Why Exponential Vasicek Rates Do Not Give an Affine Bond Price

# How to show that the exponential Vasicek model is not an affine term-structure model?


# How to show that the exponential Vasicek model is not an affine term-structure model?












From the pricing formula, we know that the value at time $t\in [0,T]$ of a zero coupon bond maturing at time $T$ is $$ B(t,T)=E\left(\exp{\left(-\int_{t}^{T}r_sds\right)}\bigg|\mathcal{F}_t\right). $$ Moreover, we say that $B(t,T)$ has an affine term-structure, if $$ B(t,T)=\exp{\left(A(t,T)-C(t,T)r_t\right)}\;\;\ \text{for} \;\;\ t\in[0,T], $$ where $A$ and $C$ are deterministic functions.

My question is the following :

For exponential Vasicek model defined by $$ r_t=\exp{(X_t)}\;\;\ \text{with}\;\;\ dX_t=k(\theta-X_t)dt+\sigma dW_t, $$ where k, $\theta$, $\sigma>0$ and $W$ is a Brownian motion under the risk-neutral measure.

How to show that this model is not an affine term-structure model?

## Answer by Hans (score 2, accepted)

https://quant.stackexchange.com/a/25814

Here is a general proof for all parameters in an open domain.

$$dr = adt+bdW:=r\big(k(\theta-x)+\frac12\sigma^2\big)dt+\sigma rdW.$$ Let $$u(r(s),s):=e^{-\int_t^sr}B(r(s),s,T)=:\phi(s) B.$$ Then $$u(r(t),t)=\mathbf E\big[u(r(s),s)\big|r(t)\big],\, \forall t<s. \tag{1}$$ So, by Ito's Lemma, \begin{align} du(r(s),s) &= Bd\phi +\phi dB \\ &= \phi \bigg(-rB+\frac{\partial B}{\partial s}ds+\frac{\partial B}{\partial r}dr+\frac12\frac{\partial^2 B}{\partial r^2}(dr)^2\bigg) \\ &= \phi \bigg[\bigg(-rB+\frac{\partial B}{\partial s}+\frac{\partial B}{\partial r}a+\frac12\frac{\partial^2 B}{\partial r^2}b^2\bigg)ds+\frac{\partial B}{\partial r}bdW\bigg] \\ &=: \phi\,(fds+gdW_s). \end{align} We see from Eq. (1) $\mathbf E\big[u(r(s),s)\big|r(t)\big]$ is constant with respect to $s$. So $$0=\frac{d\mathbf E\big[u(r(s),s)\big|r(t)\big]}{ds}\bigg|_{s=t}=f(r(t),t) \tag{2}$$ by Equation (1).

Suppose $B$ is affine. Substitute into $\frac{f}{B}$ the affine expression for $B(r,t,T)$ and the expression of $a$ and $b$, we have by Equation (2) $$A'-\Big(C'+\Big(k\theta+\frac{\sigma^2}2\Big)C-1\Big)e^{X_t}+kCX_te^{X_t}+\frac{(\sigma C)^2}{2}e^{2X_t}=0,\quad\forall X_t\in R,$$ where $'$ denotes partial derivative with respect to $t$ (denoting the first variable). By taking derivatives with respect to $X_t$ or Taylor expanding $e^{X_t}$, we see $(1,e^{X_t},X_te^{X_t},e^{2X_t})$ is linearly independent. So all factors in front of those terms vanish. This is possible only when $k=\sigma=0,\,C(t,s)=s-t$ and $A(t,s)=0$.

## Answer by Hans (score 1)

https://quant.stackexchange.com/a/25811

We shall prove this by contradiction. Let $\theta=0$ and $\sigma=0$. $X_t=X_0e^{-kt}$ and $$B(0,t)=\exp\Big(-\int_0^te^{X_0e^{-ks}}ds\Big).$$ Suppose the contrary that $B(0,t)$ is affine. We should have $$ B(0,t)=\exp{\left(A(0,t)-C(0,t)e^{X_0}\right)}\;\;\ \forall (t,X_0), \tag{1} $$ Differentiate the logarithm of Equation (1) with respect to $t$ side, $$e^{X_0e^{-kt}}=C'(0,t)e^{X_0}.$$ Take logarithm of the above equation, we get $$(1-e^{-kt})X_0=-\ln C'(0,t),\quad \forall X_0$$ which is only possible for $k=0$. Therefore, this model is not affine in general.

Alternatively and more generally, one can write out either the PDE or SPDE for both expressions and compare the coefficients of similar differential terms.