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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?

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  • $\begingroup$ Solve for $X_t$ or equivalently $r_t$ and see if $B(t,T)$ could possibly have the affine structure. $\endgroup$ – Hans May 3 '16 at 21:27
  • $\begingroup$ Thank you for your comment, but I tried to solve with Itô's formula, unfortunately, I couldn't resolve the problem. $\endgroup$ – M. A. Kacef May 3 '16 at 23:53
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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 $A(t,s)=0$.

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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.

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  • $\begingroup$ Thank you for your against-example, however, your approach considers a zero volatility and therefore you remove the random effect of Brownian motion, knowing that in this type of model all the parameters are assumed to be strictly positive. I also wanted to know if it is possible to estimate $B(0,t)$ in a lognormal model like this. $\endgroup$ – M. A. Kacef May 3 '16 at 23:28
  • $\begingroup$ For arbitrary positive volatility, you can take its limit to zero, and reach the same contradiction, unless you want to make the volatility above some positive threshold. In the latter case, you can apply the procedure similar to the one in my answer and reach the contradiction. Or more generally, you can take up the alternative method I mentioned in the last paragraph of my answer. I will write out the detail later. $\endgroup$ – Hans May 4 '16 at 0:29
  • $\begingroup$ @M.A.Kacef: Have you seen my proof below for the general case? $\endgroup$ – Hans May 4 '16 at 20:45
  • $\begingroup$ Hi @Hans, thank you again for your help. I understand your approach, which seems to be similar to the approach given by S. ROGEMAR MAMON (2004) in his article "Three ways to solve for bond prices in the Vasicek model." Unfortunately for me I should leave out the exponontielle Vasicek model because I want a model where the price of zero coupon can be calculated continuously. Knowing that I work on calculating the price of a double barriers option in a model where the underlying prices are correlated with interest rates on the market. $\endgroup$ – M. A. Kacef May 5 '16 at 18:49
  • $\begingroup$ No worries. That article you cited seems interesting. I first tried and found it hard to apply what the author called the first way to prove the conclusion. My approach is the PDE way, basically his second way. I will try his third way HJM method. What do you mean by "calculated continuously"? Do you mean calculated analytically? $\endgroup$ – Hans May 6 '16 at 3:30

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