Using the Vasicek Model to Find a Future Bond Price Probability

# Vasicek Model, zero coupon bond question


# Vasicek Model, zero coupon bond question












I am trying to solve questions in the Vasicek model. Can anyone help me to solve this question...

In the Vasicek model with parameters $\theta = 0.08$, $k$ = 2.5, $\sigma = 0.2$, assuming to be already under the risk neutral probability and that $r_0 = 0.1$, price a zero coupon bond with nominal $F = 100$ euro and maturity 2 year.

What is the probability that the bond just described is worth more than 96 euros after 1 year?

## Answer by Kevin (score 2, accepted)

https://quant.stackexchange.com/a/50553

Let $P(t,T)$ denote the time $t$ price of a zero-coupon bond (with unit face value) maturing at time $T$.

Firstly, recall that for every $s\leq t$, we have \begin{align*} r_t = r_s e^{-\kappa(t-s)}+\theta\left(1-e^{-\kappa(t-s)}\right)+\sigma \int_s^t e^{-\kappa(t-u)}\mathrm{d}W_u. \end{align*} Thus, the short rate $(r_t)$ is normally distributed for every time point $t$ with \begin{align*} \mathbb{E}^\mathbb{Q}[r_t|\mathcal{F}_s] &= r_se^{-\kappa(t-s)}+\theta\left(1-e^{-\kappa(t-s)}\right), \\ \mathbb{V}\mathrm{ar}[r_t|\mathcal{F}_s] &= \frac{\sigma^2}{2\kappa}\left(1-e^{-2\kappa(t-s)}\right). \end{align*}

Secondly, the Vasicek model is an affine term structure model, i.e. the bond price is given by $P(t,T)=e^{A(t,T)+B(t,T)r_t}$, where \begin{align*} A(t,T) &= \left(\frac{\sigma^2}{2\kappa^2}-\theta\right)\big( T-t-B(t,T)\big)-\frac{\sigma^2}{4\kappa}B(t,T)^2, \\ B(t,T)&=\frac{1}{\kappa}\left(e^{-\kappa(T-t)}-1\right). \end{align*} In particular, the zero-coupon bond price $P(t,T)$ is log-normally distributed for every time point $t$.

We finally compute the (unconditional, risk-neutral) probability that the time $t$ price of a zero-coupon bond is above a constant $c>0$. \begin{align*} \mathbb{Q}[\{P(t,T)>c\}] &= 1- \mathbb{Q}[\{P(t,T)\leq c\}]\\ &= 1- \mathbb{Q}[\{e^{A(t,T)+B(t,T)r_t}\leq c\}] \\ &= 1- \mathbb{Q}\left[\left\{r_t\leq \frac{\ln(c)-A(t,T)}{B(t,T)}\right\}\right] \\ &= 1- \mathbb{Q}\left[\left\{m_t+ s_tZ\leq \frac{\ln(c)-A(t,T)}{B(t,T)}\right\}\right] \\ &= 1- \mathbb{Q}\left[\left\{Z\leq \frac{\ln(c)-A(t,T)-m_tB(t,T)}{s_tB(t,T)}\right\}\right] \\ &= 1- \Phi\left(\frac{\ln(c)-A(t,T)-m_tB(t,T)}{s_tB(t,T)}\right), \end{align*} where $Z\sim N(0,1)$ and $m_t$ and $s_t^2$ are the unconditional mean and variance of $r_t$. Finally, $\Phi$ is the cumulative distribution function of a standard normally distributed random variable.

In your case, $c=0.96$, $t=1$ and $T=2$. Thus, \begin{align*} m_1 &= r_0e^{-\kappa}+\theta\left(1-e^{-\kappa}\right), \\ s_1 &= \sqrt{\frac{\sigma^2}{2\kappa}\left(1-e^{-2\kappa}\right)}, \\ A(1,2) &= \left(\frac{\sigma^2}{2\kappa^2}-\theta\right)\big( 1-B(1,2)\big)-\frac{\sigma^2}{4\kappa}B(1,2)^2, \\ B(1,2)&=\frac{1}{\kappa}\left(e^{-\kappa}-1\right). \end{align*}