0
$\begingroup$

I don't understand how can I conclude that F(t, T0, T1) < L(t, T1) < L(t, T0) base on the given answer.

enter image description here

$\endgroup$
6
  • 1
    $\begingroup$ The Correct solution contains a proof. What exactly there is not clear ? Intuitively, the spot rate for the longer period $L(t,T_1)$ is an average of $L(t,T_0)$ and the forward rate for the adjacent period. $\endgroup$
    – Kurt G.
    May 12 at 16:58
  • $\begingroup$ Last two lines of the proof. Where do they come from ? What exactly P(t, T0)L(t, T1) means ? and how the proof relates to the answer ? Thanks. $\endgroup$ May 13 at 2:47
  • $\begingroup$ $P(t,T)$ is a common notation for a discount factor. Somewhere in that reference where that question is from they must define this. If you find that out the rest is staightforward. I find the question how the proof relates to the answer unbelievable. Please consider what exactly that proof shows by looking the most left hand side and the most right hand side of the chain of equations/inequalities. $\endgroup$
    – Kurt G.
    May 13 at 5:37
  • $\begingroup$ One thing is probably worth thinking about: The proof uses the definition $P(t,T_0)=\frac{1}{1+(T_0-t)L(t.T_0)}$ and claims that $P(t,T_0)<1$ . Question: in the recent interest rate environment, can we say that this claim is always true ? $\endgroup$
    – Kurt G.
    May 13 at 6:02
  • $\begingroup$ According to the previous lecture notes, 𝑃(𝑑,𝑇) is the price at time t of a zero-coupon bond with maturity T. How can we conclude that the third line is greater than the second line of thr proof ? Why should we use the 𝑃(𝑑,𝑇0)L(𝑑,𝑇1) to compare the second line of the proof ? $\endgroup$ May 13 at 10:48

1 Answer 1

0
$\begingroup$

Too long for a comment.

To answer the outstanding question of yours in the comment why the third line is greater than the second line:

Apparently the source of this exercise assumes $t<T_0<T_1$ and $0\color{red}{<}L(t,T_1)<L(t,T_0)$. The red $\color{red}<$-sign they assume without mentioning it. Then \begin{align} &F(t,T_0,T_1)\\&=\frac{1}{T_1-T_0}\Big(\frac{P(t,T_0)}{P(t,T_1)}-1\Big)\\ &=\frac{1}{T_1-T_0}\Bigg(\frac{1-(T_1-t)L(t,T_1)}{1-(T_0-t)L(t,T_0)}-1\Bigg)\\ &=\frac{1}{1-(T_0-t)L(t,T_0)}\frac{L(t,T_1)(T_1-t)-L(t,T_0)(T_0-t)}{T_1-T_0}\\[3mm]\tag{2nd line} &=P(t,T_0)\frac{L(t,T_1)(T_1-t)-L(t,T_0)(T_0-t)}{T_1-T_0}\\[3mm] &=\underbrace{P(t,T_0)\frac{L(t,T_1)(T_1-t)}{T_1-T_0}}_{(*)}-\underbrace{P(t,T_0)\frac{L(t,T_0)(T_0-t)}{T_1-T_0}}_{(**)}. \end{align} Because $L(t,T_0)(T_0-t)>0$ the term ($**$) is strictly greater than zero. Dropping this term we see that the (2nd line) is less than $$\tag{$*$} P(t,T_0)\frac{L(t,T_1)(T_1-t)}{T_1-T_0}. $$ In turn, since $t<T_0$, we have $\frac{T_1-t}{T_1-T_0}<1$ so that ($*$) is less than $$\tag{3rd line} P(t,T_0)L(t,T_1)\,. $$ Together we have $$ P(t,T_0)\frac{L(t,T_1)(T_1-t)-L(t,T_0)(T_0-t)}{T_1-T_0}<P(t,T_0)L(t,T_1)\,. $$ $$\tag*{$\Box$} \quad $$

$\endgroup$
3
  • $\begingroup$ The assumptions make sense. However, it doesn't mean the third line is greater than the second line of the proof even if 𝐿(𝑑,𝑇0)(𝑇0βˆ’π‘‘) > 0 since we cannot determine if 𝐿(𝑑,𝑇0)(𝑇0βˆ’π‘‘) is greater or smaller than 𝐿(𝑑,𝑇1)(𝑇1βˆ’π‘‘). $\endgroup$ May 13 at 13:52
  • $\begingroup$ Updated (to make it clearer what was there before). Please tell me where exactly I am wrong in your opinion. $\endgroup$
    – Kurt G.
    May 13 at 14:07
  • $\begingroup$ This is very clear. I understand what you meant now. Thank you! $\endgroup$ May 13 at 15:51

Not the answer you're looking for? Browse other questions tagged or ask your own question.