# Why does the future price dominate the forward price and why doesn't the long rate fall?

There are two questions left about the book Term Structure: A graduate Course by Damir Filipovic, which bother me. The first one is about the Theorem, that the long rate never falls (p. 108). Why is the following equation true:

$$p(s)=\lim_{T\to\infty}E_{Q^t}[P(t,T)|\mathcal{F}_s]^{\frac{1}{T}}$$

We have that $E_{Q^t}[P(t,T)|\mathcal{F}_s]=\frac{P(s,T)}{P(s,t)}$. We know that $P(s,T)=e^{-\int_s^Tf(s,u)du}$ and $P(s,t)=e^{-\int_s^tf(s,u)du}$. Therefore

$$(\frac{P(s,T)}{P(s,t)})^{\frac{1}{T}}=e^{-\frac{1}{T}\int_t^Tf(s,u)du}$$ I know that $\int_s^Tf(s,u)du=(T-s)R(s,T)$, but why is $\int_t^Tf(s,u)du=(T-s)R(s,T)$?

The second question is about the relationship between forward and future price on page 120/121. There they have stated the following equation

$$F(t,T;S(T))=f(t,T;S(T))e^{\int_t^T(v(s,T)-\rho(s))v(s,T)ds}$$

Why does $S(t)P(t,T)\rho(t)v(t,T)\le 0$ imply that the future price dominates the forward price. Somehow I must conclude that

$$(v(s,T)-\rho(s))v(s,T)\le 0$$

But I don't see how this is possible from the above assumption.

-
Can you please find a better title for your question? – SRKX Feb 11 at 15:36
And even maybe split it into 2 separate questions – Alexey Kalmykov Feb 11 at 22:16
@AlexeyKalmykov I agree, but now you answered the second part on this topic so he should create another question for the first one and rename this one. – SRKX Feb 11 at 22:47
Hulik, could you please formulate the title of the post as a question? It helps the site to be referenced properly on search engines and hence can attract more users. This is also valid for your previous question. – SRKX Feb 12 at 8:18

You can try to argue in the following manner. Using the fact that $E_{Q^t}[P(t,T)|\mathcal{F}_s]P(s,t)=P(s,T)$, then:

$$p(s)=\lim_{T\to\infty}P(s,T)^{\frac{1}{T}}=\lim_{T\to\infty}(P(s,t)E_{Q^t}[P(t,T)|\mathcal{F}_s])^{\frac{1}{T}}=\lim_{T\to\infty}E_{Q^t}[P(t,T)|\mathcal{F}_s]^{\frac{1}{T}}$$

as $\lim_{T\to\infty}P(s,t)^{\frac{1}{T}}=1$

Assume that you are long the futures and co-variance of underlying and interest rates is negative. If the price of the underlying goes down, then you have a positive mark-to-market PNL and you can reinvest it at a higher interest rate. If the price goes up, you loose but you can finance your looses by borrowing at a lower interest rate. The situation is reversed for a short futures position. Therefore, in case of negative co-variance, futures price must be higher than forward as this nice dynamics doesn't apply to forwards (their only cashflow accrues at the maturity).

Formally speaking I would argue that $S(t)P(t,T)\rho(t)v(t,T)\le 0$ is equivalent to saying $\rho(t)v(t,T)\le 0$ which guarantees you that your the integral in the exponential stays positive and the whole exponential remains greater than 1.

You may also want to read "The relation between forward prices and futures prices" by Cox, Ingersoll, Ross.

-
 Stupid me! sorry I did a very bad error in reasoning. But thanks for answering the second question :) – hulik Feb 12 at 7:24 Very nice answer. Actually Steven Shreve at Carnegie Mellon in his Stochastic Calculus book II has an excellent page and explanation of the forward-future relationship in his book but I think Alexey made it pretty clear already. I think you should award the bounty to him. – Freddy Feb 14 at 18:07

Regarding your first question, let me reformulate $p(s)$:

$p(s)=(\frac{P(s,T)}{P(s,t)})^\frac{1}{T}= \frac{e^{-\frac{(T-s)R(s,T)}{T}}}{e^{⁻\frac{(t-s)R(s,t)}{T}}}.$

Now, because $R(s,t)$ is finite, the term $e^{⁻\frac{t-s}{T}R(s,t)}$ converges to 1 for $T\to\infty$. On the other hand, the term $e^-\frac{(T-s)R(s,T)}{T}$ converges to $\lim_{T\to\infty}e^{-R(s,T)}$ which gives the definition of $p(s)$:

$p(s)=e^{-R_{\infty}(s)}$

So from my feeling, its a question of the limit to infinity and you dont need the integrals for $\int_{t}^{T}f(s,u)du$ (whose meaning I cant understand intuitively).

Probably, this is just a reformulation and not the answer you were looking for, but it was too long for a comment.

-