Deriving a Forward Rate Bound from Simple Rates

Summary

The answer explains an inequality involving a simple forward rate and two simple rates for maturities ordered as t < T₀ < T₁. It rewrites the forward rate using discount factors, then expresses those discount factors in terms of the simple rates. Under the assumed conditions that both rates are positive and the earlier-maturity rate exceeds the later-maturity rate, the expression separates into a positive contribution and a term that is subtracted.

The subtracted term is positive, so removing it gives an upper bound. The remaining expression is itself below the later simple rate because the maturity ratio (T₁ − t)/(T₁ − T₀) is less than one. This establishes the stated comparison with the later simple rate. The reasoning depends on the positivity and ordering assumptions, which the answer says the exercise had left unstated; it is not a general result for arbitrary rates or maturities.

Key ideas

• The derivation assumes t < T₀ < T₁ and positive simple rates with L(t,T₁) below L(t,T₀).
• The forward rate is rewritten using discount factors and then expressed in terms of the simple rates.
• A positive subtracted term makes the forward rate smaller than the remaining contribution.
• The maturity ratio in that contribution is below one, yielding an upper bound by the later simple rate.
• The inequality depends on assumptions that must be stated explicitly.

Tags

# Simple Forward Rate


# Simple Forward Rate












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

## Answer by Kurt G. (score 0, accepted)

https://quant.stackexchange.com/a/70875

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 $$ P(t,T_0)\frac{L(t,T_1)(T_1-t)-L(t,T_0)(T_0-t)}{T_1-T_0}<P(t,T_0)L(t,T_1)\,. $$ $$\tag*{$\Box$} \quad $$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.