate be $r$. From the diagram and simple financial basis, it is true that
$$(1+r)^{(t_2-t_1)} (1+i_1)^{t_1} = (1+i_2)^{t_2}\tag{1}$$
$$v(t) = \frac{1}{(1+i)^t}$$
Thus $$v(t_1) = \frac{1}{(1+i_1)^{t_1}}$$
$$v(t_2) = \frac{1}{(1+i_2)^{t_2}}$$
We also know that $$A(t_1,t_2) = (1+r)^{(t_2-t_1)}$$
Rearranging (1)
$$\frac{1}{(1+i_2)^{t_2}} = \frac{1}{(1+i_1)^{t_1}}.\frac{1}{(1+r)^{(t_2-t_1)}}$$
$$v(t_2) = v(t_1).\frac{1}{A(t_1,t_2)}$$
Goodluck
Satish