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Which of those processes indicate an arbitrage free family of bonds in terms of r? $T^*=1$. The process $r_t$ is given by:

a) $r_t = 1$ for all $t \in [0,1]$

b) $r_t = B_t$ for all $t \in [0,1]$

c) $r_t = B_1$ for all $t \in [0,1]$

d) $r_t = t + B_t$ for all $t \in [0,1]$

I know that I should check if:

  1. $B(T,T)=1$
  2. there exists a probility meassure such that: $\frac{B(t,T)}{B_t}$ is a martingale

What's is more:

$$\frac{B(t,T)}{B_t} = \mathbb{E}(\frac{1}{B_T}|\mathcal{F}_t)$$ and $\mathbb{E}(X|\mathcal{F})$ is a martingale for X - measurable, integrable random variable

To be honest I am not quite sure if $B_t$ donotes Brownian motion or a banking account process.

Ad a) yes

$B_t = \mathbb{E}(\exp(\int \limits_0^t r_u du)) = e^t$

$\frac{B(t,T)}{B_t} = \mathbb{E}(\frac{1}{B_T}|\mathcal{F}_t)= e^{-T}$ - a martingale

$B(1,1) = 1$

Please check, if my solution to point a) is correct and help me with others.

Elizabeth_Banks
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  • What is $B_t$? You said that you are not sure whether it is a Brownian motion, or not. But in your answer , you assume that $B_t$ is "like" a bank account – Canardini Dec 21 '16 at 18:24
  • Actually, in my answer, $B_t$ is the process of a bank account. But I am not sure, how should I understand it in points (a)-(d), because sometimes $B_t$ denotes Brownian motion- and for me, here it would make some sense. – Elizabeth_Banks Dec 21 '16 at 19:30
  • Would you clarify that point ? To make sure that a short-rate model is non-arbitrage, one must make sure that $\frac{B(t,T)}{N_t}$ is a martingale under the measure related to the numeraire $N_t$, where $B(t,T)$ is the bond price at time $t$, for a maturity $T$ – Canardini Dec 21 '16 at 19:51
  • The process of a bank account satisfies: $dB_t = r_t B_t dt$, $B_0=0$, where $r_t$ denotes short-time interest rate. – Elizabeth_Banks Dec 21 '16 at 19:57

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