# Vasicek Model, zero coupon bond question [closed]

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?

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*}

• Thank you so much for the solution, I have a doubt, So let's say i got the price of a zero-coupon bond with unit face value. But my nominal is 100 euro. So is it enough to multiply the price by 100 euro? Or if not, what should I do? Jan 5 '20 at 11:44
• @saimurari Yes, it is enough to multiply the bond price by 100. (100 bonds with £1 face value are the same as 1 bond with £100 face value) Jan 5 '20 at 12:31
• Thank you so much @KeSchn :D Jan 5 '20 at 12:47