In the classic Vasicek model, the market's short rate process $(r_t)_{t \geq 0 }$ is given through the SDEs:

$$ dr_t=\alpha \left( \bar{\mu} - r_t\right) dt+\sigma d W^{\mathbb{P}}(t), $$ $$ dr_t=\alpha \left( \mu - r_t\right) dt+\sigma d W^{\mathbb{Q}}(t), $$

where $W^{\mathbb{P}}$ is a Wiener process under the objective, real-world probability measure $\mathbb{P}$, and $W^{\mathbb{Q}}$ is a Wiener process under the risk-neutral measure $\mathbb{Q}$ (measure equivalent to $\mathbb{P}$).

A standard result of this set-up is that the time $t$ price of a T-maturity zero-coupon bond is equal to:


where $A(t,T)$ and $B(t,T)$ are deterministic functions of the risk-neutral parameters $\alpha, \mu$ and $\sigma$.

Let's say that the current time is $t=0$ and we want to plot one trajectory of $P(t,T)$, i.e. $t \mapsto P(t,T)$ for $t \in \left[0,T\right]$. Thus, we want to calculate $P(\Delta_k , T)$, where $\Delta_k =k \times T/n$, and $k=0, 1, \dots, n$.

We have that $$ P( \Delta_k , T)= A(\Delta_k ,T)e^{-r( \Delta_k )B(\Delta_k ,T)}.$$

My question is: Should we simulate $r(\Delta_k )$ under the risk-neutral or under the real-world measure. That is, should we simulate it as:

$$r( \Delta_k ) \sim r(\Delta_{k-1} )+\alpha (\bar{\mu}-r(\Delta_{k-1} ))\Delta_1 + \sigma \Delta_1 \mathcal{N}(0,1), $$ or as $$r( \Delta_k ) \sim r(\Delta_{k-1} )+\alpha (\mu-r(\Delta_{k-1} ))\Delta_1 + \sigma \Delta_1 \mathcal{N}(0,1). $$

From my understanding, we should do it under the real-world measure, but I have some resources that go on talking about the trajectory of $P(t,T)$ without even touching the real-world dynamics of the short-rate, so I don't quite understand if I am right, what are they even getting when they only use the risk-neutral measure to simulate the short-rate.

  • $\begingroup$ To get to (and therefore to be able to use) the relationship you mention, you are assuming a probability space equipped with the risk neutral measure $\Bbb{Q}$, so that $P(t,T) = \Bbb{E}_t^\Bbb{Q} [e^{-\int_t^T r(s) ds} ] = A(t,T) e^{-r(t)B(t,T)}$. So, you should be simulating under $\Bbb{Q}$. $\endgroup$
    – Quantuple
    Jul 6, 2017 at 14:19
  • $\begingroup$ Thanks for your comment. Though, there is one thing I still don't quite understand. Namely, if we assume that the current time is $t=0$, in the expression $P(t,T)=\mathbb{E}^{\mathbb{Q}}\left[ e^{-\int_t^T r(s)ds} \right]$, we are considering the $\mathbb{Q}$-path of $r(s)$ on the interval $\left [ t,T \right]$, but shouldn't the starting value of the short-rate process $r(t)$ be the one observed in the market at time $t$, and we can get that one by simulating $r$ under the measure $\mathbb{P}$ on the interval $\left [ 0 ,t \right]$? $\endgroup$
    – Milan
    Jul 6, 2017 at 14:39
  • $\begingroup$ When changing measures, you do not alter the states of the world themselves (i.e. same $\Omega$), only their probability of occurring (i.e. different probability measures). Here, when conditioning with respect to $t$, the state of the world is known at $t$: $r(t)=r_t$ $\Bbb{P}$-a.s. ("almost surely" i.e. abusing the notation $\Bbb{P}(r(t)=r_t)=1$. But since $\Bbb{Q}$ is equivalent to $\Bbb{P}$ then by definition you also have $r(t)=r_t$ $\Bbb{Q}$-a.s. $\endgroup$
    – Quantuple
    Jul 6, 2017 at 14:52
  • $\begingroup$ But you are right it depends on what you want to do: if you want to simulate bond prices under $\Bbb{P}$ for "economic scenario generation" then you simulate the short rate under $\Bbb{P}$ and compute the price of the bond using the $\Bbb{Q}$ expectation (your formula). But if you want the risk neutral dynamics of the bonds you do everything under $\Bbb{Q}$. Rem: under $\Bbb{Q}$, $P(t,T)/e^{\int_t^T r(s) ds}$ should be a martingale, which is not the case under $\Bbb{P}$. $\endgroup$
    – Quantuple
    Jul 6, 2017 at 14:54
  • $\begingroup$ Yes, that's exactly what I'm looking to do - simulate bond prices under $\mathbb{P}$. It's all definitely a little bit clearer to me now. Thank you for your help, I really appreciate it. $\endgroup$
    – Milan
    Jul 6, 2017 at 15:00


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