assume $c(t)0tc(t)=\frac{c(t-\mathrm{.}5)(1-d(t))}{1-d(t-\mathrm{.}5)+\mathrm{.}5d(t)c(t-\mathrm{.}5)},c(t)>c(t-\mathrm{.}5)f(t)-c(t-\mathrm{.}5)t-\mathrm{.}5tt-\mathrm{.}5tt-\mathrm{.}5t-\mathrm{.}5t-\mathrm{.}50$ and $c(t)0$ for all $t.$

$(a)$ Derive the recursive relationship

$c(t)=\frac{c(t-\mathrm{.}5)(1-d(t))}{1-d(t-\mathrm{.}5)+\mathrm{.}5d(t)c(t-\mathrm{.}5)},$

which allows $usto$ compute the par curve "one period $at$ a time".

$(b)$ Show that $c(t)>c(t-\mathrm{.}5)if$ and only $iff(t)-c(t-\mathrm{.}5).$ Thus, the par curve

will $be$ increasing between $t-\mathrm{.}5$ and $tif$ and only $if$ the rate $(agreedto$ today$)$

for a loan starting $att-\mathrm{.}5$ and maturing $att$ exceeds the par rate $att-\mathrm{.}5.In$

other words, $ifwe$ determine today that the par rate $att-\mathrm{.}5is$ "too low" for $a$

six month forward loan made $att-\mathrm{.}5.$