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The Setup:

Suppose I know the yield curve of a Bond satisfies: f (0, t) = 0.04 for t ≥ 0 and f (ω, 1, t) = 0.06, t ≥ 1, ω = ω 1 , 0.02, t ≥ 1, ω = ω 2 , where Ω = {ω 1 , ω 2 } with P[ω i ] > 0, i = 1, 2.

And suppose that I know the matrix: $\begin{pmatrix} P (0, 1) & P (0, 2) & P (0, 3)\\ P (ω_1 , 1, 1) & P (ω_1 , 1, 2)& P (ω_1 , 1, 3) \\ P (ω_2 , 1, 1) & P (ω_2 , 1, 2) & P (ω_2 , 1, 3) \end{pmatrix} = \begin{pmatrix} e^{-0.04}&e^{-0.04}&e^{-0.04}\\ 1&e^{-0.06}&e^{-0.12} \\ 1&e^{-0.02}&e^{-0.04} \end{pmatrix} $ is invertible (for example, I have shown that it's determinant is roughly equal t$0.0004258212$).

The Question:

How can I show that there exists an arbitrage opportunity using the above matrix and the following value process:\ \begin{align} V (ω_1 , 1) = & 1 = V (ω_2 , 1)\\ V(ω_1 ,0) = & 0 = V(ω_2 ,0). \end{align}

Context: This example is supposed to help motivate why parallel shifts along yield curves yields arbitrage opportunities.

Improve this question
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  • $\begingroup$ The notation is not clear. For example, what does $V (ω_1 , 1)$ mean. It is better to define the notations before you use them. $\endgroup$
    – Gordon
    Jan 11, 2016 at 18:58
  • $\begingroup$ $V:\Omega\times [0,\infty)\rightarrow \mathbb{R}$ is a stochastic process, so the first component is the outcome in the probability space $\Omega$ and the second is the time variable, the rest is the usual stuff. $\endgroup$
    – AIM
    Jan 11, 2016 at 19:02
  • $\begingroup$ What is $P (ω 1 , 1, 1)$? $\endgroup$
    – Gordon
    Jan 11, 2016 at 19:08
  • $\begingroup$ P is the price of the bond: ie: $P(\omega,t,T)=e^{-\int_t^T f(\omega,u,T)du}$, where $T$ is the time of maturity of the Bond and $\omega \in \Omega$. $\endgroup$
    – AIM
    Jan 11, 2016 at 20:01
  • $\begingroup$ Just wanted to point out that the 2nd matrix is incorrect; P(t,T)=-f(t,T) * (T-t) $\endgroup$
    – learningIR
    Mar 21, 2017 at 23:45

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